passing a pointer of different type to a function (C)

Question

Suppose that a function from a generic library would like to use long ints (say 64 bit) for any case, but in my program I would like to use short ones (say 32 bit). Then I encounter a situation as follows:

void f(long unsigned int *a) {
  *a = 10;
}

void main(void) {
  unsigned int b;
  f(&b);
  return;
}

Do I understand correctly that this, in fact, is not a good idea and that the function f will overwrite 32 more bits (the ones following b in memory), writing 0 into them (as the compiler casts from unsigned int * to long unsigned int *)? If I imagine correctly, the following, however, will not overwrite (as the compiler casts from long unsigned int to unsigned int):

long unsigned int f(void) {
  return 10;
}

void main(void) {
  unsigned int b;
  b = f();
  return;
}

Is it correct?

This second implementation has a drawback in case it is convenient for me that the function returns some other part of the calculation, and that 10 is just an additional detail of the calculation...

Answer 1

Do I understand correctly that this, in fact, is not a good idea

Yes, that is correct.

... and that the function f will overwrite 32 more bits (the ones following b in memory), writing 0 into them (as the compiler casts from unsigned int * to long unsigned int *)?

That's one possibility. An unsigned long int* may also have stricter alignment requirements than an unsigned int* so it could crash before even getting that far.

What you'd typically do if you have an interface like that is to provide a variable of the correct type and then assign it to a variable of the type you want:

void f(unsigned long int *a) {
  *a = 10;
}

void main(void) {
  unsigned long int tmp;
  f(&tmp);
  // if (tmp > UINT_MAX) ... // possible error check
  unsigned int b = tmp;
}