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>> SPEAKER: Ifakkar li fis-C, id-data kollha
tipi għandhom daqs partikolari.

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Iżda li d-daqs tajjeb ħafna jistgħu jvarjaw bbażati
fuq il-kompjuter li fuqu inti qed tuża

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C. Issa, aħna jiġri li tkun qed tuża C ġewwa
ta 'l-appliance CS50, hekk ġewwa ta'

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l-appliance, kemm hu kbir huwa char?

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Kif big huwa doppja?

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00:00:15,270 --> 00:00:16,430
Kif big huwa int?

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Kif big huwa float?

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>> Well, ejja tagħti ħarsa
ma 'xi kodiċi.

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Inkludi io.h standard, null prinċipali int.

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U issa ejja tibda tiddikjara
ftit varjabbli li

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00:00:29,100 --> 00:00:31,330
daqsijiet aħna ser mbagħad print -

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char c, d doppja, float f, u i int.

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Issa, jien mhux ser taħżen il-valuri
f'dawn varjabbli, minħabba I biss

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kura dwar id-daqs partikolari tagħhom.

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>> Biex tara d-daqs tagħhom, I ser tuża print f,
kif ukoll bħala operatur C imsejjaħ daqs

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ta ', li tkun risposta eżattament
din il-kwistjoni.

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Ejja tagħti ħarsa.

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Stampa f char, segwit minn kolon,
mija 1, backslash n.

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Fi kliem ieħor, nixtieq li jistampa
kolon char, segwit mill-daqs tagħha.

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>> So I ser jinkludi comma
segwit minn daqs ta 'c.

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Ejja issa jagħmlu dan mill-ġdid għal doppja.

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Stampa f, double, mija i, backslash
n, kwotazzjoni qrib, comma, id-daqs ta 'd.

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Issa avviż, jien ser tkompli tuża
mija i, u b'hekk li tfisser li

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int, għaliex dak li mhux dejjem tinbidel l-
unitajiet li fihom jien li jkejlu l-

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daqs ta 'dawn il-varjabbli.

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>> Fil-fatt, id-daqs ta 'se jirritorna xi
numru ta 'bytes, forsi wieħed, forsi

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tnejn, forsi aktar.

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Iżda f'kull każ, se
tabilħaqq tagħti me int.

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U b'tali mod li d-detentur post nixtieq
għall-użu, irrispettivament mit-tip,

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li d-daqs jien jkollna.

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Ejja nagħmlu żewġ print aktar f'S.

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>> Float, mija i, backslash
n, id-daqs ta 'f.

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Stampa f, int, mija i,
backslash n, id-daqs tal-i.

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Ejja issa jiffranka, tiġbor,
u run dan il-programm.

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Għamla, id-daqs ta ', dot slash, id-daqs ta', u
naraw li fl-appliance CS50, a

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char huwa byte waħda, double huwa tmien
bytes, float huwa 4 bytes, u int

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huwa erba 'bytes, kif ukoll.

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