I wrote a little wrapper around malloc ...
I wrote a little wrapper around malloc, but it doesn't work:
#include lt;stdio.h>
#include lt;stdlib.h>
mymalloc(void *retp, size_t size)
{
retp = malloc(size);
if(retp == NULL) {
fprintf(stderr, "out of memory\n");
exit(EXIT_FAILURE);
}
}
✍: Guest
Are you sure the function initialized what you thought it did? Remember that arguments in C are passed by value. In the code above, the called function alters only the passed copy of the pointer. To make it work as you expect, one fix is to pass the address of the pointer (the function ends up accepting a pointer-to-a-pointer; in this case, we're essentially simulating pass by reference):
void f(ipp)
int **ipp;
{
static int dummy = 5;
*ipp = &dummy;
}
...
int *ip;
f(&ip);
Another solution is to have the function return the pointer:
int *f()
{
static int dummy = 5;
return &dummy;
}
...
int *ip = f();
2016-06-10, 1757👍, 0💬
Related Topics:
In a call to malloc, what does an error like
In a call to malloc, what does an error like ``Cannot convert `void *' to `int *''' mean? It means you're using a C++ compiler instead of a C compiler. 2016-06-21, 2054👍, 0💬