我和c中的size_t搞混了,我知道它是由sizeof操作符返回的。但它究竟是什么?它是数据类型吗?

假设我有一个for循环:

for(i = 0; i < some_size; i++)

我应该用int I吗?或者size_t i;?


当前回答

Size_t是无符号整数数据类型。在使用GNU C库的系统上,这将是unsigned int或unsigned long int。Size_t通常用于数组索引和循环计数。

其他回答

如果你是经验主义者,

echo | gcc -E -xc -include 'stddef.h' - | grep size_t

输出Ubuntu 14.04 64位GCC 4.8:

typedef long unsigned int size_t;

注意,stddef.h是由GCC提供的,而不是GCC 4.2中src/ GCC /ginclude/stddef.h下的glibc。

有趣的C99外观

Malloc以size_t作为参数,因此它决定了可以分配的最大大小。 由于它也是由sizeof返回的,我认为它限制了任何数组的最大大小。 请参见:C语言中数组的最大大小是多少?

Size_t或任何无符号类型可能被视为循环变量,因为循环变量通常大于或等于0。

当使用size_t对象时,必须确保在使用它的所有上下文中(包括算术)只需要非负值。例如,下面的程序肯定会给出意想不到的结果:

// C program to demonstrate that size_t or
// any unsigned int type should be used 
// carefully when used in a loop

#include<stdio.h>
int main()
{
const size_t N = 10;
int a[N];

// This is fine
for (size_t n = 0; n < N; ++n)
a[n] = n;

// But reverse cycles are tricky for unsigned 
// types as can lead to infinite loop
for (size_t n = N-1; n >= 0; --n)
printf("%d ", a[n]);
}

Output
Infinite loop and then segmentation fault

这是一个平台特定的类型定义。例如,在特定的机器上,它可能是unsigned int或unsigned long。您应该使用这个定义来提高代码的可移植性。

根据我的理解,size_t是一个无符号整数,其位大小足以容纳本机体系结构的指针。

So:

sizeof(size_t) >= sizeof(void*)

从维基百科:

According to the 1999 ISO C standard (C99), size_t is an unsigned integer type of at least 16 bit (see sections 7.17 and 7.18.3). size_tis an unsigned data type defined by several C/C++ standards, e.g. the C99 ISO/IEC 9899 standard, that is defined in stddef.h.1 It can be further imported by inclusion of stdlib.h as this file internally sub includes stddef.h. This type is used to represent the size of an object. Library functions that take or return sizes expect them to be of type or have the return type of size_t. Further, the most frequently used compiler-based operator sizeof should evaluate to a constant value that is compatible with size_t.

作为暗示,size_t是保证保存任何数组下标的类型。