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[Powered by Google Translate] [Sehemu ya 5 - Zaidi Starehe]

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[Rob Bowden - Chuo Kikuu cha Harvard]

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[Hii ni CS50. - CS50.TV]

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>> Kama nilivyosema katika barua pepe yangu, kuna mambo mengi unaweza kutumia

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nyingine kuliko appliance kwa kweli kufanya seti tatizo.

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Tunapendekeza ya kufanya hivyo katika appliance sababu tu basi tunaweza kwa urahisi zaidi kukusaidia

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na tunajua jinsi kila kitu ni kwenda kufanya kazi.

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Lakini kama mfano mmoja wa ambapo unaweza kufanya mambo kama, kusema, wewe hawana

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kwa appliance au unataka kazi katika basement Center Sayansi -

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ambayo kwa kweli wana appliance pia -

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kama unataka kufanya kazi mahali popote.

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Mfano mmoja ni umeona / sikia SSH?

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SSH kimsingi ni kama tu kuungana na kitu.

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Kweli, sasa hivi mimi nina SSHed katika appliance.

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Mimi kamwe kazi moja kwa moja katika appliance.

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Hapa ni appliance, na kama wewe kuangalia chini hapa unaweza kuona hii anwani ya IP.

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Mimi kamwe kazi katika appliance yenyewe;

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Mimi daima kuja juu na dirisha iTerm2 dirisha / terminal.

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Unaweza SSH kwa anwani hiyo IP, ssh jharvard@192.168.129.128.

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Nakumbuka kwamba idadi urahisi sana kwa sababu ni mfano kama nice.

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Lakini hilo kuuliza mimi kwa password yangu, na sasa mimi niko katika appliance.

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Kimsingi, katika hatua hii, kama wewe ni kufungua terminal ndani ya appliance yenyewe,

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interface hii, hata hivyo ungependa kutumia hiyo, ni sawa

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kama interface mimi nina kutumia zaidi ya hapa lakini sasa uko SSHed.

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Huna kwa SSH kwa appliance.

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Mfano mmoja wa mahali pengine unaweza SSH kwa ni nina uhakika pretty una by default -

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Oh. Kubwa.

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Wote wa unapaswa kuwa na akaunti FAS default kwenye seva FAS.

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Kwa maana mimi, napenda SSH kwa rbowden@nice.fas.harvard.edu.

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Ni kwenda kuuliza kwamba mara ya kwanza, na wewe kusema ndiyo.

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Password yangu ni kwenda tu kuwa FAS yangu password.

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Na hivyo sasa, mimi nina SSHed servrar nzuri, na mimi siwezi kufanya kitu chochote nataka hapa.

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mengi ya madarasa unaweza kuchukua, kama 124, ni kwenda na wewe upload stuff kwa hapa

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kwa kweli kuwasilisha tatizo seti yako.

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Lakini kusema huna upatikanaji wa appliance yako.

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Basi unaweza kufanya mambo, kama hapa, atasema -

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Hii ni sehemu yetu ya maswali.

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Ni kuuliza wewe kufanya hivyo katika appliance.

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Badala yake mimi itabidi kufanya hivyo kwenye seva.

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Mimi naenda unzip kwamba.

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tatizo ni kwenda kuwa kuwa wewe ni kutumika kwa kutumia kitu kama gedit

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au chochote ndani ya appliance.

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Wewe hutaenda kuwa na kwamba katika server FAS.

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Ni wote kwenda tu kuwa hii interface textual.

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Hivyo unaweza aidha moja, kujaribu kujifunza mhariri Nakala kwamba hawana wana.

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Wao wana Nano.

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Nano ni kawaida pretty rahisi kutumia.

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Unaweza kutumia mishale yako na aina ya kawaida.

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Basi hiyo ni si ngumu.

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Kama unataka kupata kweli dhana unaweza kutumia Emacs,

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ambayo mimi pengine haipaswi kuwa kufunguliwa kwa sababu mimi si hata kujua jinsi ya kuifunga Emacs.

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Document X, Udhibiti C? Yeah.

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Au unaweza kutumia VIM, ambayo ni nini mimi kutumia.

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Na hivyo wale ni chaguzi yako.

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Kama hutaki kufanya hivyo, unaweza pia, kama ukiangalia manual.cs50.net--

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Oh.

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Juu ya PC, unaweza kutumia SSH putty,

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ambayo wewe ni kwenda na download tofauti.

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Mac, unaweza tu na default Terminal matumizi au unaweza kushusha iTerm2,

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ambayo ni kama Terminal nice, dhana.

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Kama wewe kwenda manual.cs50.net utaona zilizounganishwa na Notepad + +,

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ambayo ni nini unaweza kutumia PC.

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Ni inakuwezesha SFTP kutoka Notepad + +, ambayo kimsingi ni SSH.

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Nini hii basi wewe kufanya ni kuhariri faili yako ndani ya nchi,

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na kisha kila unataka kuwaokoa, itakuwa ila kwa nice.fas,

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ambapo unaweza kisha kukimbia nao.

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Na sawa juu ya Mac ni kwenda kuwa TextWrangler.

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Hivyo inakuwezesha kufanya kitu kimoja.

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Ni inakuwezesha kuhariri faili ndani ya nchi na kuwaokoa nice.fas,

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ambapo unaweza kisha kukimbia nao.

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Hivyo kama wewe ni milele kukwama bila appliance, una chaguzi hizi

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bado kufanya tatizo seti yako.

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Tatizo moja ni itakuwa kwamba wewe si kwenda kuwa na maktaba CS50

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kwa sababu nice.fas haina na default kuwa na kwamba.

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Unaweza ama download maktaba CS50 -

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Sidhani mimi haja ya kuwa katika hatua hii.

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Unaweza ama download maktaba CS50 na nakala yake juu ya nice.fas,

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au nadhani katika hatua hii sisi si kutumia tena anyway.

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Au kama sisi kufanya, unaweza kwa muda kuwa badala yake pamoja na

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utekelezaji wa majukumu katika maktaba CS50 anyway.

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Ili isiwe kwamba mengi ya kizuizi.

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Na kwamba ni kwamba.

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>> Nitakwenda nyuma appliance sasa, tutaweza kufanya kila kitu katika appliance.

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Kuangalia sehemu yetu ya maswali, mwanzoni, kama nilivyosema katika barua pepe yangu,

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sisi kuwa na majadiliano kuhusu moja mfupi walipaswa kukesha.

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Tuna imetuma Mabomba & na maswali haya matatu.

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>> Ambayo mkondo gani kazi kama printf kuandika na default?

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Hivyo mkondo. Nini ni mkondo?

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mkondo ni kimsingi kama ni baadhi tu -

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Si hata chanzo cha 1s na sekunde 0.

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mkondo ni kuuliza kwa hapa ni kiwango nje.

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Na hivyo kiwango nje ni mkondo kwamba wakati wewe kuandika hayo,

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inaonekana kwenye screen.

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Nje ya kiwango, na mkondo, ina maana wewe andika tu 1s na sekunde 0 kwa hiyo,

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na mwisho mwingine wa nje ya kiwango tu anayesoma kutoka mkondo huo.

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Ni kamba tu ya 1s na sekunde 0.

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Unaweza kuandika kwa mito au unaweza kusoma kutoka katika mikondo

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kutegemea na nini mkondo kweli ni.

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wengine wawili default mito ni ya kiwango katika upotofu na kiwango.

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Darasa katika wakati wowote ni huna GetString, ni kusubiri kwa wewe stuff pembejeo.

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Hivyo ni kusubiri kwa ajili yenu, ni kweli kusubiri juu ya kiwango katika,

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ambayo ni kweli nini kupata pale unapoandika kwenye keyboard.

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Wewe ni kuandika katika kiwango in

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Kosa la kimsingi ni sawa na nje ya kiwango,

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lakini ni maalumu katika kwamba wakati magazeti ya hitilafu ya kiwango,

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wewe ni walidhani tu magazeti ujumbe wa makosa kwa kuwa

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hivyo unaweza kutofautisha kati ya ujumbe wa mara kwa mara kuchapishwa kwa screen

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dhidi ya makosa ujumbe kutegemea kama walikwenda kwa kiwango nje au kosa standard.

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Files pia.

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Darasa la nje, kiwango katika, na makosa ya hali ni tu maalum ya mito,

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lakini kwa kweli faili yoyote, wakati wa kufungua faili, inakuwa mkondo ya ka

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ambapo unaweza kusoma tu kutoka mkondo huo.

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Wewe, kwa sehemu kubwa, unaweza kufikiria tu ya faili kama mkondo wa bytes.

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Basi nini mito gani wao kuandika na default? Darasa la nje.

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>> Nini tofauti kati ya> na >>?

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Je mtu yeyote kuangalia video ya zamani? Sawa.

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> Ni kwenda kuwa jinsi redirect katika files,

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na >> pia kwenda kwingine pato katika files,

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lakini ni badala ya kwenda append na faili.

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Kwa mfano, hebu kusema mimi kutokea kwa kuwa Dict haki hapa,

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na mambo tu ndani ya Dict ni paka, paka, mbwa, samaki, mbwa.

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Amri moja kuwa una mstari amri ni paka,

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ambayo ni kwenda tu na magazeti nini katika faili.

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Hivyo wakati mimi kusema paka Dict, ni kwenda magazeti paka, paka, mbwa, samaki, mbwa. Hiyo paka wote gani.

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Hiyo ina maana kwamba ni kuchapishwa kwa kiwango nje paka, paka, mbwa mbwa, samaki,.

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Kama mimi badala wanataka redirect kwamba faili, naweza kutumia> na REDIRECT kwa chochote faili ni.

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Mimi nitakuita faili faili.

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Hivyo sasa kama mimi ls, mimi utaona nina faili mpya iitwayo faili.

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Na kama mimi kufungua it up, itakuja kuwa na nini hasa paka kuweka katika mstari amri.

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Hivyo sasa kama mimi kufanya hivyo tena, basi ni kwenda kwingine pato katika faili,

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na mimi naenda kuwa sawa exact kitu.

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Basi kitaalam, kabisa overrode nini tulikuwa.

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Na tutaweza kuona kama mimi kubadili Dict, mimi alichukua nje mbwa.

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Sasa kama sisi paka Dict katika faili tena, tunakwenda na toleo mpya na mbwa kuondolewa.

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Hivyo kabisa yanapewa kipaumbele hilo.

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Badala yake, kama sisi kutumia >>, ni kwenda append faili.

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Sasa, kufungua faili, tunaona tuna tu kitu kimoja kuchapishwa mara mbili

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sababu ilikuwa huko mara moja, kisha sisi umeongezewa awali.

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Hivyo kwamba ni nini> na >> kufanya.

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Je moja ijayo kuuliza - Haina kuuliza kuhusu hilo.

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>> nyingine moja kwamba tuna ni <, ambayo kama> redirects nje ya kiwango,

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<Inaenda redirect kiwango in

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Hebu angalia kama tuna mfano.

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Siwezi kuandika moja mziki haraka.

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Hebu kuchukua faili yoyote, hello.c.

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Kiasi moja kwa moja faili.

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Mimi tu kupata kamba na kisha kuchapisha "Hello" lolote string mimi tu aliingia mara.

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Hivyo kufanya hujambo na kisha /. Hello.

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Sasa ni uvuvio yangu kuingia kitu,

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ambayo ina maana ni kusubiri juu ya mambo kuwa aliingia katika kiwango in

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Basi ingieni chochote mimi nataka katika kiwango in Sisi ni tu kwenda kusema Hello, Rob!

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Basi ni uchapishaji kwa kiwango nje Hello, Rob!

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Kama mimi kufanya. / Hujambo na kisha kuelekeza,

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kwa sasa unaweza tu REDIRECT kutoka faili.

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Hivyo kama mimi kuweka katika baadhi ya SVG, txt, na mimi kuweka Rob,

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kama mimi kukimbia hujambo na kisha kuelekeza txt faili katika. / hello, ni kwenda kusema Hello, Rob! mara moja.

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Wakati kwanza anapata GetString na ni kusubiri juu ya kiwango katika,

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kiwango katika ni tena kusubiri kwenye keyboard kwa data kupata aliingia.

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Badala yake, tuna itaelekezwa kiwango katika kusoma kutoka txt faili.

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Na hivyo ni kwenda kusoma kutoka txt SVG, ambayo ni haki Rob line,

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na kisha itakuja magazeti Hello, Rob!

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Na kama alitaka, mimi naweza pia kufanya / hujambo <txt.

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na kisha kiwango kwamba uchapishaji ni, ambayo ni Hello, Rob!,

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Naweza kuelekeza kwamba ndani ya faili yake mwenyewe.

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Mimi itabidi kuwaita faili hujambo - hakuna, mimi si, kwa sababu hiyo ni executable - txt2.

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Sasa, txt2 ataenda kuwa na pato la. / Hujambo <txt, ambayo ni kwenda kuwa Hello, Rob!

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>> Maswali?

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>> Sawa. Hivyo basi hapa tuna bomba.

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Mabomba ni kitengo cha mwisho cha redirection.

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>> Oh. Nadhani moja zaidi kitengo cha redirection ni kama badala ya> kufanya 2>,

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hiyo imetuma makosa ya hali.

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Hivyo kama kitu akaenda makosa ya hali, ingekuwa si kupata kuweka ndani ya txt2.

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Lakini taarifa kama mimi kufanya 2>, basi ni bado uchapishaji Hello, Rob! kwa mstari amri

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kwa sababu mimi nina tu imetuma makosa ya hali, mimi nina haielekezi kiwango nje.

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Kiwango makosa na nje ya kiwango ni tofauti.

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Kama alitaka kweli kuandika makosa ya hali,

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basi mimi naweza kubadili hili kuwa fprintf kwa stderr.

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Hivyo printf, kwa default, prints kwa nje standard.

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Kama mimi nataka magazeti kwa makosa ya hali manually, basi nina kutumia fprintf

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na bayana nini nataka magazeti kwa.

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Kama badala sikuwa stdout fprintf, basi, kwamba kimsingi ni sawa na printf.

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Lakini kwa kosa fprintf standard.

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Hivyo sasa, ikiwa mimi redirect hii katika txt2, Hello, Rob! bado kupata kuchapishwa katika mstari amri

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tangu ni kupata kuchapishwa kwa makosa ya hali na mimi nina tu imetuma kiwango nje.

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Kama mimi sasa redirect makosa ya hali, sasa hawakuwa kupata kuchapishwa, na txt2 ni kwenda kuwa Hello, Rob!

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Hivyo sasa, unaweza print yako makosa halisi kwa makosa ya hali

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na magazeti yako ujumbe wa mara kwa mara nje ya kiwango.

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Na hivyo wakati wewe kuendesha programu yako, unaweza kukimbia kama /. Hujambo aina hii na 2>

189
00:14:31,420 --> 00:14:33,800
hivyo kwamba programu yako ni kwenda kukimbia kwa kawaida,

190
00:14:33,800 --> 00:14:38,400
lakini ujumbe wa makosa yoyote kwamba kupata unaweza kuangalia baadaye katika upotofu logi yako,

191
00:14:38,400 --> 00:14:44,500
hivyo makosa, na kisha kuangalia baadaye na makosa yako faili itakuwa na makosa yoyote yaliyotokea.

192
00:14:45,200 --> 00:14:47,540
>> Maswali?

193
00:14:47,540 --> 00:14:58,070
>> moja ya mwisho ni bomba, ambayo unaweza kufikiria kama kuchukua kiwango kutoka amri moja

194
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na kuifanya ya kiwango katika amri ya pili.

195
00:15:01,210 --> 00:15:05,570
mfano hapa ni mwangwi ni mstari amri kitu

196
00:15:05,570 --> 00:15:11,840
kwamba ni kwenda tu echo chochote mimi kuweka kama hoja yake. Mimi sitaweza quotes.

197
00:15:11,840 --> 00:15:16,150
Echo blah, blah, blah ni kwenda tu na magazeti blah, blah, blah.

198
00:15:16,150 --> 00:15:20,600
Kabla ya hapo, mimi nina alisema alikuwa na kuweka ndani ya Rob faili txt

199
00:15:20,600 --> 00:15:28,830
sababu naweza tu redirect files txt, badala yake, / kama mimi echo Rob

200
00:15:28,830 --> 00:15:35,520
na kisha bomba ndani / hello, ambayo pia kufanya aina moja ya kitu..

201
00:15:35,520 --> 00:15:39,160
Hii ni kuchukua pato la amri hii, echo Rob,

202
00:15:39,160 --> 00:15:43,610
na kutumia kama pembejeo kwa /. hello.

203
00:15:44,790 --> 00:15:49,560
Unaweza kufikiria kama kwanza redirect echo Rob katika faili

204
00:15:49,560 --> 00:15:54,160
na kisha kuyaingiza katika faili. / hujambo kwamba mara tu outputted.

205
00:15:54,160 --> 00:15:57,850
Lakini itachukua muda faili nje ya picha.

206
00:16:01,890 --> 00:16:04,460
>> Maswali juu ya hilo?

207
00:16:04,460 --> 00:16:07,150
>> Swali la pili ni kwenda kuhusisha hii.

208
00:16:07,150 --> 00:16:15,310
Nini bomba unaweza kutumia ili kupata idadi ya majina ya kipekee katika faili inayoitwa names.txt?

209
00:16:15,310 --> 00:16:24,160
amri tunakwenda wanataka kutumia hapa ni ya kipekee, hivyo uniq, na kisha WC.

210
00:16:24,160 --> 00:16:28,840
Unaweza kufanya mtu uniq kwa kweli kuangalia nini kwamba gani,

211
00:16:28,840 --> 00:16:34,840
na ni kwenda tu kwa chujio mistari karibu vinavyolingana kutoka pembejeo.

212
00:16:34,840 --> 00:16:40,690
Na mtu WC ni kwenda magazeti newline, neno, na makosa Byte kwa kila faili.

213
00:16:40,690 --> 00:16:43,760
Na moja ya mwisho tunakwenda wanataka kutumia ni aina ya,

214
00:16:43,760 --> 00:16:47,410
ambayo ni kwenda tu aina ya mistari ya faili txt.

215
00:16:47,410 --> 00:16:58,080
Kama mimi kufanya baadhi ya faili txt, names.txt, na ni Rob, Tommy, Joseph, Tommy, Joseph, RJ, Rob,

216
00:16:58,080 --> 00:17:03,910
nini nataka kufanya hapa ni kupata idadi ya majina ya kipekee katika faili hii.

217
00:17:03,910 --> 00:17:08,750
Hivyo kile lazima jibu kuwa? >> [Mwanafunzi] 4. >> Yeah.

218
00:17:08,750 --> 00:17:13,780
Ni lazima 4 tangu Rob, Tommy, Joseph, RJ ni pekee tu majina katika faili hii.

219
00:17:13,780 --> 00:17:20,180
hatua ya kwanza, kama mimi tu kufanya neno juu ya kuhesabu names.txt,

220
00:17:20,180 --> 00:17:24,290
hii ni kweli akiniambia kila kitu.

221
00:17:24,290 --> 00:17:32,560
Hii ni kweli uchapishaji - basi s kuona, mtu WC - newlines, maneno na kuhesabu Byte.

222
00:17:32,560 --> 00:17:38,270
Kama mimi tu huduma ya juu mistari, naweza tu kufanya WC-l names.txt.

223
00:17:41,730 --> 00:17:44,300
Basi hiyo ni hatua ya 1.

224
00:17:44,300 --> 00:17:50,510
Lakini mimi sitaki names.txt WC-l kwa sababu tu ina names.txt majina yote,

225
00:17:50,510 --> 00:17:54,170
na mimi nataka kuchuja wale yoyote yasiyo ya kipekee.

226
00:17:54,170 --> 00:18:01,200
Hivyo kama mimi kufanya uniq names.txt, kwamba hana kabisa nipe nini nataka

227
00:18:01,200 --> 00:18:03,760
kwa sababu majina duplicated bado kuna.

228
00:18:03,760 --> 00:18:07,690
Kwa nini? Kwa nini uniq si kufanya nipendavyo?

229
00:18:07,690 --> 00:18:10,500
[Mwanafunzi] marudio si [inaudible] >> Yeah.

230
00:18:10,500 --> 00:18:16,370
Kumbuka ukurasa mtu kwa uniq anasema mistari chujio vinavyolingana karibu.

231
00:18:16,370 --> 00:18:19,680
Hawako karibu, hivyo itakuwa si kuchuja yao.

232
00:18:19,680 --> 00:18:31,100
Kama mimi aina yao ya kwanza, aina names.txt ni kwenda kuweka mistari yote duplicate pamoja.

233
00:18:31,100 --> 00:18:34,450
Hivyo sasa ni kwamba aina names.txt.

234
00:18:34,450 --> 00:18:40,550
Mimi nina kwenda kutaka kutumia kama pembejeo kwa uniq, ambayo ni | uniq.

235
00:18:40,550 --> 00:18:43,390
Anipaye Joseph, RJ, Rob, Tommy,

236
00:18:43,390 --> 00:18:49,260
na mimi nataka kuwa matumizi kama pembejeo kwa WC-l,

237
00:18:49,260 --> 00:18:52,740
ambayo ni kwenda nipe 4.

238
00:18:52,740 --> 00:18:56,930
Kama inavyosema hapa, nini bomba unaweza kutumia?

239
00:18:56,930 --> 00:19:01,390
Unaweza kufanya mambo mengi kama vile kutumia mfululizo wa amri

240
00:19:01,390 --> 00:19:05,130
ambapo matumizi pato kutoka amri moja kama mchango wa amri ya pili.

241
00:19:05,130 --> 00:19:08,780
Unaweza kufanya mambo mengi, mengi ya mambo wajanja.

242
00:19:08,780 --> 00:19:11,440
>> Maswali?

243
00:19:12,910 --> 00:19:14,600
Sawa.

244
00:19:14,600 --> 00:19:17,880
Hiyo ni kwa ajili ya mabomba na redirection.

245
00:19:18,370 --> 00:19:24,090
>> Sasa tunaenda kwenye mambo halisi, stuff coding.

246
00:19:24,090 --> 00:19:29,100
Ndani ya hii PDF, utaona amri hii,

247
00:19:29,100 --> 00:19:32,950
na wewe utakuwa unataka kukimbia amri hii katika appliance yako.

248
00:19:36,240 --> 00:19:42,250
wget ni amri kwa ajili tu ya kupata kitu kutoka mtandao, kimsingi,

249
00:19:42,250 --> 00:19:45,180
hivyo wget na hii URL.

250
00:19:45,180 --> 00:19:49,110
Kama wewe akaenda URL hiki katika kivinjari chako, ingekuwa download kwamba faili.

251
00:19:49,110 --> 00:19:52,510
Mimi tu clicked juu yake, hivyo ni kupakuliwa faili kwa ajili yangu.

252
00:19:52,510 --> 00:19:55,650
Lakini kuandika wget ya kitu ndani ya terminal

253
00:19:55,650 --> 00:19:58,620
ni kwenda tu kwa kushusha ni ndani ya terminal yako.

254
00:19:58,620 --> 00:20:02,750
Nina section5.zip, na utasikia wanataka unzip section5.zip,

255
00:20:02,750 --> 00:20:06,520
ambayo ni kwenda kukupa folder aitwaye section5,

256
00:20:06,520 --> 00:20:11,550
ambayo ni kwenda wote wa files tunakwenda kutumia leo ndani yake.

257
00:20:33,380 --> 00:20:37,710
Kama faili programu hizi 'majina kupendekeza, wao uko Buggy kidogo,

258
00:20:37,710 --> 00:20:40,990
hivyo lengo lako ni kwa takwimu kwa nini kutumia gdb.

259
00:20:40,990 --> 00:20:44,560
Je, kila mtu kuwa wao kupakuliwa / kujua jinsi ya kupata yao kupakuliwa

260
00:20:44,560 --> 00:20:47,480
katika appliance zao? Sawa.

261
00:20:47,480 --> 00:20:56,400
>> Mbio ./buggy1, itakuwa kusema Mkato kosa (msingi kutupwa),

262
00:20:56,400 --> 00:21:00,500
ambayo wakati wowote kupata segfault, ni kitu mbaya.

263
00:21:00,500 --> 00:21:03,810
Chini ya mazingira gani unayoipata segfault?

264
00:21:03,810 --> 00:21:08,210
[Mwanafunzi] Dereferencing pointer null. >> Yeah. Hivyo kwamba ni mfano mmoja.

265
00:21:08,210 --> 00:21:11,580
Dereferencing pointer null utaenda kupata segfault.

266
00:21:11,580 --> 00:21:16,720
Nini segfault njia ni wewe ni kugusa kumbukumbu unapaswa kuwa kugusa.

267
00:21:16,720 --> 00:21:21,350
Hivyo dereferencing pointer null ni kugusa anuani 0,

268
00:21:21,350 --> 00:21:28,060
na kimsingi, kila kompyuta siku hizi wanasema kwamba anuani 0 ni kumbukumbu unapaswa kuwa kugusa.

269
00:21:28,060 --> 00:21:31,920
Hivyo ndiyo sababu dereferencing matokeo pointer null katika segfault.

270
00:21:31,920 --> 00:21:37,210
Wakati wewe kutokea si initialize pointer, basi ina thamani ya takataka,

271
00:21:37,210 --> 00:21:41,520
na hivyo wakati wewe kujaribu dereference hivyo, katika uwezekano wote wewe ni kugusa kumbukumbu

272
00:21:41,520 --> 00:21:43,540
kwamba katika katikati ya pa.

273
00:21:43,540 --> 00:21:45,650
Kama kutokea kwa kupata bahati na thamani ya takataka

274
00:21:45,650 --> 00:21:48,440
kilichotokea kwa uhakika na mahali fulani kwenye stack au kitu,

275
00:21:48,440 --> 00:21:50,820
basi wakati wewe dereference kwamba pointer, ambayo wewe si initialized,

276
00:21:50,820 --> 00:21:52,730
hakuna kitu kwenda vibaya.

277
00:21:52,730 --> 00:21:55,480
Lakini kama ni akizungumzia, kusema, mahali fulani kati ya stack na chungu,

278
00:21:55,480 --> 00:21:59,850
au ni akizungumzia tu mahali fulani kwamba haijawahi kutumika kwa mpango wako bado,

279
00:21:59,850 --> 00:22:02,240
basi wewe ni kugusa kumbukumbu unapaswa kuwa kugusa na wewe segfault.

280
00:22:02,240 --> 00:22:06,370
Wakati kuandika kazi ya kujirudia na recurses mara nyingi sana

281
00:22:06,370 --> 00:22:08,720
na stack yenu inakua kubwa mno na collides stack katika mambo

282
00:22:08,720 --> 00:22:12,270
kwamba ni lazima kuwa colliding na, wewe ni kugusa kumbukumbu unapaswa kuwa na kugusa,

283
00:22:12,270 --> 00:22:14,810
hivyo segfault.

284
00:22:14,810 --> 00:22:17,010
Hiyo ni nini segfault ni.

285
00:22:17,010 --> 00:22:21,810
>> Ni sababu hiyo hiyo pia kwamba kama una kamba kama -

286
00:22:21,810 --> 00:22:23,930
hebu kwenda nyuma ya mpango uliopita.

287
00:22:23,930 --> 00:22:28,530
Katika hello.c-I'm just kwenda kufanya kitu kingine.

288
00:22:28,530 --> 00:22:33,770
Char * s = "hello dunia!";

289
00:22:33,770 --> 00:22:42,310
Kama mimi kutumia * s = kitu au s [0] = 'X';

290
00:22:42,310 --> 00:22:47,290
hivyo kufanya hello, /. hello, kwa nini kuwa segfault?

291
00:22:48,410 --> 00:22:51,250
Kwa nini hii segfault?

292
00:22:55,660 --> 00:22:57,890
Gani, unatarajia nini?

293
00:22:57,890 --> 00:23:06,640
Kama mimi printf ("% s \ n", s); gani, unatarajia kuchapishwa?

294
00:23:06,640 --> 00:23:09,930
[Mwanafunzi] X hello. >> Yeah.

295
00:23:09,930 --> 00:23:15,140
Tatizo ni kwamba wakati wewe kutangaza string kama hii,

296
00:23:15,140 --> 00:23:18,190
s ni pointer kwamba kinaendelea kwenda stack,

297
00:23:18,190 --> 00:23:25,880
na kile ambacho ni akizungumzia ni hii kamba ambayo ni zilizomo katika kumbukumbu kusoma tu.

298
00:23:25,880 --> 00:23:30,560
Hivyo tu kwa jina, kumbukumbu kusoma tu, unapaswa kupata wazo

299
00:23:30,560 --> 00:23:33,010
kwamba kama wewe kujaribu kubadili nini katika kumbukumbu kusoma tu,

300
00:23:33,010 --> 00:23:36,670
wewe ni kufanya kitu unapaswa kufanya na kumbukumbu na wewe segfault.

301
00:23:36,670 --> 00:23:45,360
Hii ni kweli tofauti kubwa kati ya Char * s na Char s [].

302
00:23:45,360 --> 00:23:48,790
Hivyo Char s [], sasa string hii ni kwenda kuwekwa katika stack,

303
00:23:48,790 --> 00:23:53,960
na stack si kusoma tu, ambayo ina maana kwamba hii lazima kazi kikamilifu faini.

304
00:23:55,500 --> 00:23:57,370
Na inakuwa hivyo.

305
00:23:57,370 --> 00:24:06,250
Kumbuka kwamba wakati mimi kufanya Char * s = "hello dunia!", S yenyewe ni juu ya stack

306
00:24:06,250 --> 00:24:10,390
lakini s pointi mahali pengine, na kwamba mahali pengine hufanyika kuwa kusoma tu.

307
00:24:10,390 --> 00:24:15,640
Lakini Char s [] ni kitu tu juu ya stack.

308
00:24:17,560 --> 00:24:21,760
Basi hiyo ni mfano mwingine wa segfault kinachotokea.

309
00:24:21,760 --> 00:24:27,820
>> Tuliona kwamba ./buggy1 ilisababisha segfault.

310
00:24:27,820 --> 00:24:31,810
Katika nadharia, unapaswa kuangalia buggy1.c mara moja.

311
00:24:31,810 --> 00:24:35,170
Badala yake, tutaweza kuangalia ni kwa njia ya gdb.

312
00:24:35,170 --> 00:24:37,750
Taarifa kwamba wakati wewe kupata Mkato kosa (msingi kutupwa),

313
00:24:37,750 --> 00:24:40,850
kupata faili hii juu ya msingi hapa kuitwa.

314
00:24:40,850 --> 00:24:45,200
Kama sisi ls-l, tutaweza kuona msingi kwamba ni kawaida faili pretty kubwa.

315
00:24:45,200 --> 00:24:51,580
Hii ni idadi ya ka ya SVG, hivyo inaonekana kama ni 250-kitu kilobytes.

316
00:24:51,580 --> 00:24:56,120
Sababu hii ni kwamba kile dampo msingi kweli ni

317
00:24:56,120 --> 00:25:01,410
ni wakati mpango wako shambulio, hali ya kumbukumbu ya mpango wako

318
00:25:01,410 --> 00:25:05,230
tu anapata kunakiliwa na pasted katika faili hii.

319
00:25:05,230 --> 00:25:07,270
Ni anapata kutupwa kwenye faili hilo.

320
00:25:07,270 --> 00:25:13,060
Mpango huu, wakati ni mbio, kilichotokea kuwa na matumizi ya kumbukumbu ya kilobytes karibu 250,

321
00:25:13,060 --> 00:25:17,040
na hivyo kwamba ni nini got kutupwa kwenye faili hii.

322
00:25:17,040 --> 00:25:23,630
Sasa unaweza kuangalia faili kwamba kama sisi kufanya gdb buggy1 msingi.

323
00:25:23,630 --> 00:25:30,130
Tunaweza tu kufanya gdb buggy1, na kwamba itakuwa tu kuanzisha gdb mara kwa mara,

324
00:25:30,130 --> 00:25:33,800
kutumia buggy1 kama pembejeo faili wake.

325
00:25:33,800 --> 00:25:38,260
Lakini kama wewe kufanya gdb buggy1 msingi, basi ni hasa kwenda kuanza up gdb

326
00:25:38,260 --> 00:25:40,330
kwa kuangalia faili kwamba msingi.

327
00:25:40,330 --> 00:25:45,560
Na wewe kusema buggy1 njia gdb anajua kwamba kwamba faili msingi linatokana na mpango buggy1.

328
00:25:45,560 --> 00:25:49,580
Hivyo gdb buggy1 msingi ni kwenda mara moja kutuletea

329
00:25:49,580 --> 00:25:52,060
kwa mpango ambapo kilichotokea kusitisha.

330
00:25:57,720 --> 00:26:02,340
Tunaona hapa Programu terminated pamoja na ishara ya 11, Mkato kosa.

331
00:26:02,340 --> 00:26:10,110
Sisi kutokea kwa kuona mistari ya kanisa, ambayo pengine ni hayasaidii sana.

332
00:26:10,110 --> 00:26:15,360
Lakini kama wewe aina bt au backtrace, ambayo inaenda kuwa kazi

333
00:26:15,360 --> 00:26:19,430
kwamba inatupa orodha ya muafaka wetu wa sasa stack.

334
00:26:19,430 --> 00:26:23,150
Hivyo backtrace. Inaonekana kama sisi tu muafaka stack mbili.

335
00:26:23,150 --> 00:26:26,310
kwanza ni stack yetu kuu frame,

336
00:26:26,310 --> 00:26:29,810
na pili ni sura stack kwa kazi hii ya kuwa sisi kutokea kwa kuwa katika,

337
00:26:29,810 --> 00:26:34,440
ambayo inaonekana kama sisi tu code kanisa kwa ajili ya.

338
00:26:34,440 --> 00:26:38,050
Basi hebu kwenda nyuma katika kazi yetu kuu,

339
00:26:38,050 --> 00:26:42,300
na kwa kufanya hivyo tunaweza kufanya sura 1, na nadhani tunaweza pia kufanya chini,

340
00:26:42,300 --> 00:26:45,160
lakini mimi kamwe karibu kufanya chini - au juu. Yeah.

341
00:26:45,160 --> 00:26:50,710
Juu na chini. Hadi huleta you up moja stack frame, chini huleta wewe chini frame stack.

342
00:26:50,710 --> 00:26:53,240
Mimi huwa kamwe kutumia.

343
00:26:53,240 --> 00:26:59,120
Mimi tu kusema hasa sura 1, ambayo ni kwenda frame labeled 1.

344
00:26:59,120 --> 00:27:01,750
Frame 1 anaenda kutufanya frame kuu stack,

345
00:27:01,750 --> 00:27:05,570
na anasema haki hapa mstari wa kanuni sisi kutokea kwa kuwa katika.

346
00:27:05,570 --> 00:27:07,950
Kama sisi alitaka mistari michache zaidi ya kificho, tunaweza kusema orodha,

347
00:27:07,950 --> 00:27:11,280
na kwamba kinaendelea kutupa mistari yote ya maadili ya kuzunguka.

348
00:27:11,280 --> 00:27:13,360
mstari sisi segfaulted saa ilikuwa 6:

349
00:27:13,360 --> 00:27:17,360
kama (strcmp ("CS50 miamba", argv [1]) == 0).

350
00:27:17,360 --> 00:27:24,130
Kama si dhahiri bado, unaweza kupata moja kwa moja kutoka hapa tu kwa kufikiri kwa nini segfaulted.

351
00:27:24,130 --> 00:27:28,800
Lakini tunaweza kuchukua hatua moja zaidi na kusema, "Kwanini argv [1] segfault?"

352
00:27:28,800 --> 00:27:38,830
Magazeti Hebu argv [1], na inaonekana kama 0x0 ni, ambayo ni pointer null.

353
00:27:38,830 --> 00:27:44,750
Sisi ni strcmping miamba CS50 na null, na ili kwenda segfault.

354
00:27:44,750 --> 00:27:48,280
Na kwa nini ni argv [1] null?

355
00:27:48,640 --> 00:27:51,280
[Mwanafunzi] sababu sisi hakutoa ni hoja amri yoyote-line.

356
00:27:51,280 --> 00:27:53,390
Yeah. Sisi hatukukubaliana ni hoja amri yoyote-line.

357
00:27:53,390 --> 00:27:58,460
Hivyo ./buggy1 ni kwenda tu kuwa argv [0] kuwa ./buggy1.

358
00:27:58,460 --> 00:28:02,100
Ni si kwenda kuwa na argv [1], ili kwenda segfault.

359
00:28:02,100 --> 00:28:07,450
Lakini kama, badala yake, mimi kufanya tu CS50, ni kwenda kusema Unaweza kupata D

360
00:28:07,450 --> 00:28:09,950
kwa sababu hiyo ni nini ni zinatakiwa kufanya.

361
00:28:09,950 --> 00:28:15,240
Kuangalia buggy1.c, ni walidhani magazeti "Unaweza kupata D" -

362
00:28:15,240 --> 00:28:20,820
Kama argv [1] ni si "CS50 miamba", "Unaweza kupata D", mwingine "Unaweza kupata!"

363
00:28:20,820 --> 00:28:25,660
Hivyo kama tunataka, tunahitaji hii kulinganisha kama kweli,

364
00:28:25,660 --> 00:28:28,710
ambayo ina maana kwamba inalinganishwa na 0.

365
00:28:28,710 --> 00:28:31,100
Hivyo argv [1] inahitaji "kuwa CS50 miamba".

366
00:28:31,100 --> 00:28:35,660
Kama unataka kufanya hivyo juu ya mstari amri, unahitaji kutumia \ kutoroka nafasi.

367
00:28:35,660 --> 00:28:41,690
Hivyo CS50 \ miamba na Wewe kupata A!

368
00:28:41,690 --> 00:28:44,060
Kama huwezi kufanya backslash, kwa nini hii si kazi?

369
00:28:44,060 --> 00:28:47,190
[Mwanafunzi] Ni hoja mbili tofauti. >> Yeah.

370
00:28:47,190 --> 00:28:52,540
Argv [1] ni kwenda kuwa CS50, na argv [2] ni kwenda kuwa miamba. Sawa.

371
00:28:52,540 --> 00:28:56,470
>> Sasa ./buggy2 ni kwenda segfault tena.

372
00:28:56,470 --> 00:29:01,880
Badala ya kufungua faili kwa msingi wake, tutaweza tu kufungua buggy2 moja kwa moja,

373
00:29:01,880 --> 00:29:05,000
hivyo gdb buggy2.

374
00:29:05,000 --> 00:29:09,590
Sasa kama sisi tu kuendesha programu yetu, basi ni kwenda kusema Programu ya kupokea signal SIGSEGV,

375
00:29:09,590 --> 00:29:15,530
ambayo ni segfault signal, na hii ni mahali ambapo ilivyotokea kutokea.

376
00:29:15,530 --> 00:29:21,250
Kuangalia backtrace yetu, tunaona kwamba tulikuwa katika oh_no kazi,

377
00:29:21,250 --> 00:29:23,900
ambalo alikuwa amepewa na Dinky kazi, ambayo aliitwa na binky kazi,

378
00:29:23,900 --> 00:29:26,460
ambalo alikuwa amepewa na kuu.

379
00:29:26,460 --> 00:29:31,680
Tunaweza pia kuona hoja kwa kazi hizi.

380
00:29:31,680 --> 00:29:34,680
hoja kwa Dinky na binky alikuwa 1.

381
00:29:34,680 --> 00:29:44,390
Kama sisi orodha oh_no kazi, tunaona kwamba oh_no ni kufanya tu Char ** s = null;

382
00:29:44,390 --> 00:29:47,410
* S = "BOOM";

383
00:29:47,410 --> 00:29:50,330
Kwa nini kuwa kushindwa?

384
00:29:54,330 --> 00:29:58,380
[Mwanafunzi] Huwezi dereference pointer null? >> Yeah.

385
00:29:58,380 --> 00:30:06,090
Hii ni kusema tu s ni null, bila kujali kama kwamba hutokea kwa kuwa ** Char,

386
00:30:06,090 --> 00:30:12,070
ambayo, kulingana na jinsi kutafsiri, inaweza kuwa pointer pointer string

387
00:30:12,070 --> 00:30:15,550
au safu ya masharti.

388
00:30:15,550 --> 00:30:21,430
Ni s ni null, hivyo * s dereferencing pointer null,

389
00:30:21,430 --> 00:30:24,800
na hivyo hii ni kwenda kwa ajali.

390
00:30:24,800 --> 00:30:27,540
Hii ni moja ya njia ya haraka unaweza uwezekano segfault.

391
00:30:27,540 --> 00:30:31,300
Ni tu kutangaza pointer null na mara segfaulting.

392
00:30:31,300 --> 00:30:34,570
Hiyo ni nini oh_no ni kufanya.

393
00:30:34,570 --> 00:30:43,400
Kama Tunakwenda frame moja, basi sisi ni kwenda kupata ndani ya kazi ambayo inaitwa oh_no.

394
00:30:43,400 --> 00:30:44,830
Mimi haja ya kufanya kwamba chini.

395
00:30:44,830 --> 00:30:48,610
Kama huna kuingia amri na wewe tu hit Enter tena,

396
00:30:48,610 --> 00:30:52,350
itakuwa rudia tu amri ya awali kwamba wewe mbio.

397
00:30:52,350 --> 00:30:56,610
Sisi ni katika sura ya 1.

398
00:30:56,610 --> 00:31:04,650
Kuorodhesha frame hii, tunaona hapa ni kazi yetu.

399
00:31:04,650 --> 00:31:08,520
Unaweza hit orodha tena, au unaweza kufanya orodha 20 na orodha itakuwa ni zaidi.

400
00:31:08,520 --> 00:31:13,640
Dinky kazi anasema kama i ni 1, kisha kwenda kufanya kazi oh_no,

401
00:31:13,640 --> 00:31:15,960
mwingine kwenda kazi slinky.

402
00:31:15,960 --> 00:31:18,700
Na tunajua i ni 1 kwa sababu sisi kutokea kwa kuona hapa

403
00:31:18,700 --> 00:31:22,560
Dinky kwamba aliitwa na hoja 1.

404
00:31:22,560 --> 00:31:27,560
Au unaweza tu wala magazeti i na matokeo yake watasema i ni 1.

405
00:31:27,560 --> 00:31:33,770
Sisi ni sasa katika Dinky, na kama sisi kwenda mwingine frame, tunajua kuishia katika binky.

406
00:31:33,770 --> 00:31:36,600
Up. Sasa tuko katika binky.

407
00:31:36,600 --> 00:31:41,340
Kuorodhesha kazi hii - orodha kutoka kabla ya nusu kunikatilia mbali -

408
00:31:41,340 --> 00:31:52,670
ilianza kama kama i ni 0, kisha tunakwenda kuiita oh_no, mwingine piga Dinky.

409
00:31:52,670 --> 00:31:57,000
Tunajua i alikuwa 1, hivyo kuitwa Dinky.

410
00:31:57,000 --> 00:32:05,030
Na sasa tuko nyuma katika kuu, na kuu ni kwenda tu kuwa int i = rand ()% 3;

411
00:32:05,030 --> 00:32:08,790
Hiyo ni kwenda tu kukupa namba random kwamba ni aidha 0, 1, au 2.

412
00:32:08,790 --> 00:32:12,780
Ni kwenda kuwaita binky na idadi hiyo, nayo atarudi 0.

413
00:32:12,780 --> 00:32:16,700
Kuangalia hii,

414
00:32:16,700 --> 00:32:19,880
tu kutembea kupitia mpango manually bila ya mbio mara moja,

415
00:32:19,880 --> 00:32:25,400
ungependa kuweka uhakika wa mapumziko katika kuu, ambayo ina maana kwamba wakati sisi kuendesha programu ya

416
00:32:25,400 --> 00:32:31,020
mpango wako anaendesha hadi hits hatua ya mapumziko.

417
00:32:31,020 --> 00:32:35,450
Hivyo mbio mpango, itakuwa kukimbia na basi itakuwa hit kazi kuu na kuacha kuendesha.

418
00:32:35,450 --> 00:32:44,700
Sasa tuko ndani ya kuu, na hatua au ijayo anaenda kututoa kwa line ya pili ya code.

419
00:32:44,700 --> 00:32:47,050
Unaweza kufanya hatua au ijayo.

420
00:32:47,050 --> 00:32:51,800
Kupiga ijayo, sasa i umeanzishwa kwa rand ()% 3, ili tuweze kuchapa thamani ya i,

421
00:32:51,800 --> 00:32:55,280
na matokeo yake watasema i ni 1.

422
00:32:55,280 --> 00:32:58,110
Sasa ni gani jambo kama sisi kutumia ijayo au hatua.

423
00:32:58,110 --> 00:33:01,000
Mimi nadhani ni mattered katika moja ya awali, lakini sisi wanataka kutumia ijayo.

424
00:33:01,000 --> 00:33:06,000
Kama sisi kutumia hatua, sisi hatua katika kazi, ambayo ina maana ya kuangalia kitu halisi

425
00:33:06,000 --> 00:33:07,940
kwamba kinajitokeza ndani ya binky.

426
00:33:07,940 --> 00:33:10,510
Kama sisi kutumia ijayo, basi ina maana kwenda juu ya kazi

427
00:33:10,510 --> 00:33:14,070
na kwenda tu kwa line ya pili ya kificho katika kazi yetu kuu.

428
00:33:14,070 --> 00:33:17,900
Haki hapa kwenye mstari huu, nilikuwa katika ambapo alisema rand ()% 3;

429
00:33:17,900 --> 00:33:21,320
kama mimi hatua, ingekuwa kwenda katika utekelezaji wa rand

430
00:33:21,320 --> 00:33:25,110
na kuangalia nini kinatokea pale, na mimi nilikuwa hatua kupitia kazi rand.

431
00:33:25,110 --> 00:33:26,920
Lakini mimi sijali kuhusu kazi rand.

432
00:33:26,920 --> 00:33:30,190
Mimi nataka tu kwenda line ya pili ya kificho katika kuu, hivyo mimi kutumia ijayo.

433
00:33:30,190 --> 00:33:35,800
Lakini sasa mimi kufanya huduma kuhusu kazi binky, hivyo nataka hatua katika hilo.

434
00:33:35,800 --> 00:33:37,730
Sasa mimi nina katika binky.

435
00:33:37,730 --> 00:33:42,040
mstari wa kwanza wa code ni kwenda kusema kama (i == 0), mimi kuchukua hatua,

436
00:33:42,040 --> 00:33:44,930
tunaona sisi kuishia katika Dinky.

437
00:33:44,930 --> 00:33:51,620
Kama sisi orodha ya mambo, tunaona kwamba ni checked ni i = 0.

438
00:33:51,620 --> 00:33:55,470
i si sawa na 0, hivyo alikwenda kwa hali kingine,

439
00:33:55,470 --> 00:33:59,540
ambayo ni kwenda kuwaita Dinky (i).

440
00:33:59,540 --> 00:34:04,030
Unaweza kupata kuchanganyikiwa.

441
00:34:04,030 --> 00:34:07,380
Kama wewe tu kuangalia mistari haya moja kwa moja, unaweza kufikiri kama (i == 0),

442
00:34:07,380 --> 00:34:10,800
sawa, basi mimi alichukua hatua na sasa mimi nina katika Dinky (i),

443
00:34:10,800 --> 00:34:14,120
unaweza kufikiri kwamba lazima maana i = 0 au kitu.

444
00:34:14,120 --> 00:34:18,980
No Ni tu ina maana kwamba anajua unaweza fimbo moja kwa moja kwa Dinky line (i).

445
00:34:18,980 --> 00:34:23,300
Sababu i si 0, hatua inayofuata ni si kwenda mwisho katika mwingine.

446
00:34:23,300 --> 00:34:26,239
Mwingine si line ni kwenda kuacha katika.

447
00:34:26,239 --> 00:34:31,570
Ni tu kwenda kwenda line ijayo inaweza kweli nitafanya, ambayo ni Dinky (i).

448
00:34:31,570 --> 00:34:36,090
Kukanyaga katika Dinky (i), sisi kuona kama (i == 1).

449
00:34:36,090 --> 00:34:42,670
Sisi tunajua i = 1, hivyo wakati sisi hatua, tunajua tunakwenda kuishia katika oh_no

450
00:34:42,670 --> 00:34:46,489
kwa sababu i = 1 wito oh_no kazi, ambayo unaweza kujipenyeza katika,

451
00:34:46,489 --> 00:34:52,969
ambayo ni kwenda kuweka Char ** s = kwa null na mara moja "BOOM".

452
00:34:54,270 --> 00:34:59,690
Na kisha kwa kweli kuangalia utekelezaji wa buggy2,

453
00:34:59,690 --> 00:35:04,590
hii, i ni kupata tu idadi random - 0, 1, au 2 - wito binky,

454
00:35:04,590 --> 00:35:10,610
ambayo kama i ni 0 ni wito oh_no, mwingine ni wito Dinky, ambayo huja hapa juu.

455
00:35:10,610 --> 00:35:18,100
Kama i ni 1, wito oh_no, mwingine piga slinky, ambayo kuja hapa,

456
00:35:18,100 --> 00:35:20,460
kama i ni 2, piga oh_no.

457
00:35:20,460 --> 00:35:24,720
Mimi wala hata kufikiri kuna njia -

458
00:35:24,720 --> 00:35:30,030
Je, mtu yeyote kuona njia ya kufanya programu hii ambayo si segfault?

459
00:35:30,030 --> 00:35:37,530
Kwa sababu isipokuwa mimi kukosa kitu, kama i ni 0, utasikia mara segfault,

460
00:35:37,530 --> 00:35:41,250
mwingine wewe kwenda kufanya kazi ambayo kama i ni 1 wewe segfault,

461
00:35:41,250 --> 00:35:44,540
mwingine wewe kwenda kufanya kazi ambapo kama i ni 2 wewe segfault.

462
00:35:44,540 --> 00:35:46,810
Hivyo hakuna jambo gani wewe, wewe segfault.

463
00:35:46,810 --> 00:35:52,380
>> Nadhani njia moja ya fixing itakuwa badala ya kufanya Char ** s = null,

464
00:35:52,380 --> 00:35:55,610
unaweza malloc nafasi kwa kamba hiyo.

465
00:35:55,610 --> 00:36:04,230
Tunaweza kufanya malloc (sizeof) - sizeof nini?

466
00:36:09,910 --> 00:36:15,190
[Mwanafunzi] (Char) * 5? >> Je, hii wanaonekana haki?

467
00:36:15,190 --> 00:36:21,060
Mimi nina kuchukua hii kazi kama mimi kweli mbio, lakini si nini nina kuangalia kwa.

468
00:36:24,400 --> 00:36:32,940
Angalia katika aina ya s. Hebu kuongeza * int, hivyo int * x.

469
00:36:32,940 --> 00:36:35,600
Napenda kufanya malloc (sizeof (int)).

470
00:36:35,600 --> 00:36:40,490
Au kama nilitaka safu ya 5, napenda kufanya (sizeof (int) * 5);

471
00:36:40,490 --> 00:36:44,210
Nini kama mimi na ** int?

472
00:36:46,260 --> 00:36:49,140
Gani Mimi malloc?

473
00:36:49,140 --> 00:36:53,510
[Mwanafunzi] Ukubwa wa pointer. >> Yeah. (Sizeof (int *));

474
00:36:53,510 --> 00:36:56,960
Same kitu chini hapa.

475
00:36:56,960 --> 00:37:01,280
Mimi nataka (sizeof (Char *));

476
00:37:06,170 --> 00:37:12,840
Hii ni kwenda kutenga nafasi kwa pointer kwamba anaonyesha "BOOM".

477
00:37:12,840 --> 00:37:15,330
Sihitaji kutenga nafasi kwa ajili ya "BOOM" yenyewe

478
00:37:15,330 --> 00:37:17,210
kwa sababu hii ni kimsingi sawa na kile alisema kabla ya

479
00:37:17,210 --> 00:37:20,870
ya Char * x = "BOOM".

480
00:37:20,870 --> 00:37:27,950
"BOOM" tayari upo. Hufanyika kwa kuwepo katika kanda kusoma tu ya kumbukumbu.

481
00:37:27,950 --> 00:37:35,200
Lakini tayari ipo, ambayo ina maana hii mstari wa kanuni, ikiwa ni s ** Char,

482
00:37:35,200 --> 00:37:43,900
kisha * s ni * Char na wewe ni kuweka hii * Char na kumweka kwa "BOOM".

483
00:37:43,900 --> 00:37:50,040
Kama nilitaka kunakili "BOOM" ndani, basi mimi ingekuwa haja ya kutenga nafasi kwa s.

484
00:37:55,170 --> 00:38:03,900
Nitafanya * s = malloc (sizeof (Char) * 5);

485
00:38:03,900 --> 00:38:06,210
Mbona 5?

486
00:38:06,210 --> 00:38:10,860
Mbona si 4? Inaonekana kama "BOOM" ni 4 wahusika. >> [Mwanafunzi] tabia null.

487
00:38:10,860 --> 00:38:14,580
Yeah. Yote ya masharti yako ni kwenda haja ya tabia null.

488
00:38:14,580 --> 00:38:23,590
Sasa siwezi kufanya kitu kama strcat - Ni nini kazi kwa kuiga kamba?

489
00:38:23,590 --> 00:38:28,520
[Mwanafunzi] cpy? >> Strcpy.

490
00:38:28,520 --> 00:38:32,700
mtu strcpy.

491
00:38:36,120 --> 00:38:39,590
Hivyo strcpy au strncpy.

492
00:38:39,590 --> 00:38:43,410
strncpy ni kidogo salama tangu unaweza specify hasa jinsi wengi wahusika,

493
00:38:43,410 --> 00:38:46,190
lakini hapa haijalishi sababu tunajua.

494
00:38:46,190 --> 00:38:50,340
Hivyo strcpy na kuangalia katika hoja.

495
00:38:50,340 --> 00:38:53,100
Hoja ya kwanza ni ya marudio yetu.

496
00:38:53,100 --> 00:38:56,770
Hoja ya pili ni chanzo chetu.

497
00:38:56,770 --> 00:39:10,310
Sisi ni kwenda na nakala katika marudio * wetu s pointer "BOOM".

498
00:39:10,310 --> 00:39:19,820
Kwa nini wapate unataka kufanya hii na strcpy badala ya kile tu tulikuwa kabla ya

499
00:39:19,820 --> 00:39:22,800
ya * s = "BOOM"?

500
00:39:22,800 --> 00:39:28,630
Kuna sababu unaweza kutaka kufanya hivyo, lakini nini ni sababu ya kwamba?

501
00:39:28,630 --> 00:39:31,940
[Mwanafunzi] Kama unataka kubadilisha kitu katika "BOOM". >> Yeah.

502
00:39:31,940 --> 00:39:37,950
Sasa siwezi kufanya kitu kama s [0] = 'X';

503
00:39:37,950 --> 00:39:48,190
kwa sababu ya pointi chungu na kwamba nafasi kwenye chungu kwamba s ni akizungumzia

504
00:39:48,190 --> 00:39:52,320
ni pointer nafasi zaidi juu ya chungu, ambayo ni hifadhi ya "BOOM".

505
00:39:52,320 --> 00:39:55,150
Hivyo nakala hii ya "BOOM" ni kuwa kuhifadhiwa katika lundo.

506
00:39:55,150 --> 00:39:58,780
Kuna nakala mbili za kitaalam "BOOM" katika programu yetu.

507
00:39:58,780 --> 00:40:03,500
Kuna mmoja tu kwanza hiyo aliyopewa na hii mara kwa mara "BOOM" kamba,

508
00:40:03,500 --> 00:40:09,250
na nakala ya pili ya "BOOM", strcpy umba nakala ya "BOOM".

509
00:40:09,250 --> 00:40:13,100
Lakini nakala ya "BOOM" ni kuwa kuhifadhiwa kwenye chungu, na chungu wewe ni bure kubadilika.

510
00:40:13,100 --> 00:40:17,250
chungu si kusoma tu, hivyo hiyo ina maana kwamba s [0]

511
00:40:17,250 --> 00:40:20,500
ni kwenda basi unaweza kubadilisha thamani ya "BOOM".

512
00:40:20,500 --> 00:40:23,130
Ni kwenda basi wewe kubadili tabia hizo.

513
00:40:23,130 --> 00:40:26,640
>> Maswali?

514
00:40:27,740 --> 00:40:29,290
Sawa.

515
00:40:29,290 --> 00:40:35,500
>> Kuhamia kwenye buggy3, hebu gdb buggy3.

516
00:40:35,500 --> 00:40:39,840
Sisi tu kukimbia na sisi kuona sisi kupata segfault.

517
00:40:39,840 --> 00:40:46,550
Kama sisi backtrace, kuna mbili tu kazi.

518
00:40:46,550 --> 00:40:52,970
Kama sisi kwenda juu katika kazi yetu kuu, sisi kuona kwamba sisi segfaulted katika mstari huu.

519
00:40:52,970 --> 00:41:00,180
Hivyo tu kuangalia mstari huu, kwa ajili ya (int line = 0; fgets mambo haya haina sawa null;

520
00:41:00,180 --> 00:41:03,770
mstari + +).

521
00:41:03,770 --> 00:41:08,010
Frame wetu uliopita aliitwa _IO_fgets.

522
00:41:08,010 --> 00:41:10,720
Utaona kwamba mengi na kujengwa katika utendaji C,

523
00:41:10,720 --> 00:41:15,350
kwamba wakati wewe kupata segfault, kutakuwa kweli cryptic kazi majina

524
00:41:15,350 --> 00:41:18,090
kama hii _IO_fgets.

525
00:41:18,090 --> 00:41:21,770
Lakini ambayo inaenda yanahusiana na wito huu fgets.

526
00:41:21,770 --> 00:41:25,850
Mahali fulani ndani ya hapa, sisi ni segfaulting.

527
00:41:25,850 --> 00:41:30,340
Kama sisi kuangalia hoja kwa fgets, tunaweza magazeti buffer.

528
00:41:30,340 --> 00:41:41,180
Hebu magazeti kama - Oh, hakuna.

529
00:41:48,980 --> 00:41:51,900
Print si kwenda kufanya kazi hasa kama mimi unataka kwa.

530
00:41:55,460 --> 00:41:58,000
Hebu tuangalie mpango halisi.

531
00:42:02,200 --> 00:42:09,640
Buffer ni safu tabia. Ni safu tabia ya wahusika 128.

532
00:42:09,640 --> 00:42:14,980
Hivyo wakati mimi kusema magazeti buffer, ni kwenda magazeti wahusika wale 128,

533
00:42:14,980 --> 00:42:18,300
ambayo mimi nadhani ni mambo yanayotarajiwa.

534
00:42:18,300 --> 00:42:21,390
Nini nilikuwa kutafuta ni magazeti anuani ya buffer,

535
00:42:21,390 --> 00:42:23,680
lakini kwamba si kweli niambie sana.

536
00:42:23,680 --> 00:42:30,770
Hivyo wakati mimi kutokea kwa kusema hapa juu x buffer, inaonyesha mimi 0xbffff090,

537
00:42:30,770 --> 00:42:38,690
ambayo, kama wewe kumbuka kutoka mapema au fulani, Oxbffff huelekea kuwa kanda stack-ish.

538
00:42:38,690 --> 00:42:46,020
stack huelekea kuanza mahali fulani tu chini ya 0xc000.

539
00:42:46,020 --> 00:42:51,890
Tu kwa kuona anuani hii, najua kwamba buffer kinachojitokeza stack.

540
00:42:51,890 --> 00:43:04,500
Kuanzisha tena mpango wangu, kukimbia, juu, buffer tuliona ilikuwa hii mlolongo wa wahusika

541
00:43:04,500 --> 00:43:06,530
kwamba ni kiasi pretty maana.

542
00:43:06,530 --> 00:43:12,270
Kisha uchapishaji faili, nini faili kuangalia kama?

543
00:43:15,120 --> 00:43:17,310
[Mwanafunzi] Batili. >> Yeah.

544
00:43:17,310 --> 00:43:22,610
File ni ya aina * JALADA, hivyo ni pointer,

545
00:43:22,610 --> 00:43:26,610
na thamani ya pointer kwamba ni null.

546
00:43:26,610 --> 00:43:33,240
Hivyo fgets ni kwenda kujaribu kusoma kutoka pointer kwamba katika njia ya moja kwa moja,

547
00:43:33,240 --> 00:43:37,320
lakini ili kupata kwamba pointer, ina dereference yake.

548
00:43:37,320 --> 00:43:40,550
Au, ili kupata kile ni lazima akizungumzia, ni dereferences yake.

549
00:43:40,550 --> 00:43:43,810
Hivyo ni dereferencing pointer null na segfaults yake.

550
00:43:46,600 --> 00:43:48,730
Ningeweza imeweza kuanzisha pale.

551
00:43:48,730 --> 00:43:52,170
Kama sisi kuvunja katika hatua zetu kuu na kukimbia,

552
00:43:52,170 --> 00:43:57,320
mstari wa kwanza wa code ni Char * filename = "nonexistent.txt";

553
00:43:57,320 --> 00:44:00,870
Kwamba lazima kutoa pretty kubwa ladha kama kwa sababu programu hii inashindwa.

554
00:44:00,870 --> 00:44:06,080
Typing ijayo inanileta kwa mstari unaofuata, ambapo mimi kufungua faili hii,

555
00:44:06,080 --> 00:44:11,140
na kisha mimi mara moja kupata katika line yetu, ambapo mara moja mimi hit ijayo, ni kwenda segfault.

556
00:44:11,140 --> 00:44:16,880
Je, mtu yeyote unataka kutupa nje sababu sisi tupate kuwa segfaulting?

557
00:44:16,880 --> 00:44:19,130
[Mwanafunzi] Picha haipo. >> Yeah.

558
00:44:19,130 --> 00:44:22,250
Hii ni walidhani kuwa dokezo

559
00:44:22,250 --> 00:44:29,570
kwamba wakati wowote wewe ni kufungua faili unahitaji kuangalia kwamba faili kweli lipo.

560
00:44:29,570 --> 00:44:31,510
Hivyo hapa, "nonexistent.txt";

561
00:44:31,510 --> 00:44:34,700
Wakati sisi fopen filename kwa ajili ya kusoma, sisi basi haja ya kusema

562
00:44:34,700 --> 00:44:45,870
kama (faili == null) na kusema printf ("Picha haipo!"

563
00:44:45,870 --> 00:44:56,340
au - bado bora - filename); kurudi 1;

564
00:44:56,340 --> 00:45:00,300
Hivyo sasa sisi kuangalia kuona kama ni null

565
00:45:00,300 --> 00:45:03,930
kabla ya kuendelea na kweli kujaribu kusoma kutoka faili hilo.

566
00:45:03,930 --> 00:45:08,800
Tunaweza remake tu kuona kwamba matendo hayo.

567
00:45:11,020 --> 00:45:14,970
Mimi lengo ni pamoja na mstari mpya.

568
00:45:21,090 --> 00:45:25,290
Hivyo sasa nonexistent.txt haipo.

569
00:45:26,890 --> 00:45:30,040
Unapaswa daima kuangalia kwa aina hii ya kitu.

570
00:45:30,040 --> 00:45:33,870
Unapaswa daima kuangalia kuona kama fopen anarudi null.

571
00:45:33,870 --> 00:45:38,170
Unapaswa daima kuangalia kuhakikisha kwamba malloc haina kurudi null,

572
00:45:38,170 --> 00:45:41,410
au mwingine segfault.

573
00:45:42,200 --> 00:45:45,930
>> Sasa buggy4.c.

574
00:45:49,190 --> 00:45:58,440
Mbio. Mimi guessing hii ni kusubiri kwa ajili ya pembejeo au looping uwezekano usio.

575
00:45:58,440 --> 00:46:01,870
Ndiyo, ni usio looping.

576
00:46:01,870 --> 00:46:05,560
Hivyo buggy4. Inaonekana kama sisi ni usio looping.

577
00:46:05,560 --> 00:46:12,590
Tunaweza kuvunja saa kuu, kuendesha programu yetu.

578
00:46:12,590 --> 00:46:20,180
Katika gdb, kwa muda mrefu kama wewe kutumia abbreviation ni suala la utata

579
00:46:20,180 --> 00:46:23,420
au maalum vifupisho kwamba wao kutoa kwa ajili yenu,

580
00:46:23,420 --> 00:46:29,020
basi unaweza kutumia n kutumia ijayo badala ya kuwa na aina nje karibu njia yote.

581
00:46:29,020 --> 00:46:33,730
Na kwa kuwa sasa nimepata hit n mara moja, siwezi tu hit Enter kuendelea ijayo

582
00:46:33,730 --> 00:46:36,640
badala ya kuwa na hit n Ingiza, n Ingiza, n Ingiza.

583
00:46:36,640 --> 00:46:44,630
Inaonekana kama mimi niko katika baadhi ya aina ya kwa kitanzi ambacho kuweka safu [i] kwa 0.

584
00:46:44,630 --> 00:46:50,510
Inaonekana kama mimi kamwe kuvunja nje ya hili kwa kitanzi.

585
00:46:50,510 --> 00:46:54,780
Kama mimi magazeti i, hivyo i ni 2, kisha nitakwenda ijayo.

586
00:46:54,780 --> 00:46:59,250
Mimi itabidi magazeti i, i ni 3, kisha nitakwenda ijayo.

587
00:46:59,250 --> 00:47:05,360
Mimi itabidi magazeti i na i ni 3. Next, magazeti i, i ni 4.

588
00:47:05,360 --> 00:47:14,520
Kweli, magazeti sizeof (safu), hivyo ukubwa wa safu ni 20.

589
00:47:16,310 --> 00:47:32,870
Lakini inaonekana kama kuna baadhi maalum gdb amri kwa ajili ya kwenda mpaka kitu kinachotokea.

590
00:47:32,870 --> 00:47:37,620
Ni kama kuweka hali juu ya thamani ya kutofautiana. Lakini sikumbuki ni nini.

591
00:47:37,620 --> 00:47:44,100
Hivyo kama sisi kuendelea -

592
00:47:44,100 --> 00:47:47,120
Nini walikuwa wewe kusema? Nini kuleta?

593
00:47:47,120 --> 00:47:50,500
[Mwanafunzi] Je kuonyesha i kuongeza - >> Yeah. Hivyo kuonyesha i inaweza kusaidia.

594
00:47:50,500 --> 00:47:54,530
Kama sisi tu kuonyesha i, itakuwa kuweka hapa nini thamani ya i ni

595
00:47:54,530 --> 00:47:56,470
hivyo mimi hawana magazeti ni nje kila wakati.

596
00:47:56,470 --> 00:48:02,930
Kama sisi tu kuendelea ijayo, tunaona 0, 1, 2, 3, 4, 5, 0, 1, 2, 3, 4, 5, 0, 1, 2, 3, 4, 5.

597
00:48:02,930 --> 00:48:08,530
Kitu ni kwenda vibaya sana, na i ni kuwa upya kwa 0.

598
00:48:13,330 --> 00:48:22,220
Kuangalia buggy4.c, tunaona yote ambayo hufanyika ni int safu [5];

599
00:48:22,220 --> 00:48:26,200
kwa ajili ya (i = 0; i <= sizeof (safu); i + +)

600
00:48:26,200 --> 00:48:28,550
safu [i] = 0;

601
00:48:28,550 --> 00:48:31,390
Nini sisi kuona kwamba ni makosa hapa?

602
00:48:31,390 --> 00:48:39,480
Kama dokezo, wakati mimi nilikuwa kufanya gdb buggy4 - hebu kuvunja kuu, kukimbia -

603
00:48:39,480 --> 00:48:45,980
Sikuweza magazeti sizeof (safu) tu kuona nini hali ni ambapo mimi lazima hatimaye kuvunja nje.

604
00:48:47,690 --> 00:48:51,100
Ambapo Mimi ni nani? Je, mimi kukimbia?

605
00:48:51,100 --> 00:48:54,280
Sikuweza kutangaza bado.

606
00:48:54,280 --> 00:48:58,680
Hivyo magazeti sizeof (safu) na kwamba 20,

607
00:48:58,680 --> 00:49:06,690
ambayo inatarajiwa tangu safu yangu ni ya kawaida 5 na ni ya integers 5,

608
00:49:06,690 --> 00:49:12,410
hivyo suala zima lazima 5 * sizeof (int) ka, ambapo sizeof (int) huelekea kuwa 4.

609
00:49:12,410 --> 00:49:14,780
Hivyo sizeof (safu) ni 20.

610
00:49:14,780 --> 00:49:17,420
Nini hii?

611
00:49:17,420 --> 00:49:21,720
[Mwanafunzi] Kugawanywa na sizeof (int). >> Yeah, / sizeof (int).

612
00:49:21,720 --> 00:49:30,630
Inaonekana kama bado kuna tatizo hapa. Nadhani hii lazima tu kuwa <

613
00:49:30,630 --> 00:49:36,960
tangu ni kiasi pretty daima <na kamwe <=.

614
00:49:36,960 --> 00:49:44,860
Sasa hebu fikiria nini hii ilikuwa kweli kuvunjwa.

615
00:49:44,860 --> 00:49:53,370
Je, mtu yeyote kuwa guesses kwanini i upya kwa njia ya kila 0 iteration ya kitanzi?

616
00:50:01,300 --> 00:50:09,350
Kitu pekee ndani ya hapa kwamba kinachotokea ni kwamba safu [i] ni kuwa kuweka 0.

617
00:50:09,350 --> 00:50:15,350
Hivyo kwa namna fulani, hii mstari wa kanuni ni kusababisha int yetu i kuwekwa kwa 0.

618
00:50:16,730 --> 00:50:23,130
[Mwanafunzi] Inawezekana kuwa kwa sababu ni kuu ya kumbukumbu ya sehemu hii ya i

619
00:50:23,130 --> 00:50:27,970
wakati anadhani ni kipengele kingine cha safu? >> [Bowden] Ndiyo.

620
00:50:27,970 --> 00:50:33,880
Wakati sisi ni kwenda nje ya mwisho wa safu yetu,

621
00:50:33,880 --> 00:50:39,870
namna fulani kwamba nafasi ya kuwa sisi ni kuu ni kuu thamani ya i.

622
00:50:39,870 --> 00:50:48,030
Na hivyo kama sisi kuangalia ndani buggy4, kuvunja kuu, kukimbia,

623
00:50:48,030 --> 00:50:53,120
hebu magazeti anuani ya i.

624
00:50:53,120 --> 00:50:57,280
Inaonekana kama ni bffff124.

625
00:50:57,280 --> 00:51:03,930
Sasa hebu magazeti anuani ya safu [0]. 110.

626
00:51:03,930 --> 00:51:06,290
Je kuhusu [1]? 114.

627
00:51:06,290 --> 00:51:07,920
[2], 118.

628
00:51:07,920 --> 00:51:14,530
11c, 120. safu [5] ni bfff124.

629
00:51:14,530 --> 00:51:26,990
Hivyo safu [5] ina anuani hiyo kama i, ambayo ina maana kwamba safu [5] ni i.

630
00:51:26,990 --> 00:51:30,720
Kama wana anuani hiyo, wao ni kitu kimoja.

631
00:51:30,720 --> 00:51:38,410
Hivyo wakati sisi kuweka safu [5] kwa 0, sisi ni kuweka i kwa 0.

632
00:51:38,410 --> 00:51:46,070
Na kama wewe kufikiri juu ya hili katika suala la stack,

633
00:51:46,070 --> 00:51:55,590
int i ni alitangaza kwanza, ambayo ina maana i anapata baadhi nafasi kwenye stack.

634
00:51:55,590 --> 00:52:04,730
Kisha safu [5] ni zilizotengwa, hivyo basi ka 20 zinatolewa kwa stack.

635
00:52:04,730 --> 00:52:08,400
Hivyo i anapata zilizotengwa kwanza, basi ka hizi 20 kupata zilizotengwa.

636
00:52:08,400 --> 00:52:11,400
Hivyo i kinachotokea haki kabla ya safu,

637
00:52:11,400 --> 00:52:19,230
na kwa sababu ya njia, kama nilivyosema wiki iliyopita, ambapo kitaalam stack hukua chini,

638
00:52:19,230 --> 00:52:28,520
wakati wewe index katika safu, sisi ni uhakika kwamba nafasi 0 katika safu

639
00:52:28,520 --> 00:52:31,970
daima hufanyika kabla ya nafasi ya kwanza katika safu.

640
00:52:31,970 --> 00:52:35,900
Hii ni aina ya jinsi mimi akauchomoa wiki iliyopita.

641
00:52:35,900 --> 00:52:42,210
Ona kwamba chini tuna anuani 0 na saa ya juu tuna anuani Max.

642
00:52:42,210 --> 00:52:44,880
stack daima ni kukua chini.

643
00:52:48,100 --> 00:52:53,500
Hebu sema sisi kutenga i.

644
00:52:53,500 --> 00:52:59,680
Sisi kutenga integer i, ambayo ina maana wacha tu kusema hapa juu integer i anapata zilizotengwa.

645
00:52:59,680 --> 00:53:06,420
Kisha sisi kutenga safu yetu ya integers 5, ambayo ina maana kwamba chini ya kwamba,

646
00:53:06,420 --> 00:53:11,230
tangu stack ni kuongezeka chini, integers wale 5 kupata zilizotengwa.

647
00:53:11,230 --> 00:53:15,900
Lakini kwa sababu ya jinsi arrays kazi, sisi ni uhakika kwamba nafasi ya kwanza katika safu

648
00:53:15,900 --> 00:53:22,260
daima ina anuani chini ya Jambo la pili katika safu.

649
00:53:22,260 --> 00:53:28,270
Hivyo safu nafasi 0 daima ina kutokea kwanza katika kumbukumbu,

650
00:53:28,270 --> 00:53:30,700
ambapo safu msimamo 1 kitokee baada ya kuwa

651
00:53:30,700 --> 00:53:33,310
na safu msimamo 2 kitokee baada ya kuwa,

652
00:53:33,310 --> 00:53:37,900
ambayo ina maana kwamba nafasi safu 0 kingetokea mahali fulani hapa chini,

653
00:53:37,900 --> 00:53:40,690
safu msimamo 1 kingetokea hapo juu kwamba

654
00:53:40,690 --> 00:53:45,530
sababu ukienda juu ina maana anwani ya juu tangu anuani upeo ni hapa juu.

655
00:53:45,530 --> 00:53:50,490
Hivyo safu [0] chini hapa, safu [1] hadi hapa, safu [2] hadi hapa, safu [3] hadi hapa.

656
00:53:50,490 --> 00:53:55,620
Angalia jinsi kabla sisi zilizotengwa integer i wote njia ya juu hapa,

657
00:53:55,620 --> 00:54:01,040
kama sisi hoja zaidi na zaidi katika safu yetu, sisi ni kupata karibu na karibu na integer yetu i.

658
00:54:01,040 --> 00:54:07,640
Ni hivyo tu hutokea kwamba safu [5], ambayo ni moja nafasi zaidi ya safu yetu,

659
00:54:07,640 --> 00:54:13,010
ni hasa ambapo integer i kilichotokea kuwa zilizotengwa.

660
00:54:13,010 --> 00:54:16,920
Basi hiyo ni mahali ambapo sisi kutokea kwa kuwa kupiga nafasi kwenye stack

661
00:54:16,920 --> 00:54:21,680
zilizotengwa kwa ajili ya integer i, na sisi ni kuweka kwamba kwa 0.

662
00:54:21,680 --> 00:54:26,160
>> Hiyo ni jinsi ya kuwa kazi. Maswali? Yeah.

663
00:54:26,160 --> 00:54:30,710
[Mwanafunzi] Kamwe akili. Sawa.

664
00:54:30,710 --> 00:54:33,090
[Mwanafunzi] Jinsi gani unaweza kuepuka hayo aina ya makosa?

665
00:54:33,090 --> 00:54:41,190
Hizi aina ya makosa? Usitumie C kama programu lugha yako.

666
00:54:41,190 --> 00:54:45,840
Kutumia lugha ambayo ina mipaka safu kuangalia.

667
00:54:45,840 --> 00:54:55,900
Muda mrefu kama wewe ni makini, wewe tu haja ya kuepuka kwenda nyuma mipaka ya safu yako.

668
00:54:55,900 --> 00:54:58,300
[Mwanafunzi] Basi hapa wakati sisi akaenda nyuma mipaka ya safu yako -

669
00:54:58,300 --> 00:55:01,840
[Bowden] Hiyo ambapo mambo kuanza kwenda vibaya. >> [Mwanafunzi] Oh, okay.

670
00:55:01,840 --> 00:55:05,730
Muda mrefu kama wewe kukaa ndani ya kumbukumbu zilizotengwa kwa ajili ya safu yako, wewe ni faini.

671
00:55:05,730 --> 00:55:12,400
Lakini C gani hakuna kuangalia makosa. Kama mimi kufanya safu [1000], ni furaha tu kurekebisha chochote kinachotokea -

672
00:55:12,400 --> 00:55:16,500
Unaendelea na mwanzo wa safu, basi huenda nafasi 1000 baada na huiweka kwa 0.

673
00:55:16,500 --> 00:55:20,000
Haina kufanya yoyote ya kuangalia kwamba oh, hii si kweli kuwa mambo 1000 ndani yake.

674
00:55:20,000 --> 00:55:22,750
1000 ni njia zaidi ya kile mimi lazima kubadilisha,

675
00:55:22,750 --> 00:55:26,940
ambapo Java au kitu utapata safu nje ya mipaka index

676
00:55:26,940 --> 00:55:29,820
au index nje ya ubaguzi mipaka.

677
00:55:29,820 --> 00:55:33,950
Hiyo ni kwa nini mengi ya lugha za ngazi za juu kuwa na mambo haya

678
00:55:33,950 --> 00:55:37,340
ambapo kama kwenda zaidi ya mipaka ya safu, wewe kushindwa

679
00:55:37,340 --> 00:55:40,070
hivyo kwamba huwezi kubadili mambo kutoka chini ya wewe

680
00:55:40,070 --> 00:55:42,590
na kisha mambo kwenda mbaya zaidi kuliko kupata tu isipokuwa

681
00:55:42,590 --> 00:55:44,940
akisema kwamba alikwenda ng'ambo ya mwisho wa safu.

682
00:55:44,940 --> 00:55:50,970
[Mwanafunzi] Na hivyo tusingeli tu iliyopita <= tu <? >> [Bowden] Yeah.

683
00:55:50,970 --> 00:55:54,800
Ni lazima kuwa <sizeof (safu) / sizeof (int);

684
00:55:54,800 --> 00:55:59,560
tangu sizeof (safu) ni 20, lakini sisi tu wanataka 5. >> [Mwanafunzi] Haki.

685
00:55:59,560 --> 00:56:04,060
Zaidi maswali? Sawa.

686
00:56:04,060 --> 00:56:07,380
>> [Mwanafunzi] nina swali. >> Yeah.

687
00:56:07,380 --> 00:56:16,440
[Mwanafunzi] Nini halisi safu variable?

688
00:56:16,440 --> 00:56:20,000
[Bowden] Kama kile ni safu?

689
00:56:20,000 --> 00:56:24,930
Array yenyewe ni ishara.

690
00:56:24,930 --> 00:56:31,490
Ni haki ya anuani ya kuanza ya ka 20 kwamba sisi ni referencing.

691
00:56:31,490 --> 00:56:38,070
Unaweza kufikiria kama pointer, lakini ni pointer mara kwa mara.

692
00:56:38,070 --> 00:56:44,140
Haraka kama mambo kupata compiled, safu variable haipo tena.

693
00:56:44,140 --> 00:56:48,210
[Mwanafunzi] Hivyo jinsi gani kupata ukubwa wa safu?

694
00:56:48,210 --> 00:56:54,130
Ukubwa wa safu inahusu ukubwa wa block kwamba ishara kwamba inahusu.

695
00:56:54,130 --> 00:57:01,240
Wakati mimi kufanya kitu kama printf ("% p \ n", safu);

696
00:57:01,240 --> 00:57:05,140
hebu kukimbia.

697
00:57:12,960 --> 00:57:15,530
Nini mimi tu kufanya makosa?

698
00:57:15,530 --> 00:57:19,220
'Safu' Array alitangaza hapa.

699
00:57:20,820 --> 00:57:23,200
Oh, hapa juu.

700
00:57:23,200 --> 00:57:31,250
Clang ni wajanja, na hutokea kwa taarifa kwamba mimi alitangaza safu kama vipengele 5

701
00:57:31,250 --> 00:57:34,540
lakini nina Indexing katika nafasi 1000.

702
00:57:34,540 --> 00:57:38,450
Ni anaweza kufanya hivyo kwa sababu hizi ni baadhi tu ya constants.

703
00:57:38,450 --> 00:57:43,370
Ni inaweza tu kwenda mbali katika noticing kwamba mimi nina kwenda zaidi ya mipaka ya safu.

704
00:57:43,370 --> 00:57:46,880
Lakini taarifa kabla ya wakati tulikuwa i kuwa sio sahihi,

705
00:57:46,880 --> 00:57:51,040
haiwezi uwezekano kuamua jinsi wengi maadili i inaweza kuchukua,

706
00:57:51,040 --> 00:57:55,540
hivyo hawezi kuamua kwamba i alikuwa anaenda ng'ambo ya mwisho wa safu.

707
00:57:55,540 --> 00:57:59,430
Hiyo tu Clang kuwa wajanja.

708
00:57:59,430 --> 00:58:03,340
>> Lakini sasa kufanya buggy4. Hivyo kile kingine mimi kufanya makosa?

709
00:58:03,340 --> 00:58:05,970
Implicitly kutangaza maktaba kazi 'printf'.

710
00:58:05,970 --> 00:58:14,960
Mimi naenda wanataka # pamoja <stdio.h>.

711
00:58:14,960 --> 00:58:18,710
Sawa. Sasa mbio buggy4.

712
00:58:18,710 --> 00:58:24,840
Uchapishaji thamani ya safu kama mimi hapa, uchapishaji ni kama pointer

713
00:58:24,840 --> 00:58:30,060
prints kitu ambacho inaonekana kama hii - bfb8805c - ambayo ni baadhi anuani

714
00:58:30,060 --> 00:58:33,450
kwamba katika mkoa stack-ish.

715
00:58:33,450 --> 00:58:41,820
Array yenyewe ni kama pointer, lakini si pointer halisi,

716
00:58:41,820 --> 00:58:45,410
tangu pointer mara kwa mara tunaweza kubadilisha.

717
00:58:45,410 --> 00:58:54,700
Array ni baadhi tu ya mara kwa mara. Vitalu 20 ya kumbukumbu ya kuanza saa 0xbfb8805c anuani.

718
00:58:54,700 --> 00:59:09,020
Hivyo bfb8805c kupitia anuani hii +20--au mimi nadhani -20 -

719
00:59:09,020 --> 00:59:17,400
yote ni ya kumbukumbu zilizotengwa kwa ajili ya safu hii.

720
00:59:17,400 --> 00:59:20,350
Array, variable yenyewe si kuhifadhiwa mahali popote.

721
00:59:20,350 --> 00:59:27,660
Wakati wewe ni kuandaa, compiler - mkono wimbi saa yake -

722
00:59:27,660 --> 00:59:33,060
lakini compiler tu kutumia ambapo anajua kuwa safu.

723
00:59:33,060 --> 00:59:36,090
Ni anajua ambapo safu kwamba kuanza,

724
00:59:36,090 --> 00:59:40,910
na hivyo inaweza daima tu kufanya mambo katika suala la offsets tangu mwanzo kwamba.

725
00:59:40,910 --> 00:59:43,960
Haina haja ya variable yenyewe kuwakilisha safu.

726
00:59:43,960 --> 00:59:53,730
Lakini wakati mimi kufanya kitu kama int safu * p =; sasa p ni pointer ambayo inaelekeza kwamba safu,

727
00:59:53,730 --> 00:59:57,830
na sasa p kweli gani zipo juu ya stack.

728
00:59:57,830 --> 01:00:01,950
Mimi nina uhuru wa kubadilisha p. Naweza kufanya p = malloc.

729
01:00:01,950 --> 01:00:06,500
Hivyo awali alisema kwa safu; sasa inaelekeza katika nafasi baadhi ya chungu.

730
01:00:06,500 --> 01:00:09,620
Mimi siwezi kufanya safu = malloc.

731
01:00:09,620 --> 01:00:13,710
Kama Clang ni wajanja, itakuwa yell saa yangu haki mbali bat.

732
01:00:17,000 --> 01:00:21,430
Kweli, mimi nina uhakika pretty GCC bila kufanya hii pia.

733
01:00:21,430 --> 01:00:25,010
Hivyo safu aina ya 'int [5]' si assignable.

734
01:00:25,010 --> 01:00:28,040
Huwezi hawawajui kitu aina safu

735
01:00:28,040 --> 01:00:30,500
kwa sababu ni safu tu mara kwa mara.

736
01:00:30,500 --> 01:00:34,760
Ni ishara ambayo marejeo wale ka 20. Siwezi kuibadilisha.

737
01:00:34,760 --> 01:00:37,690
>> [Mwanafunzi] Na pale ambapo ni kawaida ya safu kuhifadhiwa?

738
01:00:37,690 --> 01:00:40,670
[Bowden] Ni si kuhifadhiwa mahali popote. Ni wakati ni kuandaa.

739
01:00:40,670 --> 01:00:46,310
Hivyo ambapo ni kawaida ya safu kuhifadhiwa?

740
01:00:46,310 --> 01:00:51,870
Unaweza tu kutumia sizeof (safu) ndani ya kazi kwamba safu ya kutangazwa yenyewe.

741
01:00:51,870 --> 01:01:03,150
Hivyo kama mimi kufanya baadhi ya kazi, foo, na mimi kufanya (int safu [])

742
01:01:03,150 --> 01:01:10,450
printf ("% d \ n", sizeof (safu));

743
01:01:10,450 --> 01:01:21,330
na kisha chini hapa mimi wito foo (safu);

744
01:01:21,330 --> 01:01:24,840
ndani ya kazi hii - wacha kukimbia.

745
01:01:34,200 --> 01:01:36,840
Hii ni Clang kuwa wajanja tena.

746
01:01:36,840 --> 01:01:43,890
Ni akiniambia kwamba sizeof kwenye safu parameter kazi

747
01:01:43,890 --> 01:01:46,690
atarudi ukubwa wa '* int'.

748
01:01:46,690 --> 01:01:55,150
Hii itakuwa kosa kama si nini nilitaka kutokea.

749
01:01:55,150 --> 01:01:58,960
Hebu kweli kugeuka mbali Werror.

750
01:02:14,950 --> 01:02:17,590
Onyo. Maonyo ni faini.

751
01:02:17,590 --> 01:02:19,960
Itakuwa bado kukusanya muda mrefu kama ina onyo.

752
01:02:19,960 --> 01:02:22,910
. / A.out ni kwenda magazeti 4.

753
01:02:22,910 --> 01:02:28,650
onyo kwamba alikuwa yanayotokana ni kiashiria wazi ya nini potoka.

754
01:02:28,650 --> 01:02:34,120
Hii safu int ni kwenda tu na magazeti sizeof (int *).

755
01:02:34,120 --> 01:02:39,790
Hata kama mimi kuweka safu [5] katika hapa, ni bado tu kwenda magazeti sizeof (int *).

756
01:02:39,790 --> 01:02:47,440
Hivyo kwa haraka kama wewe kupita ndani ya kazi, tofauti kati ya arrays na kuyatumia

757
01:02:47,440 --> 01:02:49,670
haipo.

758
01:02:49,670 --> 01:02:52,640
Hii hutokea kwa kuwa safu kwamba ilitangazwa siku ya stack,

759
01:02:52,640 --> 01:02:58,300
lakini kwa haraka kama sisi kupita kwamba thamani, kwamba 0xbf blah, blah, blah katika kazi hii,

760
01:02:58,300 --> 01:03:03,350
kisha pointer hii inaelekeza kwamba safu ya stack.

761
01:03:03,350 --> 01:03:08,310
Hivyo hiyo ina maana kwamba sizeof tu inatumika katika kazi kwamba safu ilitangazwa,

762
01:03:08,310 --> 01:03:11,230
ambayo ina maana kwamba wakati wewe ni kuandaa kazi hii,

763
01:03:11,230 --> 01:03:17,330
wakati Clang huenda kwa njia ya kazi hii, anaona safu safu int ya kawaida 5.

764
01:03:17,330 --> 01:03:20,640
Hivyo basi anaona sizeof (safu). Naam, hiyo ni 20.

765
01:03:20,640 --> 01:03:26,440
Hiyo ni kweli jinsi sizeof kimsingi kazi kwa karibu kesi zote.

766
01:03:26,440 --> 01:03:31,150
Sizeof si kazi; ni operator.

767
01:03:31,150 --> 01:03:33,570
Huwezi kuwaita kazi sizeof.

768
01:03:33,570 --> 01:03:38,280
Sizeof (int), compiler tu kutafsiri kwamba kwa 4.

769
01:03:41,480 --> 01:03:43,700
Got it? Sawa.

770
01:03:43,700 --> 01:03:47,520
>> [Mwanafunzi] Basi ni nini tofauti kati ya sizeof (safu) katika kuu na katika foo?

771
01:03:47,520 --> 01:03:52,840
Hii ni kwa sababu sisi ni kusema sizeof (safu), ambayo ni ya aina * int,

772
01:03:52,840 --> 01:03:57,120
ambapo safu chini hapa ni si ya aina * int, ni safu int.

773
01:03:57,120 --> 01:04:04,540
>> [Mwanafunzi] Hivyo kama wewe alikuwa parameter katika safu [] badala ya safu int *,

774
01:04:04,540 --> 01:04:09,230
ingekuwa kwamba maana kwamba unaweza bado kubadili safu kwa sababu sasa ni pointer?

775
01:04:09,230 --> 01:04:14,250
[Bowden] Kama hili? >> [Mwanafunzi] Yeah. Je, unaweza kubadili safu ndani ya kazi sasa?

776
01:04:14,250 --> 01:04:18,420
[Bowden] Unaweza kubadilisha safu katika kesi zote.

777
01:04:18,420 --> 01:04:23,130
Katika wote wa kesi hizi ni huru kusema safu [4] = 0.

778
01:04:23,130 --> 01:04:26,590
[Mwanafunzi] Lakini unaweza kufanya safu uhakika na kitu kingine?

779
01:04:26,590 --> 01:04:30,230
[Bowden] Oh. Yeah. Katika kesi aidha - >> [mwanafunzi] Yeah.

780
01:04:30,230 --> 01:04:38,410
[Bowden] tofauti kati ya safu [] na int safu *, hapana.

781
01:04:38,410 --> 01:04:42,570
Unaweza pia kupata baadhi ya safu multidimensional katika hapa

782
01:04:42,570 --> 01:04:47,050
kwa syntax baadhi urahisi, lakini bado tu pointer.

783
01:04:47,050 --> 01:04:56,400
Hii ina maana kwamba mimi ni huru kufanya safu = malloc (sizeof (int)); na sasa kumweka mahali pengine.

784
01:04:56,400 --> 01:04:59,610
Lakini tu kama jinsi hii matendo milele na daima,

785
01:04:59,610 --> 01:05:03,210
kubadilisha hii safu kwa kufanya hivyo uhakika na kitu kingine

786
01:05:03,210 --> 01:05:07,570
haibadiliki hii safu chini hapa kwa sababu ni nakala ya hoja,

787
01:05:07,570 --> 01:05:10,780
si pointer hoja kwamba.

788
01:05:10,780 --> 01:05:16,070
Na kweli, kama dalili zaidi kwamba ni sawa -

789
01:05:16,070 --> 01:05:21,100
sisi tayari kuona nini prints safu uchapishaji -

790
01:05:21,100 --> 01:05:31,410
nini kama sisi magazeti anuani ya safu au anuani ya anuani ya safu

791
01:05:31,410 --> 01:05:36,290
aidha ya wale?

792
01:05:41,770 --> 01:05:45,220
Hebu kupuuza hii moja.

793
01:05:48,140 --> 01:05:51,660
Sawa. Hii ni faini. Ni sasa mbio. / A.out.

794
01:05:51,660 --> 01:06:00,220
Uchapishaji safu, basi uchapishaji anuani ya safu, ni kitu kimoja.

795
01:06:00,220 --> 01:06:02,870
Array tu haipo.

796
01:06:02,870 --> 01:06:08,190
Ni anajua wakati wewe ni uchapishaji safu, wewe ni uchapishaji ishara kwamba inahusu ka wale 20.

797
01:06:08,190 --> 01:06:11,940
Uchapishaji anuani ya safu, vizuri, safu haipo.

798
01:06:11,940 --> 01:06:17,200
Haina anuani, hivyo tu Prints anuani ya ka wale 20.

799
01:06:20,820 --> 01:06:28,150
Haraka kama wewe kukusanya chini, kama katika buggy4 yako compiled /. A.out,

800
01:06:28,150 --> 01:06:30,340
safu haipo.

801
01:06:30,340 --> 01:06:33,640
Kuyatumia zipo. Arrays hawana.

802
01:06:34,300 --> 01:06:38,060
vitalu ya kumbukumbu anayewakilisha safu bado zipo,

803
01:06:38,060 --> 01:06:43,270
lakini safu kutofautiana na vigezo ya aina ya kwamba hazipo.

804
01:06:46,260 --> 01:06:50,270
Wale ni kama tofauti kubwa kati ya arrays na kuyatumia

805
01:06:50,270 --> 01:06:55,590
ni haraka kama wewe kufanya wito kazi, hakuna tofauti.

806
01:06:55,590 --> 01:07:00,460
Lakini ndani ya kazi ambayo safu yenyewe ni alitangaza, sizeof kazi tofauti

807
01:07:00,460 --> 01:07:05,190
tangu wewe ni uchapishaji ukubwa wa vitalu badala ya ukubwa wa aina,

808
01:07:05,190 --> 01:07:08,950
na huwezi kubadilisha ni kwa sababu ni ishara.

809
01:07:08,950 --> 01:07:14,370
Uchapishaji kitu na anuani ya kitu Prints kitu kimoja.

810
01:07:14,370 --> 01:07:18,480
Na kwamba ni pretty kiasi.

811
01:07:18,480 --> 01:07:20,820
[Mwanafunzi] Unaweza kusema kwamba muda zaidi?

812
01:07:21,170 --> 01:07:24,170
Nipate kuwa amekosa kitu.

813
01:07:24,170 --> 01:07:29,260
Uchapishaji safu na anuani ya safu Prints kitu kimoja,

814
01:07:29,260 --> 01:07:33,180
lakini kama wewe magazeti pointer dhidi anuani ya pointer,

815
01:07:33,180 --> 01:07:36,010
jambo moja Prints anuani ya nini wewe akizungumzia,

816
01:07:36,010 --> 01:07:40,360
nyingine Prints anuani ya pointer juu ya stack.

817
01:07:40,360 --> 01:07:47,040
Unaweza kubadilisha pointer; huwezi kubadilisha ishara safu.

818
01:07:47,740 --> 01:07:53,270
Na sizeof pointer ni kwenda magazeti ukubwa wa aina hiyo pointer.

819
01:07:53,270 --> 01:07:57,470
Hivyo int * p sizeof (p) ataenda magazeti 4,

820
01:07:57,470 --> 01:08:04,110
lakini int safu [5] magazeti sizeof (safu) ataenda magazeti 20.

821
01:08:04,110 --> 01:08:07,480
[Mwanafunzi] Basi int safu [5] itakuwa magazeti 20? >> Ndiyo.

822
01:08:07,480 --> 01:08:13,300
Hiyo ndiyo sababu ndani ya buggy4 wakati kutumika kuwa sizeof (safu)

823
01:08:13,300 --> 01:08:16,660
hii alikuwa akifanya i <20, ambayo si kile tunachotaka.

824
01:08:16,660 --> 01:08:20,880
Tunataka i <5. >> [Mwanafunzi] Sawa.

825
01:08:20,880 --> 01:08:25,569
[Bowden] Na kisha kwa haraka kama wewe kuanza kupita katika utendaji,

826
01:08:25,569 --> 01:08:34,340
kama tulivyofanya int * p = array;

827
01:08:34,340 --> 01:08:39,779
ndani ya kazi hii, tunaweza kutumia kimsingi p na safu katika hasa njia ile ile,

828
01:08:39,779 --> 01:08:43,710
isipokuwa kwa ajili ya tatizo sizeof na tatizo kubadilisha.

829
01:08:43,710 --> 01:08:49,810
Lakini p [0] = 1; ni sawa na kusema safu [0] = 1;

830
01:08:49,810 --> 01:08:55,600
Na kwa haraka kama sisi kusema foo (safu); au foo (p);

831
01:08:55,600 --> 01:08:59,760
ndani ya kazi ya foo, huu ni mwito huo mara mbili.

832
01:08:59,760 --> 01:09:03,350
Hakuna tofauti kati ya simu hizi mbili.

833
01:09:07,029 --> 01:09:11,080
>> Kila mtu mzuri juu ya hilo? Sawa.

834
01:09:14,620 --> 01:09:17,950
Tuna dakika 10.

835
01:09:17,950 --> 01:09:28,319
>> Tutajaribu kupata kupitia mpango huu typer Hacker,

836
01:09:28,319 --> 01:09:32,350
tovuti hii, ambayo alikuja nje mwaka jana au kitu.

837
01:09:34,149 --> 01:09:41,100
Ni tu walidhani kuwa kama wewe aina nasibu na Prints nje -

838
01:09:41,100 --> 01:09:46,729
Chochote faili hutokea kwa kuwa kubeba ni nini inaonekana kama wewe ni kuandika.

839
01:09:46,729 --> 01:09:52,069
Inaonekana kama baadhi ya aina ya maadili ya mfumo wa uendeshaji.

840
01:09:53,760 --> 01:09:56,890
Hiyo ni nini tunataka kutekeleza.

841
01:10:08,560 --> 01:10:11,690
Unapaswa kuwa executable binary aitwaye hacker_typer

842
01:10:11,690 --> 01:10:14,350
kwamba inachukua katika hoja moja, faili na "aina hacker."

843
01:10:14,350 --> 01:10:16,480
Mbio executable lazima wazi screen

844
01:10:16,480 --> 01:10:20,850
na kisha magazeti nje moja tabia kutoka faili kupita katika kila wakati mtumiaji presses muhimu.

845
01:10:20,850 --> 01:10:24,990
Hivyo kila ufunguo waandishi, ni lazima kutupa mbali na badala yake magazeti tabia kutoka faili

846
01:10:24,990 --> 01:10:27,810
kwamba ni hoja.

847
01:10:29,880 --> 01:10:34,350
Mimi itabidi pretty much kukuambia nini mambo ya sisi ni kwenda haja ya kujua ni.

848
01:10:34,350 --> 01:10:36,440
Lakini tunataka kuangalia nje ya maktaba termios.

849
01:10:36,440 --> 01:10:44,840
Sijawahi kutumika hii maktaba katika maisha yangu yote, hivyo ina madhumuni ya chini sana.

850
01:10:44,840 --> 01:10:48,610
Lakini hii ni kwenda kuwa maktaba tunaweza kutumia kutupa mbali tabia hit

851
01:10:48,610 --> 01:10:52,390
wakati wewe ni kuandika katika kiwango in

852
01:10:56,970 --> 01:11:05,840
Hivyo hacker_typer.c, na sisi ni kwenda wanataka # pamoja <stdio.h>.

853
01:11:05,840 --> 01:11:12,870
Kuangalia ukurasa mtu kwa termios - I'm guessing terminal ni OS au kitu -

854
01:11:12,870 --> 01:11:16,240
Sijui jinsi ya kusoma.

855
01:11:16,240 --> 01:11:21,040
Kuangalia hii, inasema kwa pamoja na hizi files 2, hivyo tutaweza kufanya hivyo.

856
01:11:37,620 --> 01:11:46,820
>> Jambo la kwanza kwanza, tunataka kuchukua katika hoja moja, ambayo ni faili sisi inapaswa kufungua.

857
01:11:46,820 --> 01:11:52,420
Basi nini nataka kufanya? Je, mimi kuangalia kuona nina hoja moja?

858
01:11:52,420 --> 01:11:56,480
[Mwanafunzi] Kama argc sawa yake. >> [Bowden] Yeah.

859
01:11:56,480 --> 01:12:21,250
Hivyo kama (argc = 2)! Printf ("matumizi:% s [faili kufungua]").

860
01:12:21,250 --> 01:12:32,750
Hivyo sasa kama mimi hii kukimbia bila kutoa hoja ya pili - oh, nahitaji mstari mpya -

861
01:12:32,750 --> 01:12:36,240
utaona anasema matumizi: / hacker_typer,.

862
01:12:36,240 --> 01:12:39,770
na kisha hoja ya pili lazima faili mimi wanataka kufungua.

863
01:12:58,430 --> 01:13:01,260
Sasa nini mimi?

864
01:13:01,260 --> 01:13:08,490
Mimi nataka kusoma kutoka faili hii. Je, mimi kusoma kutoka faili?

865
01:13:08,490 --> 01:13:11,920
[Mwanafunzi] Wewe kufungua kwanza. >> Yeah.

866
01:13:11,920 --> 01:13:15,010
Hivyo fopen. Je fopen kuangalia kama?

867
01:13:15,010 --> 01:13:22,980
[Mwanafunzi] Filename. >> [Bowden] Filename ni kwenda kuwa argv [1].

868
01:13:22,980 --> 01:13:26,110
[Mwanafunzi] Na hapo nini unataka kufanya na hayo, hivyo - >> [Bowden] Yeah.

869
01:13:26,110 --> 01:13:28,740
Hivyo kama hukuzikumbuka, unaweza tu kufanya mtu fopen,

870
01:13:28,740 --> 01:13:32,960
ambapo ni kwenda kuwa const Char * njia ambapo njia ni filename,

871
01:13:32,960 --> 01:13:34,970
const Char * mode.

872
01:13:34,970 --> 01:13:38,660
Kama kutokea kwa si unakumbuka kile mode ni, basi unaweza kuangalia kwa mode.

873
01:13:38,660 --> 01:13:44,660
Ndani ya mtu kurasa, tabia slash ni nini unaweza kutumia ili kutafuta mambo.

874
01:13:44,660 --> 01:13:49,790
Hivyo ninapoandika / mode kutafuta mode.

875
01:13:49,790 --> 01:13:57,130
n na N ni nini unaweza kutumia mzunguko kupitia mechi ya utafutaji.

876
01:13:57,130 --> 01:13:59,800
Hapa inasema pointi mode hoja kwa kamba

877
01:13:59,800 --> 01:14:01,930
mwanzo na mmoja wa Mwandamano zifuatazo.

878
01:14:01,930 --> 01:14:06,480
Hivyo r, Open Nakala faili kwa ajili ya kusoma. Hiyo ni nini tunataka kufanya.

879
01:14:08,930 --> 01:14:13,210
Kwa kusoma, na mimi nataka kuhifadhi kwamba.

880
01:14:13,210 --> 01:14:18,720
kitu kinaenda kuwa * FILE. Sasa nini nataka kufanya?

881
01:14:18,720 --> 01:14:21,200
Nipe pili.

882
01:14:28,140 --> 01:14:30,430
Sawa. Sasa nini nataka kufanya?

883
01:14:30,430 --> 01:14:32,940
[Mwanafunzi] Angalia kama ni null. >> [Bowden] Yeah.

884
01:14:32,940 --> 01:14:38,690
Yoyote wakati wa kufungua faili, kuhakikisha kwamba wewe ni mafanikio na uwezo wa kufungua hiyo.

885
01:14:58,930 --> 01:15:10,460
>> Sasa nataka kufanya kwamba stuff termios ambapo nataka kwanza kusoma yangu mazingira ya sasa ya

886
01:15:10,460 --> 01:15:14,050
na wale kuokoa katika kitu, kisha Mimi nataka kubadili mipangilio yangu

887
01:15:14,050 --> 01:15:19,420
kutupilia mbali tabia yoyote kwamba mimi aina,

888
01:15:19,420 --> 01:15:22,520
na kisha Mimi nataka kusasisha mipangilio hizo.

889
01:15:22,520 --> 01:15:27,250
Na kisha mwisho wa mpango, nataka kubadili nyuma mipangilio yangu ya awali.

890
01:15:27,250 --> 01:15:32,080
Hivyo struct ni kwenda kuwa ya termios aina, na mimi nina kwenda kutaka mbili ya wale.

891
01:15:32,080 --> 01:15:35,600
moja ya kwanza ni kwenda kuwa current_settings yangu,

892
01:15:35,600 --> 01:15:42,010
na kisha wao wanaenda kuwa hacker_settings yangu.

893
01:15:42,010 --> 01:15:48,070
Kwanza, mimi naenda wanataka kuokoa yangu mazingira ya sasa,

894
01:15:48,070 --> 01:15:53,790
basi mimi nina kwenda kutaka update hacker_settings,

895
01:15:53,790 --> 01:16:01,570
na kisha njia mwisho wa mpango wangu, nataka kurudia mazingira ya sasa.

896
01:16:01,570 --> 01:16:08,660
Hivyo kuokoa mazingira ya sasa, njia ya kwamba kazi, termios mtu sisi.

897
01:16:08,660 --> 01:16:15,810
Sisi tunaona kwamba tuna hii tcsetattr int, int tcgetattr.

898
01:16:15,810 --> 01:16:22,960
Mimi kupita katika struct termios na pointer wake.

899
01:16:22,960 --> 01:16:30,640
njia hii itakuwa kuangalia ni - I've tayari wamesahau nini kazi iliitwa.

900
01:16:30,640 --> 01:16:34,930
Nakala na kuweka.

901
01:16:39,150 --> 01:16:45,500
Hivyo tcgetattr, kisha Mimi nataka kupita katika struct kwamba mimi nina kuokoa habari katika,

902
01:16:45,500 --> 01:16:49,650
ambayo ni kwenda kuwa current_settings,

903
01:16:49,650 --> 01:16:59,120
na hoja ya kwanza ni descriptor faili kwa ajili ya jambo nataka kuokoa sifa ya.

904
01:16:59,120 --> 01:17:04,360
Nini descriptor faili ni ni kama yoyote wakati wa kufungua faili, anapata descriptor faili.

905
01:17:04,360 --> 01:17:14,560
Wakati mimi fopen argv [1], anapata descriptor faili ambayo wewe ni referencing

906
01:17:14,560 --> 01:17:16,730
kila unataka kusoma wala kuandika na hiyo.

907
01:17:16,730 --> 01:17:19,220
Hiyo si descriptor faili nataka kutumia hapa.

908
01:17:19,220 --> 01:17:21,940
Kuna tatu faili descriptors una by default,

909
01:17:21,940 --> 01:17:24,310
ambayo ni ya kiwango katika, kiwango nje, na hitilafu ya kiwango.

910
01:17:24,310 --> 01:17:29,960
By default, nadhani ni ya kiwango katika ni 0, kiwango nje ni 1, na makosa ya hali ni 2.

911
01:17:29,960 --> 01:17:33,980
Basi je, nataka kubadili mipangilio ya?

912
01:17:33,980 --> 01:17:37,370
Nataka mabadiliko ya mazingira ya wakati mimi hit tabia,

913
01:17:37,370 --> 01:17:41,590
Mimi nataka kutupa kwamba tabia mbali badala ya uchapishaji kwa screen.

914
01:17:41,590 --> 01:17:45,960
Nini mkondo - kiwango katika, nje ya kiwango, au kosa kiwango -

915
01:17:45,960 --> 01:17:52,050
anaitikia kwa mambo wakati ninapoandika katika keyboard? >> [Mwanafunzi] darasa in >> Yeah.

916
01:17:52,050 --> 01:17:56,450
Hivyo siwezi ama kufanya 0 au siwezi kufanya stdin.

917
01:17:56,450 --> 01:17:59,380
Nina kupata current_settings ya kiwango in

918
01:17:59,380 --> 01:18:01,720
>> Sasa nataka kusasisha mipangilio hizo,

919
01:18:01,720 --> 01:18:07,200
hivyo kwanza mimi itabidi nakala katika hacker_settings nini current_settings yangu ni.

920
01:18:07,200 --> 01:18:10,430
Na jinsi structs kazi itakuwa ni nakala tu.

921
01:18:10,430 --> 01:18:14,510
Hii nakala zote za mashamba, kama ungependa kutarajia.

922
01:18:14,510 --> 01:18:17,410
>> Sasa nataka update baadhi ya mashamba.

923
01:18:17,410 --> 01:18:21,670
Kuangalia termios, ungependa kuwa na kusoma kwa njia ya kura ya hii

924
01:18:21,670 --> 01:18:24,110
tu kuona nini ungependa unataka kuangalia kwa,

925
01:18:24,110 --> 01:18:28,210
lakini bendera utaenda unataka kuangalia kwa ni mwangwi,

926
01:18:28,210 --> 01:18:33,110
hivyo ECHO Echo pembejeo wahusika.

927
01:18:33,110 --> 01:18:37,710
Kwanza nataka kuweka - I've tayari wamesahau nini mashamba ni.

928
01:18:45,040 --> 01:18:47,900
Hii ni nini struct inaonekana kama.

929
01:18:47,900 --> 01:18:51,060
Hivyo pembejeo modes nadhani tunataka kubadili.

930
01:18:51,060 --> 01:18:54,210
Tutaangalia ufumbuzi kuhakikisha kwamba ni nini tunataka kubadili.

931
01:19:04,060 --> 01:19:12,610
Tunataka kubadili lflag ili kuzuia wanaohitaji kuangalia njia zote hizi.

932
01:19:12,610 --> 01:19:14,670
Tunataka kubadili modes mitaa.

933
01:19:14,670 --> 01:19:17,710
Ungependa kuwa na kusoma kwa jambo hili zima kuelewa ambapo kila kitu ni mali

934
01:19:17,710 --> 01:19:19,320
kwamba tunataka mabadiliko.

935
01:19:19,320 --> 01:19:24,120
Lakini ni ndani ya modes mitaa ambapo tunakwenda wanataka mabadiliko hayo.

936
01:19:27,080 --> 01:19:33,110
Hivyo hacker_settings.cc_lmode ni nini ni wito.

937
01:19:39,630 --> 01:19:43,020
c_lflag.

938
01:19:49,060 --> 01:19:52,280
Hii ni wapi tunapata ndani ya waendeshaji bitwise.

939
01:19:52,280 --> 01:19:54,860
Sisi ni aina ya nje ya muda, lakini tutaweza kwenda kupitia haraka ni halisi.

940
01:19:54,860 --> 01:19:56,600
Hii ni wapi tunapata ndani ya waendeshaji bitwise,

941
01:19:56,600 --> 01:19:59,950
ambapo mimi nadhani alisema mmoja wa muda mrefu uliopita kwamba wakati wowote kuanza kushughulika na bendera,

942
01:19:59,950 --> 01:20:03,370
utaenda kutumia bitwise operator mengi.

943
01:20:03,370 --> 01:20:08,240
Kila kidogo katika bendera sambamba na aina fulani ya tabia.

944
01:20:08,240 --> 01:20:14,090
Hivyo hapa, bendera hii ina rundo la vitu mbalimbali, ambapo wote maana kitu tofauti.

945
01:20:14,090 --> 01:20:18,690
Lakini nini nataka kufanya ni tu kugeuka mbali kidogo ambayo inalingana na ECHO.

946
01:20:18,690 --> 01:20:25,440
Hivyo kugeuka kuwa mbali mimi kufanya & = ¬ ECHO.

947
01:20:25,440 --> 01:20:30,110
Kweli, nadhani ni kama tECHO au kitu. Mimi tu kwenda kuangalia tena.

948
01:20:30,110 --> 01:20:34,050
Naweza termios yake. Ni tu ECHO.

949
01:20:34,050 --> 01:20:38,440
ECHO ni kwenda kuwa kidogo moja.

950
01:20:38,440 --> 01:20:44,230
¬ ECHO ni kwenda kwa maana ya bits wote ni kuweka 1, ambayo ina maana bendera wote ni kuweka kweli

951
01:20:44,230 --> 01:20:47,140
isipokuwa kwa kidogo ECHO.

952
01:20:47,140 --> 01:20:53,830
Kwa kumaliza wangu bendera ya ndani na hii, ina maana bendera zote kwamba sasa kuweka kweli

953
01:20:53,830 --> 01:20:56,520
Bado utakuwa kuweka kweli.

954
01:20:56,520 --> 01:21:03,240
Kama ECHO bendera yangu ni kuweka kweli, basi hii ni lazima kuweka uongo juu ya bendera ya ECHO.

955
01:21:03,240 --> 01:21:07,170
Hivyo hii mstari wa kanuni tu anarudi mbali bendera ECHO.

956
01:21:07,170 --> 01:21:16,270
mistari nyingine ya kificho, na mimi itabidi nakala kwa maslahi ya muda na kisha kueleza wao.

957
01:21:27,810 --> 01:21:30,180
Katika ufumbuzi, alisema 0.

958
01:21:30,180 --> 01:21:33,880
Ni pengine bora kwa uwazi kusema stdin.

959
01:21:33,880 --> 01:21:42,100
>> Ona kwamba mimi nina pia kufanya ECHO | ICANON hapa.

960
01:21:42,100 --> 01:21:46,650
ICANON inahusu kitu tofauti, ambayo ina maana mode kisheria.

961
01:21:46,650 --> 01:21:50,280
Nini kisheria mode njia ni kawaida wakati wewe ni kuandika nje ya mstari amri,

962
01:21:50,280 --> 01:21:54,670
kiwango katika mchakato haina chochote mpaka hit newline.

963
01:21:54,670 --> 01:21:58,230
Hivyo wakati wa kufanya GetString, unapoandika rundo la vitu, wewe hit newline.

964
01:21:58,230 --> 01:22:00,590
Hiyo ni wakati ni alimtuma kiwango in

965
01:22:00,590 --> 01:22:02,680
Hiyo ni default.

966
01:22:02,680 --> 01:22:05,830
Wakati mimi kugeuka mbali mode kisheria, sasa kila tabia moja vyombo vya habari

967
01:22:05,830 --> 01:22:10,910
ni nini anapata kusindika, ambayo ni kawaida aina ya mbaya kwa sababu ni mwepesi wa mchakato mambo haya,

968
01:22:10,910 --> 01:22:14,330
ambayo ni kwa nini ni vizuri buffer ndani mistari nzima.

969
01:22:14,330 --> 01:22:16,810
Lakini Mimi nataka kila tabia kuwa processed

970
01:22:16,810 --> 01:22:18,810
tangu mimi sitaki kusubiri kwa mimi hit newline

971
01:22:18,810 --> 01:22:21,280
kabla michakato ya wahusika wote nimekuwa kuandika.

972
01:22:21,280 --> 01:22:24,760
Hii anarudi mbali mode kisheria.

973
01:22:24,760 --> 01:22:31,320
Mambo haya tu ina maana wakati ni kweli michakato wahusika.

974
01:22:31,320 --> 01:22:35,830
Hii ina maana mchakato wao mara moja; haraka kama mimi kuandika yao, mchakato wao.

975
01:22:35,830 --> 01:22:42,510
Na hii ni kazi ambayo ni uppdatering mipangilio yangu kwa kiwango katika,

976
01:22:42,510 --> 01:22:45,480
na TCSA njia kufanya hivyo hivi sasa.

977
01:22:45,480 --> 01:22:50,310
chaguzi nyingine ni kusubiri hadi hapo kila kitu sasa ni juu ya mkondo ni processed.

978
01:22:50,310 --> 01:22:52,030
Hiyo haina kweli jambo.

979
01:22:52,030 --> 01:22:56,920
Haki tu sasa kubadili mipangilio yangu kuwa chochote kwa sasa katika hacker_typer_settings.

980
01:22:56,920 --> 01:23:02,210
Mimi nadhani kuitwa ni hacker_settings, hivyo hebu mabadiliko hayo.

981
01:23:09,610 --> 01:23:13,500
Kubadili kila kitu kwa hacker_settings.

982
01:23:13,500 --> 01:23:16,870
>> Sasa mwisho wa programu yetu sisi ni kwenda kutaka revert

983
01:23:16,870 --> 01:23:20,210
kwa nini kwa sasa ndani ya normal_settings,

984
01:23:20,210 --> 01:23:26,560
ambayo ni kwenda kuangalia tu kama & normal_settings.

985
01:23:26,560 --> 01:23:30,650
Taarifa mimi si iliyopita yoyote ya normal_settings yangu tangu awali ya kupata.

986
01:23:30,650 --> 01:23:34,520
Kisha kubadili tu nyuma yao, mimi kupita yao nyuma mwishoni.

987
01:23:34,520 --> 01:23:38,390
Hii ilikuwa update. Sawa.

988
01:23:38,390 --> 01:23:43,900
>> Sasa ndani ya hapa mimi itabidi kueleza code katika maslahi ya muda.

989
01:23:43,900 --> 01:23:46,350
Siyo kwamba code sana.

990
01:23:50,770 --> 01:24:03,750
Tunaona tunasoma tabia kutoka faili. Tuliihisabia f.

991
01:24:03,750 --> 01:24:07,850
Sasa unaweza mtu fgetc, lakini jinsi fgetc ni kwenda kufanya kazi

992
01:24:07,850 --> 01:24:11,910
ni tu ni kwenda na kurudi tabia ya kuwa wewe tu kusoma au EOF,

993
01:24:11,910 --> 01:24:15,680
ambazo zinahusiana na mwisho wa faili au baadhi likitokea kosa.

994
01:24:15,680 --> 01:24:19,900
Sisi ni looping, kuendelea kusoma tabia moja kutoka faili,

995
01:24:19,900 --> 01:24:22,420
mpaka tumekuwa kukimbia nje ya wahusika kusoma.

996
01:24:22,420 --> 01:24:26,650
Na wakati sisi ni kufanya hivyo, sisi kusubiri juu ya tabia moja kutoka kiwango in

997
01:24:26,650 --> 01:24:29,090
Kila wakati unapoandika kitu katika mstari amri,

998
01:24:29,090 --> 01:24:32,820
hiyo ni ya kusoma katika tabia kutoka kiwango in

999
01:24:32,820 --> 01:24:38,330
Kisha putchar ni kwenda tu kuweka Char tunasoma juu hapa kutoka faili nje ya kiwango.

1000
01:24:38,330 --> 01:24:42,890
Unaweza mtu putchar, lakini ni kuweka tu kwa kiwango nje, ni uchapishaji kwamba tabia.

1001
01:24:42,890 --> 01:24:51,600
Unaweza pia tu kufanya printf ("% c", c); Same wazo.

1002
01:24:53,330 --> 01:24:56,670
Hiyo kwenda kufanya wingi wa kazi yetu.

1003
01:24:56,670 --> 01:25:00,300
>> Jambo la mwisho tunakwenda wanataka kufanya ni tu fclose faili wetu.

1004
01:25:00,300 --> 01:25:03,310
Kama huna fclose, hiyo leak kumbukumbu.

1005
01:25:03,310 --> 01:25:06,680
Tunataka fclose faili sisi awali kufunguliwa, na nadhani hiyo ni yake.

1006
01:25:06,680 --> 01:25:13,810
Kama sisi kufanya huo, mimi tayari got matatizo.

1007
01:25:13,810 --> 01:25:17,260
Hebu angalia.

1008
01:25:17,260 --> 01:25:19,960
Ni nini kulalamika kuhusu?

1009
01:25:19,960 --> 01:25:30,220
Inatarajiwa 'int' lakini hoja ni ya aina ya 'struct _IO_FILE *'.

1010
01:25:36,850 --> 01:25:39,370
Tutaweza kuona kama kwamba kazi.

1011
01:25:45,210 --> 01:25:53,540
Kuruhusiwa tu katika C99. Augh. Okay, kufanya hacker_typer.

1012
01:25:53,540 --> 01:25:57,760
Sasa sisi kupata maelezo muhimu zaidi.

1013
01:25:57,760 --> 01:25:59,900
Hivyo matumizi ya kitambulisho Undeclared 'normal_settings'.

1014
01:25:59,900 --> 01:26:04,170
Sikuweza kuiita normal_settings. Mimi kuitwa ni current_settings.

1015
01:26:04,170 --> 01:26:12,090
Basi hebu mabadiliko yote.

1016
01:26:17,920 --> 01:26:21,710
Sasa kupita hoja.

1017
01:26:26,290 --> 01:26:29,500
Mimi itabidi kufanya hili 0 kwa sasa.

1018
01:26:29,500 --> 01:26:36,720
Sawa. . / Hacker_typer cp.c.

1019
01:26:36,720 --> 01:26:39,590
Mimi pia hakuwa wazi screen mwanzoni.

1020
01:26:39,590 --> 01:26:42,960
Lakini unaweza kuangalia nyuma kwa kuweka mwisho tatizo kuona jinsi wewe wazi screen.

1021
01:26:42,960 --> 01:26:45,160
Ni tu uchapishaji wahusika baadhi

1022
01:26:45,160 --> 01:26:47,210
wakati hii ni kufanya nini nataka kufanya.

1023
01:26:47,210 --> 01:26:48,900
Sawa.

1024
01:26:48,900 --> 01:26:55,280
Na kufikiri kuhusu nini hii zinahitajika kuwa 0 badala ya stdin,

1025
01:26:55,280 --> 01:27:00,560
ambayo inapaswa # define 0,

1026
01:27:00,560 --> 01:27:03,890
hii ni kulalamika kwamba -

1027
01:27:13,150 --> 01:27:19,360
Kabla ya wakati mimi alisema kwamba kuna faili descriptors lakini basi wewe pia kuwa JALADA yako *,

1028
01:27:19,360 --> 01:27:23,210
descriptor faili ni tu integer moja,

1029
01:27:23,210 --> 01:27:26,970
ambapo * JALADA ana rundo zima ya mambo yanayohusiana na hayo.

1030
01:27:26,970 --> 01:27:30,380
sababu tunahitaji kusema 0 badala ya stdin

1031
01:27:30,380 --> 01:27:37,480
ni kwamba ni stdin * JALADA ambayo inaelekeza kwamba ni jambo referencing faili descriptor 0.

1032
01:27:37,480 --> 01:27:45,070
Hivyo hata hapa wakati mimi kufanya fopen (argv [1], nina kupata * JALADA nyuma.

1033
01:27:45,070 --> 01:27:51,180
Lakini mahali fulani katika kwamba * JALADA ni kitu sambamba na descriptor faili kwa ajili ya faili hilo.

1034
01:27:51,180 --> 01:27:57,430
Kama ukiangalia ukurasa mtu kwa ajili ya wazi, hivyo nadhani itabidi kufanya mtu 3 wazi - nope -

1035
01:27:57,430 --> 01:27:59,380
mtu 2 wazi - yeah.

1036
01:27:59,380 --> 01:28:06,250
Kama ukiangalia ukurasa kwa wazi, wazi ni kama fopen chini ya ngazi,

1037
01:28:06,250 --> 01:28:09,350
na ni kurudi halisi faili descriptor.

1038
01:28:09,350 --> 01:28:12,050
fopen gani rundo la stuff juu ya wazi,

1039
01:28:12,050 --> 01:28:17,640
ambayo badala ya kurudi tu kwamba faili descriptor anarudi JALADA zima * pointer

1040
01:28:17,640 --> 01:28:20,590
ndani ambayo ni faili wetu descriptor kidogo.

1041
01:28:20,590 --> 01:28:25,020
Hivyo kiwango katika inahusu kitu * JALADA,

1042
01:28:25,020 --> 01:28:29,120
ambapo 0 inahusu tu kiwango faili descriptor katika yenyewe.

1043
01:28:29,120 --> 01:28:32,160
>> Maswali?

1044
01:28:32,160 --> 01:28:35,930
[Atacheka] akapiga kupitia njia hiyo.

1045
01:28:35,930 --> 01:28:39,140
Wote haki. Sisi ni kosa. [Atacheka]

1046
01:28:39,140 --> 01:28:42,000
>> [CS50.TV]