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[Powered by Google Translate] [Sehemu ya 4 - 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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Tuna kesho quiz, katika kesi guys hawakujua kwamba.

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Ni kimsingi juu ya kila kitu unaweza tumeona katika darasa au lazima tumeona katika darasa.

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Hiyo ni pamoja na kuyatumia, ingawa wao ni mada sana hivi karibuni.

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Wewe lazima angalau kuelewa viwango vya juu yao.

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Chochote kwamba ameondoka juu katika darasa unapaswa kuelewa kwa chemsha bongo.

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Hivyo kama una maswali juu yao, unaweza kuwauliza sasa.

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Lakini hii ni kwenda kuwa kikao sana mwanafunzi inayoongozwa ambapo wewe guys kuuliza maswali,

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hivyo hopefully watu wana maswali.

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Je, mtu yeyote kuwa na maswali?

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Ndiyo. >> [Mwanafunzi] Je, unaweza kwenda zaidi ya kuyatumia tena?

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Nitakwenda zaidi ya kuyatumia.

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Yote ya vigezo yako lazima kuishi katika kumbukumbu,

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lakini kwa kawaida wewe usijali kuhusu kwamba na wewe tu kusema x 2 + na + y 3

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na compiler itakuwa takwimu nje ambapo mambo wanaishi kwa ajili yenu.

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Mara wewe ni kushughulika na kuyatumia, sasa wewe ni wazi kwa kutumia anwani wale kumbukumbu.

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Hivyo variable moja itakuwa milele tu kuishi katika anuani moja wakati wowote.

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Kama tunataka kutangaza pointer, ni nini aina kwenda kuangalia kama?

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>> Nataka kutangaza p pointer. Nini aina kuangalia kama?

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[Mwanafunzi] int * p. >> Yeah. Hivyo int * p.

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Na jinsi gani mimi kufanya hivyo kumweka kwa x? >> [Mwanafunzi] Ampersand.

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[Bowden] Basi ampersand ni halisi iitwayo anuani ya operator.

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Hivyo wakati mimi kusema & x ni kupata anuani ya kumbukumbu ya x kutofautiana.

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Hivyo sasa nina p pointer, na mahali popote katika code yangu naweza kutumia * p

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au mimi naweza kutumia x na itakuwa exact kitu.

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(* P). Nini hii kufanya? Nini nyota kwamba mean?

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[Mwanafunzi] Ina maana thamani katika hatua hiyo. >> Yeah.

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Hivyo kama sisi kuangalia yake, inaweza kuwa muhimu sana ili kuweza kuteka michoro

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ambapo hii ni sanduku kidogo ya kumbukumbu kwa ajili ya x, ambayo hufanyika kuwa thamani 4,

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basi tuna sanduku kidogo ya kumbukumbu kwa ajili ya p,

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na hivyo p pointi x, hivyo sisi kuteka mshale kutoka p kwa x.

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Hivyo wakati sisi kusema * p sisi ni kusema kwenda sanduku kwamba ni p.

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Nyota ni kufuata mshale na kisha kufanya chochote unataka na sanduku ya kwamba haki pale.

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Hivyo naweza kusema * p = 7, na kwamba watakwenda sanduku kwamba ni x na mabadiliko ambayo kwa 7.

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Au mimi naweza kusema int z = * p * 2; Hiyo utata kwa sababu ni nyota, nyota.

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nyota moja ni dereferencing p, nyota nyingine ni kuzidisha kwa 2.

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Taarifa mimi inaweza kuwa tu kama vile kubadilishwa p * kwa x.

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Unaweza kutumia yao katika njia ya sawa.

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Na kisha baadaye naweza kuwa na uhakika na kitu p mpya kabisa.

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Naweza kusema tu p = &z;

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Hivyo sasa P pointi tena kwa x; inaelekeza katika z.

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Na wakati wowote mimi kufanya * p ni sawa kama kufanya z.

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Hivyo kitu muhimu kuhusu hii ni mara moja sisi kuanza kuingia katika utendaji.

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>> Ni aina ya useless kutangaza pointer kwamba pointi kwa kitu

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na kisha wewe tu dereferencing ni

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wakati unaweza kuwa na kutumika variable awali kwa kuanzia.

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Lakini wakati wewe kupata katika utendaji - hivyo hebu sema tuna baadhi ya kazi, int foo,

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kwamba inachukua pointer na tu anafanya * p = 6;

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Kama tuliona kabla na byta, huwezi kufanya byta ufanisi na kazi tofauti

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na tu kupita integers kwa sababu kila kitu katika C daima akipita thamani.

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Hata wakati wewe ni kupita kuyatumia wewe akipita thamani.

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Ni tu hivyo hutokea kwamba wale maadili ni kumbukumbu anwani.

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Hivyo wakati mimi kusema foo (p); mimi nina kupita pointer ndani ya foo kazi

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na kisha foo ni kufanya * p = 6;

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Hivyo ndani ya kazi kwamba, * p bado ni sawa na x,

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lakini siwezi kutumia x ndani ya kazi ambayo kwa sababu si scoped ndani ya kazi hiyo.

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Hivyo * p = 6 ni njia pekee ya mimi kupata variable mitaa kutoka kazi nyingine.

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Au, vizuri, kuyatumia ni njia pekee ya mimi kupata variable mitaa kutoka kazi nyingine.

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[Mwanafunzi] Hebu sema wewe alitaka kurudi pointer. Jinsi gani hasa gani unaweza kufanya hivyo?

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[Bowden] Kurudi pointer kama katika kitu kama int y = 3; kurudi & y? >> [Mwanafunzi] Yeah.

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[Bowden] Sawa. Wewe kamwe kufanya hivyo. Hii ni mbaya.

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Nadhani nikaona katika slides hizi hotuba wewe ulianza kuona mchoro nzima ya kumbukumbu

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ambapo hadi hapa nimepata kumbukumbu anuani 0

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na chini hapa una kumbukumbu anuani gigs 4 au 2-32.

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Hivyo basi nimepata baadhi ya mambo na baadhi ya mambo na kisha una stack yako

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na nimepata lundo yako, ambayo wewe tu kuanza kujifunza kuhusu, kupanda juu.

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[Mwanafunzi] Je, si chungu juu ya stack?

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>> Yeah. lundo ni juu, si hivyo? >> [Mwanafunzi] Naam, yeye kuweka 0 juu.

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[Mwanafunzi] Oh, akaweka 0 juu. >> [Mwanafunzi] Oh, okay.

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Disclaimer: popote na CS50 utaenda kuona njia hii. >> [Mwanafunzi] Sawa.

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Ni kwamba tu wakati wewe kwanza kuona mwingi,

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kama wakati unafikiri ya stack unafikiri ya stacking mambo juu ya mtu mwingine.

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Hivyo sisi huwa na hii flip kuzunguka hivyo stack ni kupanda juu kama kawaida ingekuwa stack

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badala ya stack kunyongwa chini. >> [Mwanafunzi] Wala chungu kitaalam kukua pia, ingawa?

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Ni inategemea nini maana ya kukua.

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stack na lundo daima kukua katika maelekezo kinyume.

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stack daima ni kukua kwa maana ya kwamba ni kupanda juu

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kuelekea juu kumbukumbu anwani, na chungu ni kuongezeka chini

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kwa kuwa ni kuongezeka kwa anwani ya chini kumbukumbu.

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Hivyo ni juu 0 na chini ni high kumbukumbu anwani.

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Wao ni wote kukua, tu katika kupinga maelekezo.

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[Mwanafunzi] mimi tu maana kwamba kwa sababu wewe alisema kuweka stack juu ya chini

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sababu inaonekana Intuitive zaidi kwa sababu kwa stack kuanza saa ya juu ya chungu,

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lundo ni juu ya yenyewe pia, hivyo that's - >> Yeah.

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Wewe pia kufikiria chungu kama kupanda juu na kubwa, lakini stack zaidi ya hivyo.

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Hivyo stack ni moja kwamba sisi aina ya wanataka kuonyesha kukua.

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Lakini kila mahali ukiangalia vinginevyo ni kwenda kuonyesha anwani 0 saa ya juu

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na juu ya kumbukumbu anuani hapo chini, hivyo hii ni maoni yako ya kawaida ya kumbukumbu.

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>> Je, una swali?

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[Mwanafunzi] Unaweza kutuambia zaidi juu ya lundo?

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Yeah. Mimi itabidi kupata kwamba katika pili.

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Kwanza, kurejea kwa nini kurudi & y ni kitu mbaya,

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juu ya stack wewe kuwa na rundo la muafaka stack ambayo kuwakilisha yote ya kazi

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ambayo imekuwa kuitwa.

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Hivyo kupuuza mambo ya awali, juu ya stack wako daima ni kwenda kuwa kazi kuu

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tangu kwamba kazi hiyo ni ya kwanza kuitwa.

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Na kisha wakati wewe piga kazi nyingine, stack anaenda kukua chini.

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Hivyo kama mimi kuwaita baadhi ya kazi, foo, na anapata stack yake mwenyewe frame,

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inaweza kuwaita baadhi ya kazi, bar; anapata stack yake mwenyewe frame.

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Na bar inaweza kuwa ya kujirudia na inaweza kuita yenyewe,

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na hivyo kuwa wito pili kwa bar ni kwenda kupata stack yake mwenyewe frame.

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Na hivyo yale yanayoendelea katika muafaka haya yote ni stack ya vigezo mitaa

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na wote wa hoja kwamba kazi -

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Yoyote yaliyo ya ndani ya nchi scoped kwa kazi hii kwenda katika muafaka haya stack.

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Hivyo kwamba maana wakati mimi alisema kitu kama bar ni kazi,

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Mimi tu kwenda kutangaza integer na kisha kurudi pointer integer kwamba.

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Hivyo ambapo gani y kuishi?

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[Mwanafunzi] y anaishi katika bar. >> [Bowden] Yeah.

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Mahali fulani katika mraba huu kidogo ya kumbukumbu ni mraba littler kwamba ana y ndani yake.

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Wakati mimi kurudi & y, mimi nina kurudi pointer block hii kidogo ya kumbukumbu.

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Lakini basi atakaporudi kazi, stack yake frame anapata popped mbali stack.

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Na kwamba sababu ni kuitwa stack.

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Ni kama muundo data stack, kama unajua nini kuwa ni.

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Au hata kama stack ya trays ni daima mfano,

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kuu ni kwenda juu chini, kisha kazi kwanza wewe piga yataenda juu ya kwamba,

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na huwezi kupata nyuma kuu mpaka kurudi kutoka utendaji wote ambao wameitwa

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ambayo yamekuwa kuwekwa juu yake.

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>> [Mwanafunzi] Basi kama alivyofanya kufanya kurudi y &, thamani kwamba ni kubadilika bila ya taarifa.

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Ndiyo, it's - >> [mwanafunzi] Ni inaweza kuwa overwritten. >> Yeah.

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Ni kabisa - Kama wewe kujaribu na -

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Hii pia ingekuwa int * bar kwa sababu ni kurudi pointer,

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hivyo kurudi yake ni aina int *.

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Kama wewe kujaribu kutumia thamani ya kurudi kwa kazi hii, ni undefined tabia

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kwa sababu pointer kwamba anasema kwa kumbukumbu mbaya. >> [Mwanafunzi] Sawa.

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Basi nini kama, kwa mfano, alitangaza int * y = malloc (sizeof (int))?

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Hiyo ni bora zaidi. Ndiyo.

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[Mwanafunzi] Kuongelea jinsi wakati sisi Drag mambo bin yetu kusaga

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wao siyo kweli erased; sisi tu kupoteza kuyatumia yao.

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Hivyo katika kesi hii je sisi kweli kufuta thamani au ni bado kuna katika kumbukumbu?

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Kwa sehemu kubwa, ni ya kwenda bado kuwa huko.

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Lakini hebu sema sisi kutokea kwa kuwaita baadhi ya kazi nyingine, bazi.

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Baz ni kwenda kupata stack yake mwenyewe frame ya juu hapa.

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Ni kwenda kuwa overwriting yote ya mambo haya,

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na kisha kama baadaye kujaribu na kutumia pointer kwamba got kabla,

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si kwenda kuwa thamani sawa.

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Ni kwenda yamebadilika kwa sababu tu wewe kuitwa bazi kazi.

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[Mwanafunzi] Lakini sisi si alikuwa, ingekuwa sisi bado kupata 3?

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[Bowden] Katika uwezekano wote, ninyi wenyewe.

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Lakini huwezi kutegemea kwamba. C tu anasema tabia zisizoeleweka.

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>> [Mwanafunzi] Oh, ni gani. Sawa.

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Hivyo wakati unataka kurudi pointer, hii ni wapi malloc huja katika matumizi.

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Mimi nina maandishi kweli tu kurudi malloc (3 * sizeof (int)).

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Tutaweza kwenda juu zaidi katika malloc pili, lakini wazo la malloc ni wote wa vigezo yako ya mtaa

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daima kwenda juu ya stack.

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Chochote ambacho malloced unaendelea chungu, na itakuwa ni milele na daima kuwa katika kifusi

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mpaka waziwazi huru yake.

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Hivyo hii ina maana kwamba wakati wewe malloc kitu, ni kwenda kuishi baada ya kurudi kazi.

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[Mwanafunzi] Je, ni kuishi baada ya mpango ataacha mbio? >> No

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Okay, kwa hivyo ni kwenda kuwa huko mpaka mpango ni njia yote kufanyika mbio. >> Ndiyo.

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Tunaweza kwenda zaidi ya maelezo ya kile kinachotokea wakati mpango ataacha mbio.

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Unaweza haja ya kuwakumbusha yangu, lakini kwamba ni jambo tofauti kabisa.

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[Mwanafunzi] Basi malloc inajenga pointer? >> Yeah.

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Malloc - >> [mwanafunzi] Nadhani malloc designates block ya kumbukumbu kwamba pointer wanaweza kutumia.

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[Bowden] Mimi nataka kuwa mchoro tena. >> [Mwanafunzi] Basi kazi hii kazi, ingawa?

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[Mwanafunzi] Yeah, malloc designates block ya kumbukumbu kwamba unaweza kutumia,

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na kisha kuirudisha anuani ya kuzuia kwanza ya kumbukumbu hiyo.

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>> [Bowden] Yeah. Hivyo wakati wewe malloc, wewe ni grabbing baadhi block ya kumbukumbu

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kwamba sasa katika lundo.

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Kama chungu ni ndogo mno, kuwarundika ni kwenda tu kukua, na hukua katika mwelekeo huu.

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Basi hebu kusema chungu ni ndogo mno.

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Basi ni habari kukua kidogo kidogo na kurudi pointer hii block kwamba tu ilikua.

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Wakati mambo bure, wewe ni kufanya zaidi ya chumba katika chungu,

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hivyo basi baadaye wito kwa malloc wanaweza kutumia tena kwamba kumbukumbu kwamba hapo awali alikuwa huru.

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jambo muhimu kuhusu malloc na bure ni kwamba inakupa udhibiti kamili

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juu ya maisha ya vitalu hizi kumbukumbu.

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Vigezo Global ni daima hai.

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Vigezo Mitaa ni hai ndani ya wigo wao.

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Haraka kama wewe kwenda nyuma Brace curly, vigezo mitaa ni maiti.

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Kumbukumbu Malloced ni hai wakati unataka kuwa hai

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na kisha ni iliyotolewa wakati kuwaambia ni kutolewa.

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Wale ni kweli tu aina 3 ya kumbukumbu, kwa kweli.

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Kuna moja kwa moja ya kumbukumbu ya usimamizi, ambayo ni stack.

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Mambo kutokea kwa wewe moja kwa moja.

176
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Wakati kusema x int, kumbukumbu ni zilizotengwa kwa ajili ya x int.

177
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Wakati x huenda nje ya wigo, kumbukumbu ni reclaimed kwa x.

178
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Kisha kuna nguvu kumbukumbu usimamizi, ambayo ni nini malloc ni,

179
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ambayo ni wakati wewe kuwa na udhibiti.

180
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Wewe dynamically kuamua wakati kumbukumbu lazima na haipaswi zilizotengwa.

181
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Na kisha kuna tuli, ambayo tu ina maana kwamba anaishi milele,

182
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ambayo ni nini vigezo kimataifa ni.

183
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Wao ni tu daima katika kumbukumbu.

184
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>> Maswali?

185
00:15:14,490 --> 00:15:17,230
[Mwanafunzi] Je, unaweza kufafanua block tu kwa kutumia braces curly

186
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lakini si kuwa na kuwa kama maelezo au taarifa wakati au kitu kama hicho?

187
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Unaweza kufafanua block kama katika kazi, lakini ambayo ina braces curly pia.

188
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[Mwanafunzi] Hivyo unaweza si tu kama jozi random ya braces curly katika code yako

189
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kwamba kuwa na vigezo mitaa? >> Ndiyo, unaweza.

190
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Ndani ya bar int sisi inaweza kuwa na {int y = 3;}.

191
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Hiyo walidhani kuwa hapa hapa.

192
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Lakini hiyo kabisa amefafanua wigo wa int y.

193
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Baada brace kwamba pili curly, y haiwezi kutumika tena.

194
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Wewe karibu kamwe kufanya hivyo, ingawa.

195
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Kupata nyuma na kile kinachotokea wakati mpango mwisho,

196
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kuna aina ya uongo mbaya / nusu kwamba sisi kuwapa ili tu kufanya mambo rahisi.

197
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Sisi kukuambia kwamba wakati wewe kutenga kumbukumbu

198
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wewe ni kugawa baadhi chunk ya RAM kwa variable kwamba.

199
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Lakini wewe si kweli moja kwa moja kugusa RAM milele katika programu yako.

200
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Kama wewe kufikiria hivyo, jinsi mimi akauchomoa -

201
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Na kweli, kama wewe kwenda kupitia katika GDB utaona kitu kimoja.

202
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Bila kujali mara ngapi wewe kuendesha programu yako au nini mpango wewe ni mbio,

203
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stack daima ni kwenda kuanza -

204
00:16:59,950 --> 00:17:06,510
wewe daima kwenda kuona vigezo karibu anuani kitu oxbffff.

205
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Ni kawaida mahali fulani katika kanda hiyo.

206
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Lakini jinsi gani wanaweza 2 mipango uwezekano wa kuwa na kuyatumia kwa kumbukumbu sawa?

207
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[Mwanafunzi] Kuna baadhi ya wajibu holela wa oxbfff ambapo zinatakiwa kuwa juu ya RAM

208
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kwamba kweli anaweza kuwa katika maeneo tofauti kutegemea wakati kazi iliitwa.

209
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Yeah. mrefu ni virtual kumbukumbu.

210
00:17:35,920 --> 00:17:42,250
Wazo ni kwamba, kila mchakato wa moja, kila mpango moja kwamba ni mbio kwenye kompyuta yako

211
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ina wenyewe wake - hebu kudhani bits 32 - kujitegemea kabisa anuani nafasi.

212
00:17:49,450 --> 00:17:51,590
Hii ni nafasi anuani.

213
00:17:51,590 --> 00:17:56,220
Ina kujitegemea kabisa yake mwenyewe gigabytes 4 kutumia.

214
00:17:56,220 --> 00:18:02,220
>> Hivyo kama wewe kukimbia mipango 2 wakati huo huo, mpango huu anaona gigabytes 4 kwa yenyewe,

215
00:18:02,220 --> 00:18:04,870
mpango huu anaona gigabytes 4 kwa yenyewe,

216
00:18:04,870 --> 00:18:07,720
na ni vigumu kwa mpango huu kwa dereference pointer

217
00:18:07,720 --> 00:18:10,920
na kuishia na kumbukumbu kutoka mpango huu.

218
00:18:10,920 --> 00:18:18,200
Na nini kumbukumbu virtual ni ni ramani kutoka nafasi michakato anuani

219
00:18:18,200 --> 00:18:20,470
mambo halisi juu ya RAM.

220
00:18:20,470 --> 00:18:22,940
Hivyo ni juu ya mfumo wa uendeshaji wako na kujua kwamba,

221
00:18:22,940 --> 00:18:28,080
hey, wakati huu pointer dereferences guy oxbfff, kwamba kweli ina maana

222
00:18:28,080 --> 00:18:31,040
kwamba anataka RAM Byte 1000,

223
00:18:31,040 --> 00:18:38,150
lakini kama mpango huu dereferences oxbfff, yeye kweli anataka RAM Byte 10,000.

224
00:18:38,150 --> 00:18:41,590
Wanaweza kuwa kiholela mbali mbali.

225
00:18:41,590 --> 00:18:48,730
Hii ni hata ya kweli ya mambo ndani ya nafasi michakato anuani moja.

226
00:18:48,730 --> 00:18:54,770
Hivyo kama anaona gigabytes wote 4 kwa yenyewe, lakini tuseme -

227
00:18:54,770 --> 00:18:57,290
[Mwanafunzi] Kwani kila mchakato wa moja -

228
00:18:57,290 --> 00:19:01,350
Hebu sema wewe kuwa na kompyuta na gigabytes 4 tu ya RAM.

229
00:19:01,350 --> 00:19:06,430
Je, kila mchakato wa moja kuona nzima gigabytes 4? >> Ndiyo.

230
00:19:06,430 --> 00:19:13,060
Lakini gigabytes 4 anaona ni uongo.

231
00:19:13,060 --> 00:19:20,460
Ni tu ni anadhani ana kumbukumbu hii yote kwa sababu hajui mchakato wowote mwingine lipo.

232
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Itakuwa tu kutumia kiasi kama kumbukumbu kama kweli anahitaji.

233
00:19:28,140 --> 00:19:32,340
mfumo wa uendeshaji ni si kwenda kutoa RAM kwa mchakato huu

234
00:19:32,340 --> 00:19:35,750
kama si kutumia yoyote kumbukumbu katika mkoa huu mzima.

235
00:19:35,750 --> 00:19:39,300
Si kwenda kutoa ni kumbukumbu kwa ajili ya kanda hiyo.

236
00:19:39,300 --> 00:19:54,780
Lakini wazo ni kwamba - mimi nina ninafikiria - siwezi kufikiri ya mlinganisho.

237
00:19:54,780 --> 00:19:56,780
Analojia ni ngumu.

238
00:19:57,740 --> 00:20:02,700
Moja ya masuala ya kumbukumbu virtual au moja ya mambo ni kutatua

239
00:20:02,700 --> 00:20:06,810
ni kwamba michakato ya lazima kabisa hawajui mtu mwingine.

240
00:20:06,810 --> 00:20:12,140
Na hivyo unaweza kuandika mpango wowote kwamba tu dereferences yoyote pointer,

241
00:20:12,140 --> 00:20:19,340
kama tu kuandika mpango kwamba anasema * (ox1234),

242
00:20:19,340 --> 00:20:22,890
na kwamba ni dereferencing kumbukumbu anuani 1234.

243
00:20:22,890 --> 00:20:28,870
>> Lakini ni juu ya mfumo wa uendeshaji na kisha kutafsiri nini maana ya 1234.

244
00:20:28,870 --> 00:20:33,960
Hivyo kama 1234 hutokea kwa kuwa halali kumbukumbu anuani kwa utaratibu huu,

245
00:20:33,960 --> 00:20:38,800
kama ni juu ya stack au kitu, basi hii atarudi thamani ya anuani kwamba kumbukumbu

246
00:20:38,800 --> 00:20:41,960
mbali kama mchakato anajua.

247
00:20:41,960 --> 00:20:47,520
Lakini kama 1234 ni si anuani halali, kama hutokea kwa ardhi

248
00:20:47,520 --> 00:20:52,910
katika baadhi ya kipande kidogo ya kumbukumbu hapa ambacho ni zaidi ya stack na zaidi ya chungu

249
00:20:52,910 --> 00:20:57,200
na wewe si kweli kwamba kutumika, kisha hiyo wakati wewe kupata mambo kama segfaults

250
00:20:57,200 --> 00:21:00,260
kwa sababu wewe ni kugusa kumbukumbu kwamba unapaswa kuwa kugusa.

251
00:21:07,180 --> 00:21:09,340
Hii pia ni kweli -

252
00:21:09,340 --> 00:21:15,440
Mfumo ya 32-bit, 32 bits ina maana una bits 32 kufafanua anuani kumbukumbu.

253
00:21:15,440 --> 00:21:22,970
Ni kwa nini ni kuyatumia 8 ka sababu 32 bits ni 8 ka - au ka 4.

254
00:21:22,970 --> 00:21:25,250
Kuyatumia ni 4 ka.

255
00:21:25,250 --> 00:21:33,680
Hivyo wakati wewe kuona kama pointer oxbfffff, kwamba ni -

256
00:21:33,680 --> 00:21:40,080
Ndani ya mpango wowote unaweza tu kujenga yoyote pointer holela,

257
00:21:40,080 --> 00:21:46,330
popote kutoka ox0 kwa ng'ombe 8 f's - ffffffff.

258
00:21:46,330 --> 00:21:49,180
[Mwanafunzi] Je wewe kusema wao ni 4 ka? >> Yeah.

259
00:21:49,180 --> 00:21:52,730
[Mwanafunzi] Kisha Byte kila itakuwa na - >> [Bowden] hexadesimoli.

260
00:21:52,730 --> 00:21:59,360
Hexadesimoli - 5, 6, 7, 8. Hivyo kuyatumia utaenda daima kuona katika hexadesimoli.

261
00:21:59,360 --> 00:22:01,710
Ni tu jinsi sisi classify kuyatumia.

262
00:22:01,710 --> 00:22:05,240
Kila digits 2 ya hexadesimoli ni 1 Byte.

263
00:22:05,240 --> 00:22:09,600
Hivyo kuna kwenda kuwa 8 hexadesimoli digits kwa ka 4.

264
00:22:09,600 --> 00:22:14,190
Hivyo kila pointer moja juu ya mfumo wa 32-bit ni kwenda kuwa 4 ka,

265
00:22:14,190 --> 00:22:18,550
ambayo ina maana kwamba katika mchakato wako unaweza kujenga yoyote holela ka 4

266
00:22:18,550 --> 00:22:20,550
na kufanya pointer nje ya hayo,

267
00:22:20,550 --> 00:22:32,730
ambayo ina maana kwamba mbali kama ni kufahamu, inaweza kushughulikia 2 mzima ka 32 ya kumbukumbu.

268
00:22:32,730 --> 00:22:34,760
Hata ingawa ni kweli haina kupata kwamba,

269
00:22:34,760 --> 00:22:40,190
hata kama kompyuta yako tu ina megabaiti 512, ni anadhani ana kwamba kumbukumbu sana.

270
00:22:40,190 --> 00:22:44,930
Na mfumo wa uendeshaji ni smart kutosha kwamba itakuwa tu kutenga kile haja kweli.

271
00:22:44,930 --> 00:22:49,630
Haina tu kwenda, oh, mchakato mpya: 4 gigs.

272
00:22:49,630 --> 00:22:51,930
>> Yeah. >> [Mwanafunzi] gani ng'ombe maana? Mbona kuandika?

273
00:22:51,930 --> 00:22:54,980
Ni tu ishara kwa hexadesimoli.

274
00:22:54,980 --> 00:22:59,590
Baada ya kuona idadi ya kuanza kwa ng'ombe, mambo mfululizo ni hexadesimoli.

275
00:23:01,930 --> 00:23:05,760
[Mwanafunzi] Wewe walikuwa akielezea juu ya kile kinachotokea wakati mpango mwisho. >> Ndiyo.

276
00:23:05,760 --> 00:23:09,480
Nini kinatokea wakati mpango mwisho ni mfumo wa uendeshaji

277
00:23:09,480 --> 00:23:13,600
tu erases upangaji kwamba ina kwa anwani hizi, na hiyo ni yake.

278
00:23:13,600 --> 00:23:17,770
mfumo wa uendeshaji sasa wanaweza kutoa tu kwamba kumbukumbu kwa mpango mwingine kutumia.

279
00:23:17,770 --> 00:23:19,490
[Mwanafunzi] Sawa.

280
00:23:19,490 --> 00:23:24,800
Hivyo wakati wewe kutenga kitu juu ya lundo au vigezo stack au kimataifa au kitu chochote,

281
00:23:24,800 --> 00:23:27,010
wao wote tu kutoweka kwa haraka kama mpango mwisho

282
00:23:27,010 --> 00:23:32,120
kwa sababu ya mfumo wa uendeshaji ni sasa huru kutoa kwamba kumbukumbu kwa mchakato nyingine yoyote.

283
00:23:32,120 --> 00:23:35,150
[Mwanafunzi] Hata ingawa kuna bado pengine maadili yaliyoandikwa katika? >> Yeah.

284
00:23:35,150 --> 00:23:37,740
maadili ni uwezekano bado kuna.

285
00:23:37,740 --> 00:23:41,570
Ni tu ni kwenda kuwa vigumu kupata kwao.

286
00:23:41,570 --> 00:23:45,230
Ni vigumu zaidi kupata saa yao zaidi ni kupata saa faili ilifutwa

287
00:23:45,230 --> 00:23:51,450
kwa sababu ilifutwa faili aina ya aketiye huko kwa muda mrefu na ngumu kuendesha ni mengi zaidi.

288
00:23:51,450 --> 00:23:54,120
Hivyo ni kwenda overwrite sehemu mbalimbali za kumbukumbu

289
00:23:54,120 --> 00:23:58,640
kabla hayajatokea ili overwrite chunk ya kumbukumbu kwamba faili kwamba kutumika kuwa katika.

290
00:23:58,640 --> 00:24:04,520
Lakini kuu kumbukumbu, RAM, wewe mzunguko kwa njia ya kura kwa kasi,

291
00:24:04,520 --> 00:24:08,040
hivyo ni kwenda haraka sana kuwa overwritten.

292
00:24:10,300 --> 00:24:13,340
Maswali juu ya hili au kitu kingine?

293
00:24:13,340 --> 00:24:16,130
[Mwanafunzi] nina maswali kuhusu mada tofauti tofauti. >> Sawa.

294
00:24:16,130 --> 00:24:19,060
Je, mtu yeyote kuwa na maswali juu ya hili?

295
00:24:20,170 --> 00:24:23,120
>> Sawa. Mbalimbali ya mada. >> [Mwanafunzi] Sawa.

296
00:24:23,120 --> 00:24:26,550
Mimi alikuwa anapita katika baadhi ya vipimo vya mazoezi,

297
00:24:26,550 --> 00:24:30,480
na katika mmoja wao ni kuzungumza kuhusu sizeof

298
00:24:30,480 --> 00:24:35,630
na thamani ya kwamba kuirudisha au aina tofauti kutofautiana. >> Ndiyo.

299
00:24:35,630 --> 00:24:45,060
Na alisema kwamba wote int na ya muda mrefu kurudi 4, hivyo ni wote ka 4 mrefu.

300
00:24:45,060 --> 00:24:48,070
Je, kuna tofauti yoyote kati ya int na muda mrefu, au ni kitu kimoja?

301
00:24:48,070 --> 00:24:50,380
Ndiyo, kuna tofauti.

302
00:24:50,380 --> 00:24:52,960
C kiwango -

303
00:24:52,960 --> 00:24:54,950
Mimi pengine anaenda fujo up.

304
00:24:54,950 --> 00:24:58,800
Kiwango C ni kama yale C ni, nyaraka rasmi ya C.

305
00:24:58,800 --> 00:25:00,340
Hii ni nini anasema.

306
00:25:00,340 --> 00:25:08,650
Hivyo kiwango C tu anasema kwamba Char itakuwa milele na daima kuwa 1 Byte.

307
00:25:10,470 --> 00:25:19,040
Kila kitu baada ya kuwa - mfupi ni daima tu hufafanuliwa kama ulikuwa mkubwa kuliko au sawa na char.

308
00:25:19,040 --> 00:25:23,010
Hii inaweza kuwa madhubuti zaidi kuliko, lakini si chanya.

309
00:25:23,010 --> 00:25:31,940
int ni tu hufafanuliwa kama ulikuwa mkubwa kuliko au sawa na mfupi.

310
00:25:31,940 --> 00:25:36,210
Na muda mrefu ni tu hufafanuliwa kama ulikuwa mkubwa kuliko au sawa na int.

311
00:25:36,210 --> 00:25:41,600
Na muda mrefu, ni mkubwa kuliko au sawa na muda mrefu.

312
00:25:41,600 --> 00:25:46,610
Hivyo kitu pekee kiwango C amefafanua ni kuagiza jamaa wa kila kitu.

313
00:25:46,610 --> 00:25:54,880
kiasi halisi ya kumbukumbu kwamba mambo kuchukua ni ujumla hadi utekelezaji,

314
00:25:54,880 --> 00:25:57,640
lakini ni pretty defined vizuri katika hatua hii. >> [Mwanafunzi] Sawa.

315
00:25:57,640 --> 00:26:02,490
Hivyo kaptula ni karibu daima itakuwa 2 ka.

316
00:26:04,920 --> 00:26:09,950
Ints ni karibu daima itakuwa 4 ka.

317
00:26:12,070 --> 00:26:15,340
Hutamani muda mrefu ni karibu daima itakuwa 8 bytes.

318
00:26:17,990 --> 00:26:23,160
Na anatamani, inategemea kama unatumia 32-bit au mfumo 64-bit.

319
00:26:23,160 --> 00:26:27,450
Hiyo kwa muda mrefu ni kwenda yanahusiana na aina ya mfumo.

320
00:26:27,450 --> 00:26:31,920
Kama wewe ni kutumia mfumo wa 32-bit kama Appliance, itakavyo kuwa 4 ka.

321
00:26:34,530 --> 00:26:42,570
Kama unatumia 64-bit kama mengi ya kompyuta hivi karibuni, ni kwenda kuwa 8 bytes.

322
00:26:42,570 --> 00:26:45,230
>> Ints ni karibu daima ka 4 katika hatua hii.

323
00:26:45,230 --> 00:26:47,140
Hutamani muda mrefu ni karibu daima 8 bytes.

324
00:26:47,140 --> 00:26:50,300
Huko nyuma, ints kutumika tu 2 ka.

325
00:26:50,300 --> 00:26:56,840
Lakini taarifa kwamba hii kabisa satisfies yote ya mahusiano haya ya mkuu kuliko na kuwa sawa na.

326
00:26:56,840 --> 00:27:01,280
Hiyo kwa muda mrefu kikamilifu kuruhusiwa kuwa ukubwa sawa kama integer,

327
00:27:01,280 --> 00:27:04,030
na ni pia kuruhusiwa kuwa ukubwa sawa kama muda mrefu.

328
00:27:04,030 --> 00:27:11,070
Na tu hivyo hutokea kuwa kwamba katika 99.999% ya mifumo, ni kwenda kuwa sawa na

329
00:27:11,070 --> 00:27:15,800
aidha int au muda mrefu. Ni inategemea tu kidogo 32-au 64-bit. >> [Mwanafunzi] Sawa.

330
00:27:15,800 --> 00:27:24,600
Katika ikifungwa, ni jinsi gani ya uhakika decimal mteule katika suala la bits?

331
00:27:24,600 --> 00:27:27,160
Kama kama binary? >> Yeah.

332
00:27:27,160 --> 00:27:30,570
Huna haja ya kujua kwamba kwa CS50.

333
00:27:30,570 --> 00:27:32,960
Huna hata kujifunza kuwa katika 61.

334
00:27:32,960 --> 00:27:37,350
Huwezi kujifunza kwamba kweli katika shaka yoyote.

335
00:27:37,350 --> 00:27:42,740
Ni tu uwakilishi.

336
00:27:42,740 --> 00:27:45,440
Mimi kusahau allotments kidogo halisi.

337
00:27:45,440 --> 00:27:53,380
wazo la uhakika floating ni kwamba kutenga idadi maalum ya bits kuwakilisha -

338
00:27:53,380 --> 00:27:56,550
Kimsingi, kila kitu ni kwa nukuu ya kisayansi.

339
00:27:56,550 --> 00:28:05,600
Hivyo kutenga idadi maalum ya bits kuwakilisha idadi yenyewe, kama 1.2345.

340
00:28:05,600 --> 00:28:10,200
Siwezi kamwe kuwakilisha idadi na tarakimu zaidi ya 5.

341
00:28:12,200 --> 00:28:26,300
Kisha wewe pia kutenga idadi maalum ya bits ili huelekea kuwa kama

342
00:28:26,300 --> 00:28:32,810
unaweza tu kwenda idadi fulani, kama kwamba exponent kubwa unaweza kuwa,

343
00:28:32,810 --> 00:28:36,190
na unaweza tu kwenda chini kwa exponent fulani,

344
00:28:36,190 --> 00:28:38,770
kama hiyo ni ndogo exponent unaweza kuwa.

345
00:28:38,770 --> 00:28:44,410
>> Sikumbuki bits njia halisi ni kwa ajili ya wote wa maadili haya,

346
00:28:44,410 --> 00:28:47,940
lakini idadi fulani ya bits ni wakfu kwa 1.2345,

347
00:28:47,940 --> 00:28:50,930
idadi mwingine baadhi ya bits ni wakfu kwa exponent,

348
00:28:50,930 --> 00:28:55,670
na ni inawezekana tu kuwakilisha exponent ya ukubwa fulani.

349
00:28:55,670 --> 00:29:01,100
[Mwanafunzi] Na mara mbili? Ni kwamba kama kuelea ziada kwa muda mrefu? >> Yeah.

350
00:29:01,100 --> 00:29:07,940
Ni kitu kimoja kama kuelea ila sasa unatumia ka 8 badala ya ka 4.

351
00:29:07,940 --> 00:29:11,960
Sasa wewe utakuwa na uwezo wa kutumia tarakimu 9 au tarakimu ya 10,

352
00:29:11,960 --> 00:29:16,630
na hii itakuwa na uwezo wa kwenda hadi 300 badala ya 100. >> [Mwanafunzi] Sawa.

353
00:29:16,630 --> 00:29:21,550
Na ikifungwa ni pia 4 ka. >> Ndiyo.

354
00:29:21,550 --> 00:29:27,520
Naam, tena, labda inategemea kwa ujumla juu ya utekelezaji wa jumla,

355
00:29:27,520 --> 00:29:30,610
lakini ikifungwa ni 4 ka, DOUBLES ni 8.

356
00:29:30,610 --> 00:29:33,440
DOUBLES huitwa mara mbili kwa sababu ni mara mbili ya ukubwa wa inaelea.

357
00:29:33,440 --> 00:29:38,380
[Mwanafunzi] Sawa. Na ni pale mara mbili mara mbili? >> Kuna si.

358
00:29:38,380 --> 00:29:43,660
Nadhani - >> [mwanafunzi] Kama hutamani muda mrefu? >> Yeah. Sidhani hivyo. Ndiyo.

359
00:29:43,660 --> 00:29:45,950
[Mwanafunzi] cha mtihani mwaka jana kulikuwa na swali kuhusu kazi kuu

360
00:29:45,950 --> 00:29:49,490
kuwa na kuwa sehemu ya programu yako.

361
00:29:49,490 --> 00:29:52,310
Jibu ni kwamba hana na kuwa sehemu ya programu yako.

362
00:29:52,310 --> 00:29:55,100
Katika hali gani? Hiyo ni nini nikaona.

363
00:29:55,100 --> 00:29:59,090
[Bowden] Inaonekana - >> [mwanafunzi] Nini hali?

364
00:29:59,090 --> 00:30:02,880
Je, una tatizo? >> [Mwanafunzi] Yeah, mimi unaweza dhahiri kuvuta it up.

365
00:30:02,880 --> 00:30:07,910
Haina kuwa, kitaalam, lakini kimsingi ni kwenda kuwa.

366
00:30:07,910 --> 00:30:10,030
[Mwanafunzi] nikaona moja juu ya tofauti wa mwaka.

367
00:30:10,030 --> 00:30:16,220
Ilikuwa kama kweli au uongo: halali - >> Oh, c faili.?

368
00:30:16,220 --> 00:30:18,790
. [Mwanafunzi] yoyote c faili lazima kuwa - [wote akizungumza kwa mara moja - unintelligible]

369
00:30:18,790 --> 00:30:21,120
Sawa. Basi hiyo ni tofauti.

370
00:30:21,120 --> 00:30:26,800
>> C. Faili tu mahitaji vyenye kuharibika.

371
00:30:26,800 --> 00:30:32,400
Unaweza kukusanya faili katika code mashine, binary, chochote,

372
00:30:32,400 --> 00:30:36,620
bila kuwa executable bado.

373
00:30:36,620 --> 00:30:39,420
executable halali lazima kuwa na kazi kuu.

374
00:30:39,420 --> 00:30:45,460
Unaweza kuandika kazi 100 katika faili 1 lakini hakuna kuu

375
00:30:45,460 --> 00:30:48,800
na kisha kukusanya kwamba chini binary,

376
00:30:48,800 --> 00:30:54,460
kisha kuandika faili nyingine kwamba tu ana kuu lakini ni wito rundo la kazi hizi

377
00:30:54,460 --> 00:30:56,720
katika faili hii binary zaidi ya hapa.

378
00:30:56,720 --> 00:31:01,240
Na hivyo wakati wewe ni kufanya executable, kwamba ni nini linker gani

379
00:31:01,240 --> 00:31:05,960
ni inaunganisha hizi files binary 2 ndani ya kutekelezwa.

380
00:31:05,960 --> 00:31:11,400
Hivyo c. Faili haina haja ya kuwa na kazi kuu wakati wote.

381
00:31:11,400 --> 00:31:19,220
Na juu ya besi kubwa code utaona maelfu ya. Files c na 1 kuu faili.

382
00:31:23,960 --> 00:31:26,110
Zaidi maswali?

383
00:31:29,310 --> 00:31:31,940
[Mwanafunzi] Kulikuwa na swali lingine.

384
00:31:31,940 --> 00:31:36,710
Ni alisema kufanya ni compiler. Kweli au uongo?

385
00:31:36,710 --> 00:31:42,030
Na jibu ilikuwa ni uongo, na mimi kueleweka kwa nini siyo kama Clang.

386
00:31:42,030 --> 00:31:44,770
Lakini je, sisi kuwaita kufanya kama si?

387
00:31:44,770 --> 00:31:49,990
Matokeo ni ya kimsingi tu - naweza kuona hasa nini wito yake.

388
00:31:49,990 --> 00:31:52,410
Lakini ni anaendesha tu amri.

389
00:31:53,650 --> 00:31:55,650
Kufanya.

390
00:31:58,240 --> 00:32:00,870
Naweza kuvuta hii up. Yeah.

391
00:32:10,110 --> 00:32:13,180
Oh, yeah. Matokeo pia anafanya hivyo.

392
00:32:13,180 --> 00:32:17,170
Hii inasema madhumuni ya shirika kufanya ni kuamua moja kwa moja

393
00:32:17,170 --> 00:32:19,610
ambayo vipande vya mpango kubwa yanahitaji recompiled

394
00:32:19,610 --> 00:32:22,350
na kutoa amri kwa recompile yao.

395
00:32:22,350 --> 00:32:27,690
Unaweza kufanya kufanya files kwamba ni kabisa kubwa.

396
00:32:27,690 --> 00:32:33,210
Matokeo inaangalia mihuri wakati wa files na, kama sisi alisema kabla,

397
00:32:33,210 --> 00:32:36,930
unaweza kukusanya files binafsi chini, na si mpaka kupata kwa linker

398
00:32:36,930 --> 00:32:39,270
kwamba wao ni kuweka pamoja ndani ya kutekelezwa.

399
00:32:39,270 --> 00:32:43,810
Hivyo kama una 10 files tofauti na kufanya mabadiliko kwa 1 wao,

400
00:32:43,810 --> 00:32:47,870
basi nini kufanya ni kwenda kufanya ni tu kwamba recompile 1 faili

401
00:32:47,870 --> 00:32:50,640
na kisha relink kila kitu pamoja.

402
00:32:50,640 --> 00:32:53,020
Lakini ni kiasi dumber kuliko hiyo.

403
00:32:53,020 --> 00:32:55,690
Ni juu ya wewe kabisa kufafanua kwamba hiyo ndiyo wanapaswa kufanya.

404
00:32:55,690 --> 00:32:59,560
Ni kwa default ina uwezo wa kutambua hili muhuri wakati stuff,

405
00:32:59,560 --> 00:33:03,220
lakini unaweza kuandika faili kufanya kufanya kitu chochote.

406
00:33:03,220 --> 00:33:09,150
Unaweza kuandika kufanya faili ili pale unapoandika kufanya hivyo tu cd ya saraka nyingine.

407
00:33:09,150 --> 00:33:15,560
Mimi nilikuwa frustrated kupata kwa sababu mimi tack kila kitu ndani ya Appliance yangu

408
00:33:15,560 --> 00:33:21,740
na kisha mimi mtazamo PDF kutoka Mac.

409
00:33:21,740 --> 00:33:30,720
>> Kwa hiyo mimi kwenda Finder na siwezi kufanya Nenda, Connect na Server,

410
00:33:30,720 --> 00:33:36,950
na server mimi kuungana na ni Appliance yangu, kisha mimi kufungua PDF

411
00:33:36,950 --> 00:33:40,190
kwamba anapata ulioandaliwa na mpira.

412
00:33:40,190 --> 00:33:49,320
Lakini mimi kupata frustrated kwa sababu kila wakati mimi zinahitajika kwa ajili ya kujikumbusha PDF,

413
00:33:49,320 --> 00:33:53,900
Mimi nilikuwa na nakala yake kwa directory maalum kwamba inaweza kupata

414
00:33:53,900 --> 00:33:57,710
na ilikuwa kupata annoying.

415
00:33:57,710 --> 00:34:02,650
Hivyo badala niliandika faili kufanya, ambayo una kufafanua jinsi inafanya mambo.

416
00:34:02,650 --> 00:34:06,130
Jinsi ya kufanya katika hili ni PDF mpira.

417
00:34:06,130 --> 00:34:10,090
Tu kama faili yoyote nyingine kufanya - au mimi nadhani wewe hawajaona files kufanya,

418
00:34:10,090 --> 00:34:13,510
lakini sisi katika appliance kimataifa kufanya faili kwamba tu anasema,

419
00:34:13,510 --> 00:34:16,679
kama wewe ni kuandaa faili C, kutumia Clang.

420
00:34:16,679 --> 00:34:20,960
Na hivyo hapa katika kufanya file yangu kwamba mimi kufanya nasema,

421
00:34:20,960 --> 00:34:25,020
faili hii utaenda wanataka kukusanya na PDF mpira.

422
00:34:25,020 --> 00:34:27,889
Na hivyo ni PDF mpira kwamba anafanya kuandaa.

423
00:34:27,889 --> 00:34:31,880
Kufanya si kuandaa. Ni kukimbia tu amri hizi katika mlolongo mimi maalum.

424
00:34:31,880 --> 00:34:36,110
Hivyo ni anaendesha PDF mpira, ni nakala yake na directory mimi nataka kunakiliwa kwa,

425
00:34:36,110 --> 00:34:38,270
ni cd ya saraka na anafanya mambo mengine,

426
00:34:38,270 --> 00:34:42,380
lakini wote ni gani ni kutambua wakati mabadiliko ya SVG,

427
00:34:42,380 --> 00:34:45,489
na kama ni mabadiliko, basi itakuwa kukimbia amri kwamba ni walidhani kukimbia

428
00:34:45,489 --> 00:34:48,760
wakati mabadiliko faili. >> [Mwanafunzi] Sawa.

429
00:34:50,510 --> 00:34:54,420
Sijui ambapo kimataifa kufanya files ni kwa ajili yangu na kuangalia ni nje.

430
00:34:57,210 --> 00:35:04,290
Maswali mengine? Kitu chochote kutoka zamani Quizzes? Mambo pointer yoyote?

431
00:35:06,200 --> 00:35:08,730
Kuna mambo ya hila na kuyatumia kama -

432
00:35:08,730 --> 00:35:10,220
Mimi si kwenda kuwa na uwezo wa kupata swali Jaribio juu yake -

433
00:35:10,220 --> 00:35:16,250
lakini tu kama jambo la aina hii.

434
00:35:19,680 --> 00:35:24,060
Hakikisha kuelewa kwamba wakati mimi kusema int * x * y -

435
00:35:24,890 --> 00:35:28,130
Hii si hasa chochote hapa, mimi nadhani.

436
00:35:28,130 --> 00:35:32,140
Lakini kama * x * y, wale ni 2 vigezo kwamba ni juu ya stack.

437
00:35:32,140 --> 00:35:37,220
Wakati mimi kusema x = malloc (sizeof (int)), x bado ni kutofautiana juu ya stack,

438
00:35:37,220 --> 00:35:41,180
malloc ni baadhi block juu katika chungu, na sisi ni kuwa x uhakika na lundo.

439
00:35:41,180 --> 00:35:43,900
>> Hivyo kitu juu ya pointi stack chungu.

440
00:35:43,900 --> 00:35:48,100
Kila wewe malloc kitu chochote, wewe ni inevitably hifadhi hiyo ndani ya pointer.

441
00:35:48,100 --> 00:35:55,940
Hivyo pointer kwamba ni juu ya stack, block malloced ni juu ya lundo.

442
00:35:55,940 --> 00:36:01,240
mengi ya watu kupata kuchanganyikiwa na kusema int * x = malloc; x ni juu ya lundo.

443
00:36:01,240 --> 00:36:04,100
No nini x anaonyesha ni juu ya lundo.

444
00:36:04,100 --> 00:36:08,540
x yenyewe ni juu ya stack, isipokuwa kwa sababu yoyote una x kuwa variable kimataifa,

445
00:36:08,540 --> 00:36:11,960
katika kesi ambayo hutokea kwa kuwa katika mkoa mwingine wa kumbukumbu.

446
00:36:13,450 --> 00:36:20,820
Hivyo kuweka wimbo, hizi michoro sanduku na mshale ni pretty kawaida kwa chemsha bongo.

447
00:36:20,820 --> 00:36:25,740
Au kama si juu ya jaribio 0, itakuwa juu ya chemsha bongo 1.

448
00:36:27,570 --> 00:36:31,940
Unapaswa kujua haya yote, kwa hatua katika kuandaa

449
00:36:31,940 --> 00:36:35,740
tangu alikuwa na kujibu maswali juu ya hayo. Ndiyo.

450
00:36:35,740 --> 00:36:38,940
[Mwanafunzi] Inawezekana sisi kwenda juu ya hatua hizo - >> Uhakika.

451
00:36:48,340 --> 00:36:58,640
Kabla ya hatua na kuandaa tuna preprocessing,

452
00:36:58,640 --> 00:37:16,750
kuandaa, kukusanyika, na kuunganisha.

453
00:37:16,750 --> 00:37:21,480
Preprocessing. Je, hiyo nini?

454
00:37:29,720 --> 00:37:32,290
Ni hatua rahisi katika - vizuri, si kama -

455
00:37:32,290 --> 00:37:35,770
kwamba haina maana ni lazima kuwa wazi, lakini ni hatua rahisi.

456
00:37:35,770 --> 00:37:38,410
You guys yangeweza kutekeleza ninyi wenyewe. Yeah.

457
00:37:38,410 --> 00:37:43,410
[Mwanafunzi] Chukua yale katika yako ni pamoja na kama hii na nakala na kisha pia amefafanua.

458
00:37:43,410 --> 00:37:49,250
Inaonekana kwa mambo kama # ni pamoja na # define,

459
00:37:49,250 --> 00:37:53,800
na ni haki nakala na pastes nini wale kweli maana.

460
00:37:53,800 --> 00:37:59,240
Hivyo wakati wewe kusema ni pamoja # cs50.h, preprocessor ni kuiga na pasting cs50.h

461
00:37:59,240 --> 00:38:01,030
ndani ya mstari huo.

462
00:38:01,030 --> 00:38:06,640
Wakati kusema # define x kuwa 4, preprocessor inakwenda kupitia mpango mzima

463
00:38:06,640 --> 00:38:10,400
na nafasi ya matukio yote ya x kwa 4.

464
00:38:10,400 --> 00:38:17,530
Hivyo preprocessor inachukua halali C faili na matokeo halali C faili

465
00:38:17,530 --> 00:38:20,300
ambapo mambo wamekuwa kunakiliwa na pasted.

466
00:38:20,300 --> 00:38:24,230
Hivyo ikiandaa. Je, hiyo nini?

467
00:38:25,940 --> 00:38:28,210
[Mwanafunzi] Ni inakwenda kutoka C kwa binary.

468
00:38:28,210 --> 00:38:30,970
>> [Bowden] Ni haina kwenda njia yote ya binary.

469
00:38:30,970 --> 00:38:34,220
[Mwanafunzi] Ili code mashine, basi? >> Ni si mashine code.

470
00:38:34,220 --> 00:38:35,700
[Mwanafunzi] Bunge? >> Bunge.

471
00:38:35,700 --> 00:38:38,890
Unaendelea na Bunge kabla huenda njia yote ya code C,

472
00:38:38,890 --> 00:38:45,010
na zaidi lugha kufanya kitu kama hiki.

473
00:38:47,740 --> 00:38:50,590
Pick lugha yoyote ya ngazi ya juu, na kama wewe ni kwenda kukusanya yake,

474
00:38:50,590 --> 00:38:52,390
ni uwezekano wa kukusanya katika hatua.

475
00:38:52,390 --> 00:38:58,140
Kwanza ni kwenda kukusanya Chatu na C, basi ni kwenda kukusanya C kwa Bunge,

476
00:38:58,140 --> 00:39:01,600
na kisha Bunge ni kwenda kupata Tafsiri ya binary.

477
00:39:01,600 --> 00:39:07,800
Hivyo kuandaa anaenda kuleta kutoka C kwa Bunge.

478
00:39:07,800 --> 00:39:12,130
neno kuandaa kawaida maana kuleta kutoka ngazi ya juu

479
00:39:12,130 --> 00:39:14,340
kwa ngazi ya chini lugha ya programu.

480
00:39:14,340 --> 00:39:19,190
Hivyo hii ni hatua tu katika mkusanyiko ambapo kuanza kwa lugha ya ngazi ya juu

481
00:39:19,190 --> 00:39:23,270
na kuishia katika lugha ngazi ya chini, na kwamba sababu ya hatua inaitwa kuandaa.

482
00:39:25,280 --> 00:39:33,370
[Mwanafunzi] Wakati wa kuandaa, hebu kusema kwamba umefanya kosa # pamoja cs50.h.

483
00:39:33,370 --> 00:39:42,190
Will compiler recompile cs50.h, kama kazi ya kwamba ni huko,

484
00:39:42,190 --> 00:39:45,280
na kutafsiri kwamba katika kanuni Bunge kama vile,

485
00:39:45,280 --> 00:39:50,830
au itakuwa nakala na kuweka kitu ambacho imekuwa ni kabla ya Bunge?

486
00:39:50,830 --> 00:39:56,910
cs50.h itakuwa pretty much kamwe kuishia katika Bunge.

487
00:39:59,740 --> 00:40:03,680
Mambo kama prototypes kazi na mambo tu kwa ajili ya wewe kuwa makini.

488
00:40:03,680 --> 00:40:09,270
Ni dhamana kwamba compiler unaweza kuangalia mambo kama wewe ni wito kazi

489
00:40:09,270 --> 00:40:12,910
pamoja na haki ya kurudi aina na hoja haki na mambo ya ajabu.

490
00:40:12,910 --> 00:40:18,350
>> Hivyo cs50.h itakuwa preprocessed ndani ya faili, na kisha wakati ni kuandaa

491
00:40:18,350 --> 00:40:22,310
ni kimsingi kutupwa mbali baada inafanya kuhakikisha kwamba kila kitu ni kuitwa kwa usahihi.

492
00:40:22,310 --> 00:40:29,410
Lakini kazi defined katika maktaba CS50, ambayo ni tofauti na cs50.h,

493
00:40:29,410 --> 00:40:33,610
wale si tofauti compiled.

494
00:40:33,610 --> 00:40:37,270
Hiyo kweli kuja chini katika hatua ya kuunganisha, hivyo tutaweza kupata kwamba katika pili.

495
00:40:37,270 --> 00:40:40,100
Lakini kwanza, ni nini kukusanyika?

496
00:40:41,850 --> 00:40:44,500
[Mwanafunzi] Bunge kwa binary? >> Yeah.

497
00:40:46,300 --> 00:40:48,190
Kukusanyika.

498
00:40:48,190 --> 00:40:54,710
Hatuna kuiita kuandaa kwa sababu Bunge ni kiasi pretty tafsiri safi ya binary.

499
00:40:54,710 --> 00:41:00,230
Kuna ni kidogo sana mantiki katika kwenda Bunge kwa binary.

500
00:41:00,230 --> 00:41:03,180
Ni kama tu kuangalia juu katika meza, oh, tuna maelekezo haya;

501
00:41:03,180 --> 00:41:06,290
kwamba sambamba na binary 01,110.

502
00:41:10,200 --> 00:41:15,230
Na hivyo files kwamba kukusanyika kwa ujumla matokeo ni files o..

503
00:41:15,230 --> 00:41:19,020
Na. O files ni nini sisi walikuwa wakisema kabla,

504
00:41:19,020 --> 00:41:21,570
jinsi faili haina haja ya kuwa na kazi kuu.

505
00:41:21,570 --> 00:41:27,640
Faili yoyote inaweza compiled chini faili o. Muda mrefu kama ni halali C faili.

506
00:41:27,640 --> 00:41:30,300
Ni inaweza compiled chini. O.

507
00:41:30,300 --> 00:41:43,030
Sasa, kuunganisha ni nini hasa huleta rundo la o files. Na huleta yao kwa kutekelezwa.

508
00:41:43,030 --> 00:41:51,110
Na hivyo kile kuunganisha gani ni unaweza kufikiri ya maktaba CS50 kama faili o..

509
00:41:51,110 --> 00:41:56,980
Tayari ni compiled binary file.

510
00:41:56,980 --> 00:42:03,530
Na hivyo wakati wewe kukusanya faili yako, hello.c yako, ambayo inatoa wito GetString,

511
00:42:03,530 --> 00:42:06,360
hello.c anapata compiled chini hello.o,

512
00:42:06,360 --> 00:42:08,910
hello.o ni sasa katika binary.

513
00:42:08,910 --> 00:42:12,830
Ni anatumia GetString, hivyo mahitaji ya kwenda juu kwa cs50.o,

514
00:42:12,830 --> 00:42:16,390
na linker smooshes yao pamoja na nakala GetString ndani ya faili hili

515
00:42:16,390 --> 00:42:20,640
na hutoka nje na executable kwamba ana kazi yote ni mahitaji.

516
00:42:20,640 --> 00:42:32,620
Hivyo cs50.o si kweli faili O, lakini ni karibu kutosha kwamba hakuna tofauti ya kimsingi.

517
00:42:32,620 --> 00:42:36,880
Hivyo kuunganisha tu huleta rundo la files pamoja

518
00:42:36,880 --> 00:42:41,390
kwamba tofauti vyenye wote wa kazi mimi haja ya kutumia

519
00:42:41,390 --> 00:42:46,120
na inajenga executable kwamba mapenzi kweli kukimbia.

520
00:42:48,420 --> 00:42:50,780
>> Na hivyo pia kwamba kile sisi walikuwa wakisema kabla ya

521
00:42:50,780 --> 00:42:55,970
ambapo unaweza 1000 files c., wewe kukusanya wote wale o files.,

522
00:42:55,970 --> 00:43:00,040
ambayo pengine kuchukua muda, kisha wewe kubadili 1 c faili..

523
00:43:00,040 --> 00:43:05,480
Unahitaji tu recompile kwamba 1. C faili na kisha relink kila kitu kingine,

524
00:43:05,480 --> 00:43:07,690
kuhusisha kila kitu nyuma pamoja.

525
00:43:09,580 --> 00:43:11,430
[Mwanafunzi] Wakati sisi ni kuunganisha sisi kuandika lcs50?

526
00:43:11,430 --> 00:43:20,510
Yeah, hivyo-lcs50. Ishara kwamba bendera kwa linker kwamba unapaswa kuunganisha katika maktaba hiyo.

527
00:43:26,680 --> 00:43:28,910
Maswali?

528
00:43:41,310 --> 00:43:46,860
Je, sisi wamekwenda juu ya binary nyingine ya sekunde kwamba 5 katika hotuba ya kwanza?

529
00:43:50,130 --> 00:43:53,010
Sidhani hivyo.

530
00:43:55,530 --> 00:43:58,820
Unapaswa kujua yote ya Os kubwa kwamba tumeenda juu,

531
00:43:58,820 --> 00:44:02,670
na unapaswa kuwa na uwezo wa, ikiwa tuliyowapeni kazi,

532
00:44:02,670 --> 00:44:09,410
unapaswa kuwa na uwezo wa kusema ni kubwa O, ukali. Au vizuri, big O ni mbaya.

533
00:44:09,410 --> 00:44:15,300
Hivyo kama unaweza kuona nested kwa matanzi looping juu ya idadi sawa ya mambo,

534
00:44:15,300 --> 00:44:22,260
kama int i, i <n; int j, j <n - >> [mwanafunzi] n squared. >> Huelekea kuwa n squared.

535
00:44:22,260 --> 00:44:25,280
Kama una triple nested, huelekea kuwa n cubed.

536
00:44:25,280 --> 00:44:29,330
Hivyo kwamba aina ya kitu unapaswa kuwa na uwezo wa kumweka nje mara moja.

537
00:44:29,330 --> 00:44:33,890
Unahitaji kujua insertion aina na aina ya Bubble na kuunganisha aina na wale wote.

538
00:44:33,890 --> 00:44:41,420
Ni rahisi kuelewa kwa nini wao ni wale n squared na n logi n na yote ya kwamba

539
00:44:41,420 --> 00:44:47,810
sababu nadhani kulikuwa na juu ya chemsha bongo ya mwaka mmoja ambapo sisi kimsingi alitoa wewe

540
00:44:47,810 --> 00:44:55,050
utekelezaji wa aina Bubble na akasema, "Ni wakati wa mbio kazi hii?"

541
00:44:55,050 --> 00:45:01,020
Hivyo kama wewe kutambua kama aina Bubble, basi unaweza kusema mara moja n squared.

542
00:45:01,020 --> 00:45:05,470
Lakini kama wewe tu kuangalia ni, huna hata haja ya kutambua Bubble ni aina;

543
00:45:05,470 --> 00:45:08,990
unaweza tu kusema hii ni kufanya hii na hii. Hii ni n squared.

544
00:45:12,350 --> 00:45:14,710
[Mwanafunzi] Je, kuna mifano mgumu unaweza kuja na,

545
00:45:14,710 --> 00:45:20,370
kama wazo sawa ya kuhesabia nje?

546
00:45:20,370 --> 00:45:24,450
>> Sidhani tunataka kuwapa mifano yoyote mgumu.

547
00:45:24,450 --> 00:45:30,180
aina Bubble kitu ni kuhusu ngumu kama tunataka kwenda,

548
00:45:30,180 --> 00:45:36,280
na hata kuwa, kwa muda mrefu kama wewe kuelewa kwamba wewe ni juu ya safu iterating

549
00:45:36,280 --> 00:45:41,670
kwa kila kipengele katika safu, ambayo ni kwenda kuwa kitu ambacho n squared.

550
00:45:45,370 --> 00:45:49,940
Kuna jumla, maswali kama haki hapa tuna - Oh.

551
00:45:55,290 --> 00:45:58,530
Tu siku nyingine, Doug alidai, "nimeizua algorithm kwamba unaweza aina safu

552
00:45:58,530 --> 00:46:01,780
"Ya idadi n katika O (logi n) wakati!"

553
00:46:01,780 --> 00:46:04,900
Hivyo ni jinsi gani tunajua kwamba ni vigumu?

554
00:46:04,900 --> 00:46:08,850
[Inaudible mwanafunzi majibu] >> Yeah.

555
00:46:08,850 --> 00:46:13,710
Kwa uchache sana, una kugusa kila kipengele katika safu,

556
00:46:13,710 --> 00:46:16,210
hivyo ni vigumu kutatua safu ya -

557
00:46:16,210 --> 00:46:20,850
Kama kila kitu ni kwa utaratibu zisizochambuliwa, basi wewe ni kwenda kuwa kugusa kila kitu katika safu,

558
00:46:20,850 --> 00:46:25,320
hivyo ni vigumu kufanya hivyo katika chini ya O ya n.

559
00:46:27,430 --> 00:46:30,340
[Mwanafunzi] Wewe alituonyesha kuwa mfano wa kuwa na uwezo wa kufanya hivyo katika O ya n

560
00:46:30,340 --> 00:46:33,920
kama matumizi mengi ya kumbukumbu. >> Yeah.

561
00:46:33,920 --> 00:46:37,970
Na that's - mimi kusahau kile that's - Je, ni aina ya kuhesabu?

562
00:46:47,360 --> 00:46:51,330
Hmm. Hiyo ni integer kuchagua algorithm.

563
00:46:59,850 --> 00:47:05,100
Mimi nilikuwa kuangalia kwa jina maalum kwa ajili ya hili kwamba mimi hakuweza kukumbuka wiki iliyopita.

564
00:47:05,100 --> 00:47:13,000
Yeah. Hizi ni aina ya aina yake kuwa anaweza kukamilisha mambo katika kubwa O ya n.

565
00:47:13,000 --> 00:47:18,430
Lakini kuna mapungufu, kama unaweza tu kutumia integers hadi idadi fulani.

566
00:47:20,870 --> 00:47:24,560
Plus kama wewe ni kujaribu kutatua that's kitu -

567
00:47:24,560 --> 00:47:30,750
Kama safu yako ni 012, -12, 151, milioni 4,

568
00:47:30,750 --> 00:47:35,120
basi kwamba kipengele kimoja kwenda kabisa uharibifu wa kutenganisha nzima.

569
00:47:42,060 --> 00:47:44,030
>> Maswali?

570
00:47:49,480 --> 00:47:58,870
[Mwanafunzi] Kama una kazi ya kujirudia na hivyo hufanya tu wito kujirudia

571
00:47:58,870 --> 00:48:02,230
ndani ya taarifa ya kurudi, hiyo ni mkia kujirudia,

572
00:48:02,230 --> 00:48:07,360
na hivyo ingekuwa kwamba si kutumia zaidi ya kumbukumbu wakati wa Runtime

573
00:48:07,360 --> 00:48:12,550
au ingekuwa angalau kutumia kumbukumbu kulinganishwa kama iterative suluhisho?

574
00:48:12,550 --> 00:48:14,530
[Bowden] Ndiyo.

575
00:48:14,530 --> 00:48:19,840
Itakuwa uwezekano kuwa fulani polepole, lakini si kweli.

576
00:48:19,840 --> 00:48:23,290
Mkia kujirudia ni nzuri sana.

577
00:48:23,290 --> 00:48:32,640
Kuangalia tena katika muafaka stack, hebu sema tuna kuu

578
00:48:32,640 --> 00:48:42,920
na tuna bar int (int x) au kitu.

579
00:48:42,920 --> 00:48:52,310
Hii si kamilifu kujirudia kazi, lakini kurudi bar (x - 1).

580
00:48:52,310 --> 00:48:57,620
Hivyo ni wazi, hii ni dosari. Unahitaji kesi ya msingi na mambo ya ajabu.

581
00:48:57,620 --> 00:49:00,360
Lakini wazo hapa ni kwamba hii ni mkia kujirudia,

582
00:49:00,360 --> 00:49:06,020
ambayo ina maana wakati wito kuu bar ni kwenda kupata stack yake frame.

583
00:49:09,550 --> 00:49:12,440
Katika sura hii kuna stack kwenda kuwa block kidogo ya kumbukumbu

584
00:49:12,440 --> 00:49:17,490
kwamba sambamba na hoja x wake.

585
00:49:17,490 --> 00:49:25,840
Na hivyo hebu sema kuu kinachotokea kuwaita bar (100);

586
00:49:25,840 --> 00:49:30,050
Hivyo x ni kwenda kuanza nje kama 100.

587
00:49:30,050 --> 00:49:35,660
Kama compiler inatambua kuwa hii ni mkia kujirudia kazi,

588
00:49:35,660 --> 00:49:38,540
basi wakati bar hufanya wito wake kujirudia kwa bar,

589
00:49:38,540 --> 00:49:45,490
badala ya kufanya mpya stack frame, ambayo ni ambapo stack kuanza kukua kwa kiasi kikubwa,

590
00:49:45,490 --> 00:49:48,220
hatimaye itakuwa kukimbia katika kifusi na kisha kupata segfaults

591
00:49:48,220 --> 00:49:51,590
kwa sababu kumbukumbu kuanza colliding.

592
00:49:51,590 --> 00:49:54,830
>> Hivyo badala ya kufanya stack yake mwenyewe frame, inaweza kutambua,

593
00:49:54,830 --> 00:49:59,080
hey, mimi kwa kweli kamwe haja ya kurudi tena katika sura hii stack,

594
00:49:59,080 --> 00:50:08,040
hivyo badala mimi itabidi kuchukua nafasi ya hoja hii kwa 99 na kisha kuanza bar kote.

595
00:50:08,040 --> 00:50:11,810
Na basi itakuwa kufanya hivyo tena na itafikia kurudi bar (x - 1),

596
00:50:11,810 --> 00:50:17,320
na badala ya kufanya mpya stack frame, itakuwa tu nafasi ya hoja yake ya sasa na 98

597
00:50:17,320 --> 00:50:20,740
na kisha kuruka nyuma mwanzo sana ya bar.

598
00:50:23,860 --> 00:50:30,430
Those shughuli, kuondoa kwamba thamani 1 juu ya stack na kuruka nyuma mwanzo,

599
00:50:30,430 --> 00:50:32,430
ni pretty ufanisi.

600
00:50:32,430 --> 00:50:41,500
Hivyo si tu kuwa hii ni sawa kumbukumbu ya matumizi kama kazi tofauti ambayo ni iterative

601
00:50:41,500 --> 00:50:45,390
kwa sababu wewe ni tu kwa kutumia 1 stack frame, lakini wewe si mateso downsides

602
00:50:45,390 --> 00:50:47,240
ya kuwa na kuwaita kazi.

603
00:50:47,240 --> 00:50:50,240
Wito kazi wanaweza kuwa fulani ghali kwa sababu ina nini kuanzisha hii yote

604
00:50:50,240 --> 00:50:52,470
na teardown na mambo haya yote.

605
00:50:52,470 --> 00:50:58,160
Hivyo hii recursion mkia ni nzuri.

606
00:50:58,160 --> 00:51:01,170
[Mwanafunzi] Kwa nini ni si kujenga hatua mpya?

607
00:51:01,170 --> 00:51:02,980
Kwa sababu anatambua haina haja.

608
00:51:02,980 --> 00:51:07,800
wito kwa bar ni tu kurudi wito kujirudia.

609
00:51:07,800 --> 00:51:12,220
Hivyo hana haja ya kufanya chochote kwa thamani ya kurudi.

610
00:51:12,220 --> 00:51:15,120
Ni tu kwenda mara moja kurudi.

611
00:51:15,120 --> 00:51:20,530
Hivyo tu kwenda kuchukua nafasi ya hoja yake mwenyewe na kuanza juu.

612
00:51:20,530 --> 00:51:25,780
Na pia, kama huna mkia kujirudia version,

613
00:51:25,780 --> 00:51:31,460
basi kupata hizi baa zote ambapo wakati bar hili hurejea

614
00:51:31,460 --> 00:51:36,010
ina kurudi thamani yake kwa hii moja, basi hiyo bar mara moja anarudi

615
00:51:36,010 --> 00:51:39,620
na kuirudisha thamani yake kwa moja hii, basi tu kwenda mara moja kurudi

616
00:51:39,620 --> 00:51:41,350
na kurudi thamani yake kwa moja hii.

617
00:51:41,350 --> 00:51:45,350
Basi, wewe ni kuokoa hii popping mambo yote haya mbali ya stack

618
00:51:45,350 --> 00:51:48,730
tangu thamani ya kurudi ni kwenda tu kuwa kupita njia yote nyuma hadi anyway.

619
00:51:48,730 --> 00:51:55,400
Hivyo kwa nini siyo tu nafasi ya hoja yetu na hoja updated na kuanza juu?

620
00:51:57,460 --> 00:52:01,150
Kama kazi ni si mkia kujirudia, kama wewe kufanya kitu kama -

621
00:52:01,150 --> 00:52:07,530
[Mwanafunzi] kama bar (x + 1). >> Yeah.

622
00:52:07,530 --> 00:52:11,770
>> Hivyo kama wewe kuiweka katika hali, basi wewe ni kufanya kitu kwa thamani ya kurudi.

623
00:52:11,770 --> 00:52:16,260
Au hata kama wewe tu kufanya kurudi 2 * bar (x - 1).

624
00:52:16,260 --> 00:52:23,560
Hivyo sasa bar (x - 1) anahitaji kurudi katika utaratibu kwa ajili yake kwa mahesabu ya mara 2 kuwa thamani,

625
00:52:23,560 --> 00:52:26,140
hivyo sasa haina haja zake tofauti mwenyewe stack frame,

626
00:52:26,140 --> 00:52:31,180
na sasa, hakuna jambo jinsi ngumu wewe jaribu, wewe ni kwenda haja ya -

627
00:52:31,180 --> 00:52:34,410
Hii si mkia kujirudia.

628
00:52:34,410 --> 00:52:37,590
[Mwanafunzi] Je, mimi kujaribu kuleta recursion kwa lengo recursion mkia -

629
00:52:37,590 --> 00:52:41,450
[Bowden] Katika ulimwengu halisi, lakini katika CS50 wewe huna.

630
00:52:43,780 --> 00:52:49,280
Ili kupata mkia recursion, kwa ujumla, wewe kuanzisha hoja ya ziada

631
00:52:49,280 --> 00:52:53,550
ambapo bar itachukua x int ndani ya y

632
00:52:53,550 --> 00:52:56,990
na y sambamba na kitu mwisho unataka kurudi.

633
00:52:56,990 --> 00:53:03,650
Hivyo basi hii utaenda tunarudi bar (x - 1), 2 * y.

634
00:53:03,650 --> 00:53:09,810
Hivyo kwamba tu wa ngazi za juu jinsi kubadilisha mambo kuwa mkia kujirudia.

635
00:53:09,810 --> 00:53:13,790
Lakini hoja ya ziada -

636
00:53:13,790 --> 00:53:17,410
Na kisha katika mwisho wakati wewe kufikia msingi wako kesi, wewe tu kurudi y

637
00:53:17,410 --> 00:53:22,740
kwa sababu tumekuwa kukusanya muda wote thamani ya kurudi kwamba unataka.

638
00:53:22,740 --> 00:53:27,280
Wewe aina ya wamekuwa wakifanya hivyo iteratively lakini kutumia calls kujirudia.

639
00:53:32,510 --> 00:53:34,900
Maswali?

640
00:53:34,900 --> 00:53:39,890
[Mwanafunzi] Labda kuhusu hesabu pointer, kama wakati wa kutumia masharti. >> Uhakika.

641
00:53:39,890 --> 00:53:43,610
Pointer hesabu.

642
00:53:43,610 --> 00:53:48,440
Wakati wa kutumia masharti ni rahisi kwa sababu ya masharti ni Char nyota,

643
00:53:48,440 --> 00:53:51,860
chars ni milele na daima Byte moja,

644
00:53:51,860 --> 00:53:57,540
na hivyo pointer arithmetic ni sawa na hesabu ya mara kwa mara wakati wewe ni kushughulika na masharti.

645
00:53:57,540 --> 00:54:08,790
Hebu tu kusema Char * s = "hello".

646
00:54:08,790 --> 00:54:11,430
Hivyo tuna block katika kumbukumbu.

647
00:54:19,490 --> 00:54:22,380
Inahitaji ka 6 kwa sababu wewe daima haja Terminator null.

648
00:54:22,380 --> 00:54:28,620
Na Char * s inaenda kwa mwanzo wa safu hii.

649
00:54:28,620 --> 00:54:32,830
Hivyo anasema kuna s.

650
00:54:32,830 --> 00:54:36,710
Sasa, hii ni kimsingi jinsi safu yoyote kazi,

651
00:54:36,710 --> 00:54:40,780
bila kujali kama ni kurudi kwa malloc au kama ni juu ya stack.

652
00:54:40,780 --> 00:54:47,110
Safu yoyote kimsingi ni pointer mwanzo wa safu,

653
00:54:47,110 --> 00:54:53,640
na kisha yoyote operesheni safu, yoyote Indexing, ni kwenda tu katika safu kwamba baadhi ya kukabiliana.

654
00:54:53,640 --> 00:55:05,360
>> Hivyo wakati mimi kusema kitu kama s [3]; hii ni kwenda s na kuhesabu chars 3 in

655
00:55:05,360 --> 00:55:12,490
Hivyo s [3], tuna 0, 1, 2, 3, hivyo s [3] ni kwenda kwa kutaja l hii.

656
00:55:12,490 --> 00:55:20,460
[Mwanafunzi] Na sisi inaweza kufikia thamani sawa kwa kufanya s + 3 na kisha mabano nyota?

657
00:55:20,460 --> 00:55:22,570
Ndiyo.

658
00:55:22,570 --> 00:55:26,010
Hii ni sawa na * (s + 3);

659
00:55:26,010 --> 00:55:31,240
na kwamba ni ya milele na daima sawa bila kujali wewe kufanya.

660
00:55:31,240 --> 00:55:34,070
Kamwe haja ya kutumia syntax bracket.

661
00:55:34,070 --> 00:55:37,770
Daima unaweza kutumia * (s + 3) syntax.

662
00:55:37,770 --> 00:55:40,180
Watu huvaa kama syntax bracket, ingawa.

663
00:55:40,180 --> 00:55:43,860
[Mwanafunzi] Basi wote arrays ni kweli tu kuyatumia.

664
00:55:43,860 --> 00:55:53,630
Kuna tofauti kidogo wakati mimi kusema int x [4]; >> [mwanafunzi] Je, hiyo kujenga kumbukumbu?

665
00:55:53,630 --> 00:56:03,320
[Bowden] Hayo ni kwenda kujenga ints 4 juu ya stack, hivyo 16 ka ujumla.

666
00:56:03,320 --> 00:56:05,700
Ni kwenda kujenga ka 16 juu ya stack.

667
00:56:05,700 --> 00:56:09,190
x si kuhifadhiwa mahali popote.

668
00:56:09,190 --> 00:56:13,420
Ni haki ya alama akimaanisha mwanzo wa kitu.

669
00:56:13,420 --> 00:56:17,680
Kwa sababu wewe alitangaza safu ndani ya kazi hii,

670
00:56:17,680 --> 00:56:22,340
nini compiler anaenda kufanya ni tu nafasi matukio yote ya x variable

671
00:56:22,340 --> 00:56:26,400
na ambapo ilivyotokea kuchagua kuweka ka hizi 16.

672
00:56:26,400 --> 00:56:30,040
Ni hawezi kufanya hivyo kwa Char * ni kwa sababu ya ni pointer halisi.

673
00:56:30,040 --> 00:56:32,380
Ni bure na kisha kumweka kwa mambo mengine.

674
00:56:32,380 --> 00:56:36,140
x ni mara kwa mara. Huwezi kuwa ni hatua ya safu mbalimbali. >> [Mwanafunzi] Sawa.

675
00:56:36,140 --> 00:56:43,420
Lakini wazo hili, hii Indexing, ni sawa bila kujali kama ni safu ya jadi

676
00:56:43,420 --> 00:56:48,230
au kama ni pointer kitu au kama ni pointer safu malloced.

677
00:56:48,230 --> 00:56:59,770
Na kwa kweli, ni sawa hivyo kwamba pia ni kitu kimoja.

678
00:56:59,770 --> 00:57:05,440
Ni kweli inasababisha tu nini ndani ya mabano na nini kushoto ya mabano,

679
00:57:05,440 --> 00:57:07,970
anaongeza wote pamoja, na dereferences.

680
00:57:07,970 --> 00:57:14,710
Hivyo hii ni halali kama * (s + 3) au s [3].

681
00:57:16,210 --> 00:57:22,090
[Mwanafunzi] Je, unaweza kuwa na kuyatumia akizungumzia arrays 2-dimensional?

682
00:57:22,090 --> 00:57:27,380
>> Ni vigumu. Kijadi, hakuna.

683
00:57:27,380 --> 00:57:34,720
Safu 2-dimensional ni safu 1-dimensional na syntax baadhi rahisi

684
00:57:34,720 --> 00:57:54,110
kwa sababu wakati mimi kusema int x [3] [3], hii ni kweli tu 1 safu na maadili 9.

685
00:57:55,500 --> 00:58:03,000
Na hivyo wakati mimi index, compiler anajua nini namaanisha.

686
00:58:03,000 --> 00:58:13,090
Nikisema x [1] [2], anajua mimi nataka kwenda mstari wa pili, hivyo ni kwenda ruka 3 ya kwanza,

687
00:58:13,090 --> 00:58:17,460
na kisha anataka Jambo la pili kwa kuwa, hivyo ni kwenda kupata hii moja.

688
00:58:17,460 --> 00:58:20,480
Lakini bado tu safu moja-dimensional.

689
00:58:20,480 --> 00:58:23,660
Na hivyo kama nilitaka hawawajui pointer safu kwamba,

690
00:58:23,660 --> 00:58:29,770
Naweza kusema int * p = x;

691
00:58:29,770 --> 00:58:33,220
aina ya x ni tu -

692
00:58:33,220 --> 00:58:38,280
Ni mbaya akisema aina ya x kwa vile ni ishara tu na si kutofautiana halisi,

693
00:58:38,280 --> 00:58:40,140
lakini ni tu * int.

694
00:58:40,140 --> 00:58:44,840
x ni tu pointer mwanzo wa hii. >> [Mwanafunzi] Sawa.

695
00:58:44,840 --> 00:58:52,560
Na hivyo mimi si kuwa na uwezo wa kupata [1] [2].

696
00:58:52,560 --> 00:58:58,370
Nadhani kuna syntax maalum kwa ajili ya kutangaza pointer,

697
00:58:58,370 --> 00:59:12,480
kitu ridiculous kama int (* p [-. kitu ujinga kabisa mimi hata sijui.

698
00:59:12,480 --> 00:59:17,090
Lakini kuna syntax kwa ajili ya kutangaza kuyatumia kama kwa mabano na mambo.

699
00:59:17,090 --> 00:59:22,960
Inaweza hata basi wewe kufanya hivyo.

700
00:59:22,960 --> 00:59:26,640
Mimi naweza kuangalia nyuma katika kitu ambacho ingekuwa kuniambia ukweli.

701
00:59:26,640 --> 00:59:34,160
Nami kuangalia kwa hilo baadaye, kama kuna syntax kwa uhakika. Lakini kamwe kuona.

702
00:59:34,160 --> 00:59:39,670
Na hata hivyo ni syntax kizamani kwamba kama matumizi yake, watu watakuwa baffled.

703
00:59:39,670 --> 00:59:43,540
Arrays multidimensional ni pretty nadra kama ni.

704
00:59:43,540 --> 00:59:44,630
Wewe pretty much -

705
00:59:44,630 --> 00:59:48,490
Naam, kama wewe ni kufanya mambo tumbo si kwenda kuwa adimu,

706
00:59:48,490 --> 00:59:56,730
lakini katika C wewe ni mara chache kwenda kuwa na kutumia arrays multidimensional.

707
00:59:57,630 --> 01:00:00,470
Yeah. >> [Mwanafunzi] Hebu sema una safu kweli kwa muda mrefu.

708
01:00:00,470 --> 01:00:03,900
>> Hivyo katika kumbukumbu virtual ingekuwa kuonekana kuwa yote mfululizo,

709
01:00:03,900 --> 01:00:05,640
kama vipengele haki ya karibu na kila mmoja,

710
01:00:05,640 --> 01:00:08,770
lakini katika kumbukumbu ya kimwili, ingekuwa inawezekana kwa kuwa kwa kuwa wameigawanya? >> Ndiyo.

711
01:00:08,770 --> 01:00:16,860
Jinsi virtual kumbukumbu matendo ni kama tu anavyotenganisha -

712
01:00:19,220 --> 01:00:24,860
kitengo cha mgao ni ukurasa, ambayo huelekea kuwa 4 kilobytes,

713
01:00:24,860 --> 01:00:29,680
na hivyo wakati wa mchakato anasema, hey, nataka kutumia kumbukumbu hii,

714
01:00:29,680 --> 01:00:35,970
mfumo wa uendeshaji ni kwenda kuipa kilobytes 4 Kitalu kwamba kidogo ya kumbukumbu.

715
01:00:35,970 --> 01:00:39,100
Hata kama wewe tu kutumia moja kidogo Byte katika block nzima ya kumbukumbu,

716
01:00:39,100 --> 01:00:42,850
mfumo wa uendeshaji ni kwenda kuwapa kamili kilobytes 4.

717
01:00:42,850 --> 01:00:49,410
Basi nini maana ya hii ni mimi naweza kuwa - wacha kusema hii ni stack yangu.

718
01:00:49,410 --> 01:00:53,180
Stack Hii inaweza kutengwa. Stack yangu inaweza kuwa megabaiti na megabaiti.

719
01:00:53,180 --> 01:00:55,020
Stack yangu inaweza kuwa kubwa.

720
01:00:55,020 --> 01:01:00,220
Lakini stack yenyewe ina kuwa umegawanyika katika kurasa binafsi,

721
01:01:00,220 --> 01:01:09,010
ambayo kama sisi kuangalia juu hapa hebu kusema hii ni RAM yetu,

722
01:01:09,010 --> 01:01:16,600
kama nina 2 gigabytes ya RAM, hii ni halisi anuani 0 0 Byte kama ya RAM yangu,

723
01:01:16,600 --> 01:01:22,210
na hii ni 2 gigabytes njia yote chini hapa.

724
01:01:22,210 --> 01:01:27,230
Hivyo ukurasa huu ili yanahusiana na block hii zaidi hapa.

725
01:01:27,230 --> 01:01:29,400
Ukurasa Hii inaweza yanahusiana na block hii zaidi hapa.

726
01:01:29,400 --> 01:01:31,560
Moja hii inaweza yanahusiana na hii moja zaidi ya hapa.

727
01:01:31,560 --> 01:01:35,540
Hivyo mfumo wa uendeshaji ni bure hawawajui kumbukumbu kimwili

728
01:01:35,540 --> 01:01:39,320
yoyote ukurasa binafsi kiholela.

729
01:01:39,320 --> 01:01:46,180
Na hiyo ina maana kwamba kama mpaka hii hutokea kwa magamaga safu,

730
01:01:46,180 --> 01:01:50,070
safu kinachotokea kwa kuwa kushoto ya hii na haki ya utaratibu huu wa ukurasa,

731
01:01:50,070 --> 01:01:54,460
kisha safu kwamba ni kwenda kuwa umegawanyika katika kumbukumbu ya kimwili.

732
01:01:54,460 --> 01:01:59,280
Na kisha wakati wewe kuacha mpango, wakati mchakato kumalizika,

733
01:01:59,280 --> 01:02:05,690
haya upangaji kupata erased na basi ni huru kutumia vitalu hizi kidogo kwa ajili ya mambo mengine.

734
01:02:14,730 --> 01:02:17,410
Zaidi maswali?

735
01:02:17,410 --> 01:02:19,960
[Mwanafunzi] arithmetic pointer. >> Oh yeah.

736
01:02:19,960 --> 01:02:28,410
Strings yalikuwa rahisi, lakini kuangalia kitu kama ints,

737
01:02:28,410 --> 01:02:35,000
hivyo nyuma int x [4];

738
01:02:35,000 --> 01:02:41,810
Kama hii ni safu au kama ni pointer safu malloced ya integers 4,

739
01:02:41,810 --> 01:02:47,060
itakavyo kutibiwa njia sawa.

740
01:02:50,590 --> 01:02:53,340
[Mwanafunzi] Basi ni juu ya arrays mirundi?

741
01:03:01,400 --> 01:03:05,270
[Bowden] Arrays ni si juu ya lundo. >> [Mwanafunzi] Oh.

742
01:03:05,270 --> 01:03:08,320
>> [Bowden] Aina hii ya safu huelekea kuwa juu ya stack

743
01:03:08,320 --> 01:03:12,220
isipokuwa wewe amekiri kuwa ni saa - kupuuza vigezo kimataifa. Usitumie vigezo kimataifa.

744
01:03:12,220 --> 01:03:16,280
Ndani ya kazi nasema int x [4];

745
01:03:16,280 --> 01:03:22,520
Ni kwenda kujenga block 4-integer juu ya stack kwa safu hii.

746
01:03:22,520 --> 01:03:26,960
Lakini hii malloc (4 * sizeof (int)); yataenda juu ya lundo.

747
01:03:26,960 --> 01:03:31,870
Lakini baada ya hatua hii naweza kutumia x na p katika pretty much njia hiyo hiyo,

748
01:03:31,870 --> 01:03:36,140
nyingine kuliko isipokuwa nilivyosema kabla kuhusu unaweza reassign p.

749
01:03:36,140 --> 01:03:40,960
Kitaalam, ukubwa wao ni tofauti kwa kiasi fulani, lakini hiyo ni kabisa lisilo na maana.

750
01:03:40,960 --> 01:03:43,310
Wewe kweli kamwe kutumia ukubwa wao.

751
01:03:48,020 --> 01:03:56,810
p mimi naweza kusema p [3] = 2; au x [3] = 2;

752
01:03:56,810 --> 01:03:59,680
Unaweza kutumia yao katika njia hasa sawa.

753
01:03:59,680 --> 01:04:01,570
Hivyo pointer arithmetic sasa - Ndiyo.

754
01:04:01,570 --> 01:04:07,390
[Mwanafunzi] Je kufanya p * kama una mabano?

755
01:04:07,390 --> 01:04:11,720
mabano ni dereference thabiti. >> Sawa.

756
01:04:11,720 --> 01:04:20,200
Kweli, pia unachosema na unaweza kupata arrays multidimensional

757
01:04:20,200 --> 01:05:02,650
na kuyatumia, nini unaweza kufanya ni kitu kama, hebu sema, int ** pp = malloc (sizeof (int *) * 5);

758
01:05:02,650 --> 01:05:06,900
Mimi itabidi kuandika yote nje ya kwanza.

759
01:05:37,880 --> 01:05:41,020
Mimi sitaki kuwa moja.

760
01:05:41,020 --> 01:05:42,550
Sawa.

761
01:05:42,550 --> 01:05:48,910
Nini mimi hapa ni - Kwamba lazima pp [i].

762
01:05:48,910 --> 01:05:53,680
Hivyo pp ni pointer pointer.

763
01:05:53,680 --> 01:06:02,420
Wewe mallocing pp kwa uhakika na safu ya nyota 5 int.

764
01:06:02,420 --> 01:06:10,950
Hivyo katika kumbukumbu una juu ya pp stack

765
01:06:10,950 --> 01:06:20,150
Ni kwenda na kumweka kwa safu ya vitalu 5 ambao wote ni wenyewe kuyatumia.

766
01:06:20,150 --> 01:06:28,210
Na kisha wakati mimi malloc chini hapa, mimi malloc kwamba kila mtu binafsi ya kuyatumia those

767
01:06:28,210 --> 01:06:32,080
wanapaswa kumweka kuzuia tofauti ya ka 4 juu ya lundo.

768
01:06:32,080 --> 01:06:35,870
Hivyo hii pointi ka 4.

769
01:06:37,940 --> 01:06:40,660
Na hii pointi moja kwa ka mbalimbali 4.

770
01:06:40,660 --> 01:06:43,200
>> Na wote wanaongelea ka yao wenyewe 4.

771
01:06:43,200 --> 01:06:49,080
Hii inanipa njia ya kufanya mambo multidimensional.

772
01:06:49,080 --> 01:06:58,030
Mimi naweza kusema pp [3] [4], lakini sasa hii si kitu sawa kama arrays multidimensional

773
01:06:58,030 --> 01:07:05,390
kwa sababu arrays multidimensional ni kutafsiriwa [3] [4] ndani ya moja katika kukabiliana na safu x.

774
01:07:05,390 --> 01:07:14,790
Hii p dereferences, grossistledet index ya tatu, kisha dereferences kwamba

775
01:07:14,790 --> 01:07:20,790
na grossistledet - 4 itakuwa batili - index pili.

776
01:07:24,770 --> 01:07:31,430
Ambapo wakati tulikuwa int x [3] [4] kabla kama safu multidimensional

777
01:07:31,430 --> 01:07:35,740
na wakati wewe mara mbili bracket ni kweli tu dereference moja,

778
01:07:35,740 --> 01:07:40,490
wewe ni kufuatia pointer moja na kisha kukabiliana,

779
01:07:40,490 --> 01:07:42,850
hii ni kweli 2D rejea.

780
01:07:42,850 --> 01:07:45,840
Wewe kufuata 2 kuyatumia tofauti.

781
01:07:45,840 --> 01:07:50,420
Hivyo hii pia kitaalam inaruhusu kuwa na arrays multidimensional

782
01:07:50,420 --> 01:07:53,550
ambapo kila mtu binafsi ni safu ukubwa tofauti.

783
01:07:53,550 --> 01:07:58,000
Hivyo nadhani jagged arrays multidimensional ni nini ni wito

784
01:07:58,000 --> 01:08:01,870
tangu kweli jambo la kwanza inaweza kumweka kwa kitu ambayo ina vipengele 10,

785
01:08:01,870 --> 01:08:05,540
Jambo la pili inaweza kumweka kwa kitu ambayo ina vipengele 100.

786
01:08:05,540 --> 01:08:10,790
[Mwanafunzi] Je, kuna kikomo na idadi ya kuyatumia unaweza kuwa

787
01:08:10,790 --> 01:08:14,290
akizungumzia kuyatumia nyingine? >> No

788
01:08:14,290 --> 01:08:17,010
Unaweza kuwa na int ***** p.

789
01:08:18,050 --> 01:08:23,760
Rudi arithmetic pointer - >> [mwanafunzi] Oh. >> Yeah.

790
01:08:23,760 --> 01:08:35,649
[Mwanafunzi] Kama mimi na int *** p na kisha mimi kufanya dereferencing na mimi kusema p * ni sawa na thamani hii,

791
01:08:35,649 --> 01:08:39,560
ni hayo tu kwenda kufanya 1 ngazi ya dereferencing? >> Ndiyo.

792
01:08:39,560 --> 01:08:43,340
Hivyo kama nataka kupata kitu kwamba pointer mwisho akionyesha -

793
01:08:43,340 --> 01:08:46,210
Kisha wewe kufanya *** p. >> Sawa.

794
01:08:46,210 --> 01:08:54,080
Hivyo hii ni p pointi 1 block, pointi ya kuzuia mwingine, pointi ya kuzuia mwingine.

795
01:08:54,080 --> 01:09:02,010
Kisha kama wewe kufanya * p = kitu kingine, basi wewe ni kubadilisha huu

796
01:09:02,010 --> 01:09:13,640
kwa sasa kumweka kuzuia tofauti. >> Sawa.

797
01:09:13,640 --> 01:09:17,649
>> [Bowden] Na kama hawa walikuwa malloced, basi una sasa kuvuja kumbukumbu

798
01:09:17,649 --> 01:09:20,430
isipokuwa wewe kutokea kwa kuwa na references tofauti ya hizi

799
01:09:20,430 --> 01:09:25,270
tangu huwezi kupata nyuma wale walio kuwa wewe tu akatupa mbali.

800
01:09:25,270 --> 01:09:29,550
Pointer hesabu.

801
01:09:29,550 --> 01:09:36,310
int x [4]; ni kwenda kutenga safu ya integers 4

802
01:09:36,310 --> 01:09:40,670
ambapo x inaenda kwa mwanzo wa safu.

803
01:09:40,670 --> 01:09:50,420
Hivyo wakati mimi kusema kitu kama x [1]; nataka kumaanisha kwenda integer pili katika safu,

804
01:09:50,420 --> 01:09:53,319
ambayo itakuwa ni hii moja.

805
01:09:53,319 --> 01:10:04,190
Lakini kwa kweli, hilo ni 4 ka katika safu hii tangu integer inachukua hadi ka 4.

806
01:10:04,190 --> 01:10:08,470
Hivyo kukabiliana ya 1 kweli maana yake kukabiliana ya 1

807
01:10:08,470 --> 01:10:12,030
ukubwa wa mara kila aina ya safu ni.

808
01:10:12,030 --> 01:10:17,170
Hii ni safu ya integers, hivyo anajua kufanya 1 ukubwa wa mara int wakati anataka kukabiliana.

809
01:10:17,170 --> 01:10:25,260
syntax nyingine. Kumbuka kuwa hii ni sawa na * (x + 1);

810
01:10:25,260 --> 01:10:35,250
Wakati mimi kusema pointer + 1, nini kwamba anarudi ni anuani ya kwamba pointer ni hifadhi ya

811
01:10:35,250 --> 01:10:40,360
plus 1 ukubwa wa mara ya aina ya pointer.

812
01:10:40,360 --> 01:10:59,510
Hivyo kama x = ox100, basi x + 1 = ox104.

813
01:10:59,510 --> 01:11:19,750
Na unaweza vibaya hii na kusema kitu kama Char * c = (Char *) x;

814
01:11:19,750 --> 01:11:23,050
na sasa c ni kwenda kuwa anuani hiyo kama x.

815
01:11:23,050 --> 01:11:26,040
c ni kwenda kuwa sawa na ox100,

816
01:11:26,040 --> 01:11:31,490
lakini c + 1 ni kwenda kuwa sawa na ox101

817
01:11:31,490 --> 01:11:38,030
tangu pointer arithmetic hutegemea aina ya pointer kwamba wewe ni kuongeza.

818
01:11:38,030 --> 01:11:45,390
Hivyo c + 1, inaangalia c, ni pointer Char, hivyo ni kwenda kuongeza 1 ukubwa wa mara Char,

819
01:11:45,390 --> 01:11:48,110
ambayo ni daima itakuwa 1, ili kupata 101,

820
01:11:48,110 --> 01:11:54,890
lakini kama mimi kufanya x, ambayo pia bado 100, x + 1 ni kwenda kuwa 104.

821
01:11:56,660 --> 01:12:06,340
[Mwanafunzi] Je, unaweza kutumia c + + ili kuendeleza pointer yako na 1?

822
01:12:06,340 --> 01:12:09,810
Ndiyo, unaweza.

823
01:12:09,810 --> 01:12:16,180
Huwezi kufanya hivyo kwa sababu x x ni ishara tu, ni mara kwa mara; huwezi kubadilisha x.

824
01:12:16,180 --> 01:12:22,610
>> Lakini c hutokea tu kuwa pointer, hivyo c + + ni kikamilifu halali na itakuwa na increment 1.

825
01:12:22,610 --> 01:12:32,440
Kama c walikuwa tu * int, basi c + + itakuwa 104.

826
01:12:32,440 --> 01:12:41,250
+ + Gani pointer hesabu tu kama c + 1 ingekuwa kufanyika pointer hesabu.

827
01:12:43,000 --> 01:12:48,870
Hii ni kweli jinsi mambo mengi kama aina kuunganisha -

828
01:12:49,670 --> 01:12:55,710
Badala ya kujenga nakala za mambo, unaweza badala kupita -

829
01:12:55,710 --> 01:13:02,400
Kama kama mimi alitaka kupita hii nusu ya safu - basi s kufuta baadhi ya hii.

830
01:13:04,770 --> 01:13:10,520
Hebu sema mimi alitaka kupita upande huu wa safu ndani ya kazi.

831
01:13:10,520 --> 01:13:12,700
Gani mimi kupita kwa kazi hiyo?

832
01:13:12,700 --> 01:13:17,050
Kama mimi kupita x, mimi kupita anwani hii.

833
01:13:17,050 --> 01:13:23,780
Lakini nataka kupita anwani hii hasa. Basi nini lazima mimi kupita?

834
01:13:23,780 --> 01:13:26,590
[Mwanafunzi] Pointer + 2?

835
01:13:26,590 --> 01:13:29,350
[Bowden] Basi x + 2. Ndiyo.

836
01:13:29,350 --> 01:13:31,620
Hiyo itakuwa anwani hii.

837
01:13:31,620 --> 01:13:42,810
Utapata pia mara nyingi sana kuona kama x [2] na kisha anuani ya kwamba.

838
01:13:42,810 --> 01:13:47,850
Hivyo haja ya kuchukua anuani yake kwa sababu ni bracket dereference thabiti.

839
01:13:47,850 --> 01:13:53,250
x [2] unahusu thamani ya kwamba ni katika sanduku hili, na kisha unataka anuani ya sanduku kwamba,

840
01:13:53,250 --> 01:13:56,850
hivyo unaweza kusema & x [2].

841
01:13:56,850 --> 01:14:02,880
Basi hiyo ni jinsi ya kitu katika aina kuunganisha ambapo unataka kupita orodha nusu kwa kitu

842
01:14:02,880 --> 01:14:08,790
wewe kweli tu kupita & x [2], na sasa kama mbali kama wito kujirudia ni wasiwasi,

843
01:14:08,790 --> 01:14:12,510
safu yangu mpya kuanza huko.

844
01:14:12,510 --> 01:14:15,130
Dakika ya mwisho maswali.

845
01:14:15,130 --> 01:14:20,050
[Mwanafunzi] Kama hatuwezi kuweka au ampersand - nini kwamba uitwao? >> Nyota?

846
01:14:20,050 --> 01:14:23,200
[Mwanafunzi] Star. >> Kitaalam, dereference operator, lakini - >> [mwanafunzi] Dereference.

847
01:14:23,200 --> 01:14:29,310
>> Kama hatuwezi kuweka nyota au ampersand, kile kinachotokea kama mimi tu kusema y = x x na ni pointer?

848
01:14:29,310 --> 01:14:34,620
Je, ni aina ya y? >> [Mwanafunzi] mimi itabidi tu kusema ni pointer 2.

849
01:14:34,620 --> 01:14:38,270
Hivyo kama wewe tu kusema y = x, sasa x na y uhakika na kitu kimoja. >> [Mwanafunzi] Point kwa kitu kimoja.

850
01:14:38,270 --> 01:14:45,180
Na kama x ni pointer int? >> Ingekuwa kulalamika kwa sababu huwezi hawawajui kuyatumia.

851
01:14:45,180 --> 01:14:46,540
[Mwanafunzi] Sawa.

852
01:14:46,540 --> 01:14:51,860
Kumbuka kuyatumia kwamba, hata ingawa sisi kuteka yao kama mishale,

853
01:14:51,860 --> 01:15:02,010
kweli yote ya hazina wao - int * x - kweli x wote ni hifadhi ni kitu kama ox100,

854
01:15:02,010 --> 01:15:06,490
ambayo sisi kutokea kwa kuwakilisha kama akizungumzia kuzuia kuhifadhiwa katika 100.

855
01:15:06,490 --> 01:15:19,660
Hivyo wakati mimi kusema int * y = x; Mimi tu kuiga ox100 ndani ya y,

856
01:15:19,660 --> 01:15:24,630
ambayo sisi ni kwenda tu kuwakilisha kama y, pia akizungumzia ox100.

857
01:15:24,630 --> 01:15:39,810
Na kama nasema int i = (int) x; kisha i ni kwenda kuhifadhi chochote thamani ya ox100 ni

858
01:15:39,810 --> 01:15:45,100
ndani yake, lakini sasa ni kwenda kufasiriwa kama integer badala ya pointer.

859
01:15:45,100 --> 01:15:49,310
Lakini unahitaji kutupwa au pengine itakuwa kulalamika.

860
01:15:49,310 --> 01:15:53,300
[Mwanafunzi] Basi nini maana kuwafukuza -

861
01:15:53,300 --> 01:16:00,290
Je, ni kwenda kuwa akitoa int ya x au akitoa int ya y?

862
01:16:00,290 --> 01:16:03,700
[Bowden] Nini?

863
01:16:03,700 --> 01:16:07,690
[Mwanafunzi] Sawa. Baada ya mabano haya ni kuna kwenda kuwa x au ay huko?

864
01:16:07,690 --> 01:16:11,500
>> [Bowden] Ama. x na y ni sawa. >> [Mwanafunzi] Sawa.

865
01:16:11,500 --> 01:16:14,390
Kwa sababu uko wote kuyatumia. >> Yeah.

866
01:16:14,390 --> 01:16:21,050
[Mwanafunzi] Hivyo itakuwa kuhifadhi 100 hexadesimoli katika fomu integer? >> [Bowden] Yeah.

867
01:16:21,050 --> 01:16:23,620
Lakini si thamani ya chochote ni pointi.

868
01:16:23,620 --> 01:16:29,940
[Bowden] Yeah. >> [Mwanafunzi] Basi tu anuani katika fomu integer. Sawa.

869
01:16:29,940 --> 01:16:34,720
[Bowden] Kama alitaka kwa sababu baadhi ya ajabu,

870
01:16:34,720 --> 01:16:38,900
unaweza peke kukabiliana na kuyatumia na kamwe kukabiliana na integers

871
01:16:38,900 --> 01:16:49,240
na tu kuwa kama int * x = 0.

872
01:16:49,240 --> 01:16:53,000
Kisha utaenda kupata kweli kuchanganyikiwa mara moja pointer arithmetic kuanza kinachotokea.

873
01:16:53,000 --> 01:16:56,570
Hivyo idadi ya kwamba wao ni kuhifadhi maana.

874
01:16:56,570 --> 01:16:58,940
Ni jinsi tu wewe kuishia kutafsiri yao.

875
01:16:58,940 --> 01:17:02,920
Hivyo mimi nina bure nakala ox100 kutoka * int kwa int,

876
01:17:02,920 --> 01:17:07,790
na mimi nina bure hawawajui - you're pengine kwenda kupata yelled saa kwa si akitoa -

877
01:17:07,790 --> 01:17:18,160
Mimi nina bure hawawajui kitu kama (int *) ox1234 katika * huu holela int.

878
01:17:18,160 --> 01:17:25,480
Hivyo ox123 ni tu kama halali anuani kumbukumbu kama ni & y.

879
01:17:25,480 --> 01:17:32,060
& Y kinachotokea kurudi kitu ambacho ni kiasi pretty ox123.

880
01:17:32,060 --> 01:17:35,430
[Mwanafunzi] Laiti kuwa njia ya kweli ya baridi kwenda kutoka hexadesimoli kwa ajili ya decimal,

881
01:17:35,430 --> 01:17:39,230
kama kama una pointer na wewe kuwatupia kama int?

882
01:17:39,230 --> 01:17:44,860
[Bowden] Unaweza kweli tu magazeti kutumia kama printf.

883
01:17:44,860 --> 01:17:50,300
Hebu sema nina int y = 100.

884
01:17:50,300 --> 01:18:02,700
Hivyo printf (% d \ n - kama unapaswa tayari kujua - magazeti kwamba kama integer,% x.

885
01:18:02,700 --> 01:18:05,190
Tutaweza tu magazeti ni kama hexadesimoli.

886
01:18:05,190 --> 01:18:10,760
Hivyo pointer si kuhifadhiwa kama hexadesimoli,

887
01:18:10,760 --> 01:18:12,960
na integer si kuhifadhiwa kama decimal.

888
01:18:12,960 --> 01:18:14,700
Kila kitu ni kuhifadhiwa kama binary.

889
01:18:14,700 --> 01:18:17,950
Ni tu kwamba sisi huwa na kuonyesha kuyatumia kama hexadesimoli

890
01:18:17,950 --> 01:18:23,260
kwa sababu tunafikiri ya mambo haya katika vitalu 4-Byte,

891
01:18:23,260 --> 01:18:25,390
na kumbukumbu anwani huwa na kuwa familiar.

892
01:18:25,390 --> 01:18:28,890
Sisi ni kama, ikiwa ni kuanza na bf, basi hutokea kwa kuwa juu ya stack.

893
01:18:28,890 --> 01:18:35,560
Hivyo ni tu tafsiri yetu ya kuyatumia kama hexadesimoli.

894
01:18:35,560 --> 01:18:39,200
Sawa. Maswali yoyote ya mwisho?

895
01:18:39,200 --> 01:18:41,700
>> Nitakuwa hapa kwa kidogo baada ya kama una kitu kingine chochote.

896
01:18:41,700 --> 01:18:46,070
Na kwamba mwisho wa.

897
01:18:46,070 --> 01:18:48,360
>> [Mwanafunzi] Yay! [Applause]

898
01:18:51,440 --> 01:18:53,000
>> [CS50.TV]