我错过了什么？ [英] what I miss?

问题描述

#include< stdio.h>

#include< stdlib.h>

#include< limits.h>

int main（）

{

unsigned char * c;

c = malloc（sizeof（unsigned char））;

printf（" unsigned char的大小：％d \ n"，sizeof（unsigned char））;

printf（" c的大小：％d \ n"，sizeof（c））;

返回0;

}


执行此操作时，它表示大小unsigned char是1是1。 &安培;尺寸

的c是4。这不是很奇怪吗？


（gcc版本4.0.0 20050519（Red Hat 4.0.0-8））

#include <stdio.h>
#include <stdlib.h>
#include <limits.h>

int main()
{
unsigned char *c;
c = malloc(sizeof(unsigned char));
printf("size of unsigned char: %d\n", sizeof(unsigned char));
printf("size of c: %d\n", sizeof(c));
return 0;
}

When I execute this, it says that size of "unsigned char" is "1" & size
of "c" is "4". isn''t that strange?

( gcc version 4.0.0 20050519 (Red Hat 4.0.0-8) )

推荐答案

Parahat Melayev < PA ***** @ gmail.com>在消息中写道

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#include< stdio.h>
#include< stdlib.h>
#include< limits.h>

int main（）
{
unsigned char * c;
c = malloc（sizeof（unsigned char））;
printf（" unsigned char的大小：％d \ n"，sizeof（unsigned） char））;
printf（c：％d \ n的大小，sizeof（c））;
返回0;
}

我执行此操作，它表示unsigned char的大小。是1是1。 &安培; c的大小
是4。这不奇怪吗？

#include <stdio.h>
#include <stdlib.h>
#include <limits.h>

int main()
{
unsigned char *c;
c = malloc(sizeof(unsigned char));
printf("size of unsigned char: %d\n", sizeof(unsigned char));
printf("size of c: %d\n", sizeof(c));
return 0;
}

When I execute this, it says that size of "unsigned char" is "1" & size
of "c" is "4". isn''t that strange?

不。 C不是char，它是指针（对于char）。


Alex

Nope. C isn''t a char, it''s a pointer (to a char).

Alex

哦是的，那是对:)恐慌


tnx

oh yeah that is right :) panic

tnx

Parahat Melayev写道：

Parahat Melayev wrote:

#include< stdio.h>
#include< stdlib.h>
#include< limits.h>

int main（）
{
unsigned char * c;
c = malloc（sizeof（unsigned char））;
printf（" unsigned char的大小：％d \ n"，sizeof（unsigned char））;
printf（c：％d \ n的大小，sizeof（c））;
返回0;
}
当我执行此操作时，它表示unsigned char的大小。是1是1。 &安培; c的大小
是4。这不奇怪吗？


什么是c ????

c是指针，它将保存

char的地址类型。


malloc将返回已分配内存的地址。

等等c保存地址，即int。

c = malloc（sizeof（unsigned char））;


HTH

ranjeet

（gcc 4.0版。 0 20050519（Red Hat 4.0.0-8））

#include <stdio.h>
#include <stdlib.h>
#include <limits.h>

int main()
{
unsigned char *c;
c = malloc(sizeof(unsigned char));
printf("size of unsigned char: %d\n", sizeof(unsigned char));
printf("size of c: %d\n", sizeof(c));
return 0;
}

When I execute this, it says that size of "unsigned char" is "1" & size
of "c" is "4". isn''t that strange?
what is c ????
c is the pointer which is going to hold the address which is of the
char type.

malloc will return the address of the allocated memory.
and so c holds the address, which is int.

c = malloc(sizeof(unsigned char));

HTH
ranjeet

( gcc version 4.0.0 20050519 (Red Hat 4.0.0-8) )

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