How can I allocate memory and return it (via a pointer-parameter) to the calling function?

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You want to use a pointer-to-pointer:

void someFunction (int **data) {
  *data = malloc (sizeof (int));
}

void useData (int *data) {
  printf ("%p", data);
}

int main () {
  int *data = NULL;

  someFunction (&data);

  useData (data);

  return 0;
}

Why? Well, you want to change your pointer data in the main function. In C, if you want to change something that’s passed in as a parameter (and have that change show up in the caller’s version), you have to pass in a pointer to whatever you want to change. In this case, that “something you want to change” is a pointer — so to be able to change that pointer, you have to use a pointer-to-pointer…

Note that on top of your main problem, there was another bug in the code: sizeof(data) gives you the number of bytes required to store the pointer (4 bytes on a 32-bit OS or 8 bytes on a 64-bit OS), whereas you really want the number of bytes required to store what the pointer points to (an int, i.e. 4 bytes on most OSes). Because typically sizeof(int *)>=sizeof(int), this probably wouldn’t have caused a problem, but it’s something to be aware of. I’ve corrected this in the code above.

Here are some useful questions on pointers-to-pointers:

How do pointer to pointers work in C?

Uses for multiple levels of pointer dereferences?

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