The argument tab
of the function sth
is a copy of what is passed and change to that won't affect what is passed. Therefore, free(tab);
in the main()
function means free(NULL);
. This is defined to do nothing and it won't contribute for avoiding memory leak. Pass pointers to what should be modified to have functions modify what are passed.
#include <stdio.h>
#include <stdlib.h>
struct x{
int a;
char b;
};
void allocate( struct x **tab,int *size)
{
*size = 1+2*(*size);
*tab= realloc(*tab, (size_t) (*size) * sizeof (**tab));
}
void sth (struct x **tab, int *size) // receive a pointer of struct x*
{
//do something here
// allocate(&(*tab), size);
allocate(tab, size);
}
int main(void)
{
int size=0;
struct x *tab=NULL;
sth(&tab, &size); // pass a pointer to what should be modified
//do sth here with tab
free(tab);
return 0;
}
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