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Old Apr 22, 2005, 17:40   #1
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Wink задачка для ассов c/c++
с виду тривиальная задачка ...

#include <stdio.h>
int main (int argc, unsigned char **argv) {
int i = 0;
i = i++;
printf ("i = %d\n", i);
return 0;
}

какое значение i будет выведенно ?
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Old Apr 22, 2005, 17:44   #2
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i=i++ eto kaneshna kruto
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Old Apr 22, 2005, 17:47   #3
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Old Apr 22, 2005, 17:49   #4
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rezul'tat i++ vozvrashaet 0. znachit tipa vse v poryadke. a potom etot rezul'tat pripisivaetsya k i. znachit budet 0
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Old Apr 22, 2005, 17:52   #5
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poprobui eshe i=printf("%d", i); ili eshe chego.
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Old Apr 22, 2005, 17:57   #6
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btw ас пишется с одним с :]
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Old Apr 22, 2005, 17:59   #7
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а можете обьяснить почему ?
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Old Apr 22, 2005, 18:05   #8
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ya vrode obyasnil. rezul'tat operacii i++ vozvrashaet 0
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Old Apr 22, 2005, 18:05   #9
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потому что приоритет унарного оператора выше чем оператора присвоения
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Old Apr 22, 2005, 18:06   #10
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не обьяснил
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Old Apr 22, 2005, 18:14   #11
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imeyushie glaza da uvidyat...
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Old Apr 22, 2005, 18:15   #12
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Danyer
пойми меня правильно
я имею ввиду по стандарту
а не логикой ... твоей логикой
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Old Apr 22, 2005, 18:21   #13
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oi, sorry ya ne zametil, dlya accov...

acc eto je latinica... sorry. molchu.

xotya zombie toje neslabo skazal
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Old Apr 22, 2005, 18:23   #14
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Quote:
Originally Posted by z0mbie
btw ас пишется с одним с :]
на русском на других языках это может быть не так.
ну извини, какой есть так и формулирует.

а вообще-то постфиксный оператор инкремента(декремента) выполняется _после_ выполнения _всей_ операции в которой он записан.

посему имеем --
i=i;
i++;
что и наблюдаем на экране при выполнении
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Old Apr 22, 2005, 18:29   #15
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я часто делаю ошибки
ну и что ?
Кстате читай беспредел ... там для тебя сообщение ...
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