Question

int do_memory()
{
  int * empty_ptr = (int *) malloc(sizeof(int));
  *empty_ptr = 5;
  return *empty_ptr;
}
...
int b = do_memory();
free(&b); //obviously not valid

When b goes out of scope, I would be right in assuming the memory in empty_ptr still exists? But is impossible to free and therefore this is bad code?

Answer 1

The “int *empty_ptr” (the pointer to the allocated memory block) is never released, you just get the return value from do_memory.

If you want no leaks, use this

int* do_memory()
{
  int * empty_ptr = (int *) malloc(sizeof(int));
  *empty_ptr = 5;
   return empty_ptr;
}
...
int* b = do_memory();
int b_local = *b;
free(b); // valid

or this (no leaks, no allocations except the var on stack, no performance hit):

void do_memory(int* RetValue)
{
  *RetValue = 5;
}
...
/// b is allocated locally on stack
int b;
do_memory(&b);
// no free() calls are needed