sizeof void或者函数

最新推荐文章于 2021-03-13 10:09:30 发布

原创最新推荐文章于 2021-03-13 10:09:30 发布 · 3.8k 阅读

0 ·

CC 4.0 BY-SA版权

C/C++ 专栏收录该内容

46 篇文章

订阅专栏

本文详细介绍了GCC环境下sizeof函数在void和函数类型上的使用方式，特别指出其大小为1，并提供了相关代码示例进行验证。

sizeof的标准规定：

https://en.cppreference.com/w/cpp/language/sizeof

sizeof cannot be used with function types, incomplete types, or bit-field glvalues.

When applied to a reference type, the result is the size of the referenced type.

When applied to a class type, the result is the size of an object of that class plus any additional padding required to place such object in an array.

The result of sizeof is always nonzero, even if applied to an empty class type.

但是，对于gcc来说：

http://gcc.gnu.org/onlinedocs/gcc/Pointer-Arith.html

6.24 Arithmetic on `void`- and Function-Pointers

In GNU C, addition and subtraction operations are supported on pointers to void and on pointers to functions. This is done by treating the size of a void or of a function as 1.

A consequence of this is that sizeof is also allowed on void and on function types, and returns 1.

The option -Wpointer-arith requests a warning if these extensions are used.

因此，对于void和函数来说，其sizeof为1.

例子：

#include <stdio.h>

int test(int i)
{
    i++;
}

int main(void)
{
    int tmp_int = 1;
    int *int_ptr = &tmp_int;
    void *void_ptr = &tmp_int;

    printf("func name size is: %u\n", sizeof(test));

    printf("void size is: %u\n", sizeof(*void_ptr));

    printf("int ptr is [%p]\n", int_ptr);
    int_ptr++;
    printf("int ptr ++ is [%p]\n", int_ptr);

    printf("void ptr is [%p]\n", void_ptr);
    void_ptr++;
    printf("void ptr ++ is [%p]\n", void_ptr++);

    return 0;
}

输出：

func name size is: 1
void size is: 1
int ptr is [0x7fff97dec65c]
int ptr ++ is [0x7fff97dec660]
void ptr is [0x7fff97dec65c]
void ptr ++ is [0x7fff97dec65d]