Question

Consider the c code:

void mycode() {
  MyType* p = malloc(sizeof(MyType));
  /* set the values for p and do some stuff with it */
  cleanup(p);
}


void cleanup(MyType* pointer) {
  free(pointer);
  pointer = NULL;
}

Am I wrong in thinking that after cleanup(p); is called, the contents of p should now be NULL? Will cleanup(MyType* pointer) properly free the memory allocation?

I am coding my college assignment and finding that the debugger is still showing the pointer to have a memory address instead of 0x0 (or NULL) as I expect.

I am finding the memory management in C to be very complicated (I hope that’s not just me). can any shed some light onto what’s happening?

Answer 1

Yes that will free the memory correctly.

pointer inside the cleanup function is a local variable; a copy of the value passed in stored locally for just that function.

This might add to your confusion, but you can adjust the value of the variable p (which is local to the mycode method) from inside the cleanup method like so:

void cleanup(MyType** pointer) {
  free(*pointer);
  *pointer = NULL;
}

In this case, pointer stores the address of the pointer. By dereferencing that, you can change the value stored at that address. And you would call the cleanup method like so:

cleanup(&p);

(That is, you want to pass the address of the pointer, not a copy of its value.)

I will note that it is usually good practice to deal with allocation and deallocation on the same logical ‘level’ of the software – i.e. don’t make it the callers responsibility to allocate memory and then free it inside functions. Keep it consistent and on the same level.