Any reasons why this can not be standard behavior of free()
?
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\n", p, p2);
safefree((void*&)p2);
printf("p=%p p2=%p\n", p, p2);
safefree((void*&)p); // safe
return 0;
}
assignment from malloc
demands cast from void*
vice versa:
safefree()
demands cast to void*&
(reference)
If it did, you would have to pass a pointer to a pointer to the function:
int * p = malloc( sizeof( int ));
free( & p );
which I'm sure many people would get wrong.