What are the operations supported by raw pointer and function pointer in C/C++?

Question

What are all operations supported by function pointer differs from raw pointer? Is > , < , <= , >=operators supported by raw pointers if so what is the use?

Answer 1

For both function and object pointers, they compile but their result is only guaranteed to be consistent for addresses to sub-objects of the same complete object (you may compare the addresses of two members of a class or array) and if you compare a function or object against itself.

Using std::less<>, std::greater<> and so on will work with any pointer type, and will give consistent results, even if the result of the respective built-in operator is unspecified:

void f() { }
void g() { }

int main() {
  int a, b;

  ///// not guaranteed to pass
  assert((&a < &b) == (&a < &b));

  ///// guaranteed to pass
  std::less<int*> lss1;
  assert(lss1(&a, &b) == lss1(&a, &b));
  // note: we don't know whether lss1(&a, &b) is true or false. 
  //       But it's either always true or always false. 

  ////// guaranteed to pass
  int c[2];
  assert((&c[0] < &c[1]) == (&c[0] < &c[1]));
  // in addition, the smaller index compares less:
  assert(&c[0] < &c[1]);

  ///// not guaranteed to pass
  assert((&f < &g) == (&f < &g));

  ///// guaranteed to pass
  assert((&g < &g) == (&g < &g));
  // in addition, a function compares not less against itself. 
  assert(!(&g < &g));

  ///// guaranteed to pass
  std::less<void(*)()> lss2;
  assert(lss2(&f, &g) == lss2(&f, &g));
  // note: same, we don't know whether lss2(&f, &g) is true or false.

  ///// guaranteed to pass
  struct test {
    int a;
  // no "access:" thing may be between these!
    int b;

    int c[1];
  // likewise here
    int d[1];

    test() {
      assert((&a < &b) == (&a < &b));
      assert((&c[0] < &d[0]) == (&c[0] < &d[0]));

      // in addition, the previous member compares less:
      assert((&a < &b) && (&c[0] < &d[0]));
    }
  } t;
}

Everything of that should compile though (although the compiler is free to warn about any code snippet it wants).

---

Since function types have no sizeof value, operations that are defined in terms of sizeof of the pointee type will not work, these include:

void(*p)() = ...;
// all won't work, since `sizeof (void())` won't work.
// GCC has an extension that treats it as 1 byte, though.
p++; p--; p + n; p - n; 

The unary + works on any pointer type, and will just return the value of it, there is nothing special about it for function pointers.

+ p; // works. the result is the address stored in p.

Finally note that a pointer to a function pointer is not a function pointer anymore:

void (**pp)() = &p;
// all do work, because `sizeof (void(*)())` is defined.
pp++; pp--; pp + n; pp - n;