为什么 free(p) 不将 p 设置为 NULL?

2021-12-31 00:00:00 c memory-management memory-leaks c++ free

为什么这不能成为 free() 的标准行为?

Any reasons why this can not be standard behavior of free()?

多个指针指向同一个对象:

multiple pointers pointing to the same object:

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

void safefree(void*& p)
{
    free(p); p = NULL;
}

int main()
{
    int *p = (int *)malloc(sizeof(int));
    *p = 1234;
    int*& p2 = p;
    printf("p=%p p2=%p
", p, p2);
    safefree((void*&)p2);
    printf("p=%p p2=%p
", p, p2);
    safefree((void*&)p); // safe

    return 0;
}

来自 malloc 的赋值要求来自 void*

assignment from malloc demands cast from void*

反之亦然:

safefree() 要求强制转换为 void*&(参考)

safefree() demands cast to void*& (reference)

推荐答案

如果是这样，你就必须传递一个指向函数指针的指针:

If it did, you would have to pass a pointer to a pointer to the function:

int * p = malloc( sizeof( int ));
free( & p );

我相信很多人都会弄错.

which I'm sure many people would get wrong.

相关文章