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[Powered by Google Translate] Ejja nitkellmu dwar structs.

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Structs tagħtina mod biex tiġbor mazz ta 'varjabbli flimkien

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fis-pakkett sbieħ.

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Huwa probabbilment aktar faċli li wieħed jara eżempju dritt bogħod,

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hekk aħna ngħidu Struct,

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imbagħad ftuħ Brace kaboċċi,

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u f'dan Struct, aħna ser ikollhom l-età int,

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isem char *,

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u li hu.

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Dan jista 'jidher stramb ma' punto virgola wara Brace kaboċċi,

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iżda huwa fil-fatt meħtieġ mal structs.

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Kwalunkwe tip valida tista 'tmur taħt id-definizzjoni Struct.

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Hawnhekk, aħna ve użat int u * char,

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imma int tista 'wkoll tuża firxa, tal jgħidu, 100 elementi

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jew saħansitra ieħor Struct.

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Meta inti qed tuża structs fis-C,

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int ħolqien ta 'tipi ġodda

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barra ta 'ġbir ta' tipi oħra.

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Hawnhekk, aħna qed jagħmlu tip ġdid

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minn integer u * char.

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Kif Ser naraw aktar tard, tip Struct

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huwa fil-lott ta 'modi ekwivalenti għal kwalunkwe tip ieħor int użati biex.

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Normalment, I ser tkun jitqabblu kif tip Struct

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huwa simili għal tip numru sħiħ.

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Filwaqt li l-kodiċi aħna kiteb huwa C valida,

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mhuwiex utli ħafna,

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u clang se tagħtina twissija.

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Ftakar kif structs u tagħha huma simili?

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Ukoll, aħna bażikament biss qal

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int,

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li ma jkunx konformi utli ħafna.

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Mela ejja attwalment tiddikjara varjabbli ta 'dak it-tip

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billi jagħtuha l-isem qabel il-virgola.

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Aħna ser sejħa-istudent varjabbli.

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Issa aħna ve iddikjarat student imsejħa varjabbli

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mat-tip mogħtija mill-Struct.

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Kif se nilħqu dawn l-varjabbli ġewwa l-Struct?

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Teknikament, l-ismijiet ta 'dawn il-varjabbli

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huma membri.

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Biex ikollok aċċess għal xi membru partikolari fi Struct student,

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inti tehmeż dot għall-isem varjabbli,

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segwit bl-isem tal-membru li trid.

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Allura hawn, l-uniċi 2 possibbiltajiet validi

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huma student.age

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. student.name u

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U nistgħu nagħmlu xi ħaġa simili

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student.age = 12

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u student.name. istudent =

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Issa dak li jekk ridna li tagħmel student tieni?

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Għandek mnejn taħseb li kopja u paste dawn il-linji

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u l-bidla istudenti għall-istudenti 2 jew xi ħaġa,

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u li se jaħdmu,

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iżda teknikament, l-istudenti u l-istudenti 2

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ma jkollhomx l-istess tip.

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Ara, inti mhux ser ikunu jistgħu jassenjaw lill xulxin.

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Dan għaliex, s'issa,

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Struct tiegħek ġie anonimu.

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Għandna bżonn li tagħtiha isem.

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Biex tagħmel dan, aħna daħħal l-isem tal-Struct

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wara l-Struct kelma.

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student,

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segwita mid-definizzjoni.

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Nistgħu xorta immedjatament tiddikjara varjabbli ta 'tip

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Struct student, bħal għamilna qabel.

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Aħna ser sejħa hija S1

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Billi tagħti l-Struct isem,

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nistgħu issa jużaw istudent Struct

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fi kważi bl-istess mod eżatt aħna se jużaw int.

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Allura nistgħu tiddikjara varjabbli ta 'student Struct tip,

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bħal

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S2 student Struct.

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Bħal arrays, structs jipprovdu sintassi inizjalizzazzjoni shortcut,

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hekk nistgħu ngħidu, Struct student S2

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ugwali

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xellug kaboċċi Brace 3, S2.

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Hawnhekk, S2.age se jkun 3,

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u S2.name se jindika S2.

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Think ta 'l-affarijiet inti tista' tagħmel ma 'tip int

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u ħafna minnhom inti tista 'tagħmel ma' tip student Struct.

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Nistgħu nużaw student Struct bħala tip ta 'parametru funzjoni.

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Nistgħu nużaw student Struct ġewwa ta 'Struct ġdid.

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Jista 'jkollna pointer li student Struct.

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Nistgħu nagħmlu daqs ta 'student Struct.

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Struct istudent huwa tip

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bħad int huwa tip.

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Nistgħu wkoll jassenja S1 sa S2

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peress li kemm huma tal-istess tip, sabiex inkunu nistgħu nagħmlu

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S1 =. S2

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X'jiġri jekk nagħmlu

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S1.age = 10?

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Il-bidla S2 fil-livelli kollha?

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Għal darb'oħra, think tal-structs biss bħala interi regolari.

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Jekk aħna tassenja xi X int sa ċertu Y int,

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bħall X = Y

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u mbagħad bidla X,

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bħal fil X + +,

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ma Y bidla fil-livelli kollha?

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Y ma tinbidilx hawn, u għalhekk lanqas ma S2 hawn fuq.

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S2.age għadu 3.

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Iżda jinnota li meta jassenjaw 1 Struct għall-ieħor,

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kollha tal-pointers għadhom jindikaw li l-istess ħaġa,

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peress li dawn kienu biss kkupjati.

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Jekk inti ma tridx l-pointers li tkun maqsuma,

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ikollok bzonn li manwalment jimmaniġġjaw dan,

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forsi billi malicking blokka waħda ta 'memorja għal wieħed mill-indikaturi għall-punt li

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u l-ikkupjar tad-data fuq.

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Jista 'jkun annoying li jkollhom biex jiktbu student Struct kullimkien.

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Permezz ta 'def tip, nistgħu nagħmlu

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Tip def

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Struct

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u aħna ser sejħa hija student.

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Issa, nistgħu nużaw l-istudenti kullimkien

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li aħna użati biex jużaw istudent Struct.

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Dan it-tip huwa def xi Struct anonimu

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u jitlob li l-istudenti.

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Imma jekk aħna wkoll iżommu l-identifikatur istudent

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ħdejn il-Struct kelma, bħal fil student Struct typedef,

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nistgħu jużaw kemm Struct student u student xulxin issa.

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Huma ma jkollhomx li jkollhom l-istess isem.

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Nistgħu tip istudenti Struct def li Bob

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u mbagħad Struct student u Bob

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Ikun tipi interkambjabbli.

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Irrispettivament mill-def tip,

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għandna bżonn l-identifikatur jmiss Struct

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jekk id-definizzjoni ta 'l-Struct

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huwa rikursivi.

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Per eżempju,

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Tip def Struct node

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u se jiġi definit bħala val int

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u din se jkollha pointer li l-punti għall-ieħor node Struct.,

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bħal fil Struct node * jmiss.

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U allura aħna ser sejħa hija node.

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Dan Struct huwa rikursivi,

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peress li d-definizzjoni ta node Struct jkun jinkludi l

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pointer għal node Struct.

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Avviż li rridu ngħidu Struct node * li jmiss

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ġewwa tad-definizzjoni ta 'l-node Struct,

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peress li l-def tip ma jkunx għadu lest biex inessu li tkun simplifikata din

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għal ftit * node. jmiss

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Int ser ikollok jitgħallmu aktar dwar structs simili għal din

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meta jittrattaw mal-listi marbuta u siġar.

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What about structs fil-funzjoni?

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Dan huwa wkoll perfettament valida.

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Aħna jista 'jkollhom

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null funzjonijiet

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li tieħu bħala argument,

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istudent s

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u ma xi ħaġa ma 'dak istudent.

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U allura nistgħu jgħaddu bħala Struct student bħal dan.

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Funzjoni tal S1 minn qabel.

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Il Struct jaġixxi

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eżatt kif integer kien meta għadda għal funzjoni.

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Funzjonijiet tirċievi kopja ta 'S1

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u għalhekk ma jistgħux jimmodifikaw S1;

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pjuttost, biss il-kopja ta 'dan li l-maħżuna S.

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Jekk trid li l-funzjoni li tkun tista 'timmodifika S1,

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funzjonijiet ser ikollok bżonn tieħu * student S,

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u inti ser ikollhom jgħaddu S1 mill-indirizz, bħal dan.

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Student * S, funzjonijiet u S1.

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Hemm raġuni oħra li jgħaddu minn indirizz hawn.

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X'jiġri jekk Struct tagħna jinsab 100 oqsma?

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Kull darba waħda aħna jgħaddu student funzjoni,

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programm tagħna jeħtieġ li kopja kollha ta 'dawk l-oqsma 100 fis S argument funzjonament, il-

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anke jekk qatt ma juża l-maġġoranza l-kbira minnhom.

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Għalhekk anki jekk funzjoni ma pjan dwar timmodifika l-istudent,

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jekk xorta tista 'tkun ta' valur li jgħaddu mill-indirizz.

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Okay, dak li jekk irridu li tinħoloq pointer għal Struct?

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Nistgħu nagħmlu xi ħaġa simili

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student * S

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ugwali malloc

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daqs tal-istudent.

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Avviż li d-daqs tal-għadu jaħdem hawn.

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Allura kif nistgħu issa jaċċessaw il-membru età

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tal-blokka li l-punti S sa?

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Inti tista 'taħseb l-ewwel li jagħmlu

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* S.age = 4,

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iżda dan mhux se pjuttost xogħol.

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Peress li dan se verament jiġi interpretat bħala li

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* S.age fil-parentesi = 4,

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li mhux saħansitra se jiġbor,

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peress S mhix Struct jew pjuttost pointer għal Struct,

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u għalhekk il-dot mhux se jaħdem hawn.

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Nistgħu nagħmlu

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(* S). Età = 4

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iżda l-parentesi jistgħu jiksbu annoying u konfuża.

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B'xorti tajba, aħna għandna operatur vleġġa speċjali

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li tidher xi ħaġa bħal

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S-> età = 4.

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Dawn il-modi 2 ta 'referenzar età

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huma ekwivalenti

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u aħna ma verament qatt bżonn l-operatur vleġġa,

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iżda jagħmel l-affarijiet ħarsa nicer.

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Peress S hija pointer għal xi blokk ta 'memorja li fiha l-Struct,

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inti tista 'taħseb età> S kif ġej il-vleġġa pointer

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u grab-membru età.

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Allura għaliex għandna qatt tuża structs?

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Huwa ċertament possibbli li jitbiegħed mal biss l-interi primittiv,

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Chars, pointers u simili

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li aħna qed jintużaw biex;

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minflok S1 u S2 qabel,

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nistgħu kellhom age1, age2, name1, u name2

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kollha fl varjabbli separati.

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Dan huwa multa ma 'biss 2 istudenti,

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imma dak jekk kellna 10 minnhom?

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U jekk dak minflok biss 2 oqsma,

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il Struct istudent kellu 100 oqsma?

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GPA, korsijiet, kulur tax-xagħar, sess, u l-bqija.

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Minflok biss 10 structs, għandna bżonn 1000 varjabbli separati.

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Ukoll, jikkunsidraw il-funzjoni

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li tieħu din Struct bi 100 oqsma l-argument biss tagħha

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u tistampa oqsma kollha.

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Jekk aħna ma użatx Struct,

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kull wieħed ħin nitolbu dik il-funzjoni,

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għandna bżonn li tgħaddi fil-varjabbli 100,

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u jekk għandna 100 fatturi varjabbli għall-istudenti 1,

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u 100 fatturi varjabbli għall-istudenti 2,

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għandna bżonn biex tkun ċert aħna ma aċċidentalment jgħaddu xi varjabbli minn student 1

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u xi varjabbli minn student 2.

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Huwa impossibbli li jagħmlu dan żball ma 'Struct,

220
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peress varjabbli kollha 100 jinsabu fil-pakkett wieħed.

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Biss ftit ta 'noti finali:

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Jekk inti ħadthom mifhum kollox sa dan il-punt, kbir.

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Il-bqija tal-video huwa biss għall-finijiet kompletezza ".

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Minħabba structs tista 'żżomm kull tip ta' pointer,

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huma jistgħu jżommu wkoll pointers funzjoni.

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Jekk int familjari mal-programmazzjoni orjentata oġġett,

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dan jipprovdi mod biex jużaw structs għall-programm bi stil orjentati oġġett.

228
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Aktar dwar pointers funzjoni f 'mument ieħor.

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Ukoll, kultant jista 'jkollok 2 structs

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definizzjonijiet li jiddependu fuq xulxin.

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Per eżempju,

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jista 'jkollna Struct A,

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li hija definita bħala

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pointer għal B Struct,

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Struct B * X,

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u issa jista 'jkollna B Struct

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li hija definita bħala indikatur

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għal Struct A,

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Struct A * Y.

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Imma dan mhux se jiġbor,

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peress Struct B ma jeżistix fil-ħin li Struct A qed tiġi miġbura.

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U jekk aħna tpartit Struct A u Struct B,

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allura aħna'd biss jitħallew bil-istess problema;

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din id-darba, ma Struct A eżistenti le.

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Biex issolvi din, nistgħu jikteb

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Struct B;

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qabel id-definizzjoni ta Struct A.

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Dan jissejjaħ dikjarazzjoni quddiem.

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Dan biss tikri l-kumpilatur jafu li

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Struct B huwa tip valida li se tkun kompletament definiti aktar tard jew x'imkien ieħor.

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Jisimni Rob Bowden, u dan huwa CS50.

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[CS50.TV]