How can a sizeof operator calculate size of return type of a function?

Question

Code-

#include <bits/stdc++.h>
using namespace std;

 int *fun()
 {  static int a; 
     ++a; 
     cout<<"value of a = "<<a<<endl;
     return &a;
  } 

 int main() {
 cout<<"simple call \n"; 
 fun(); 
 cout<<"calling inside sizeof \n"; 
  cout<<sizeof(fun())<<endl;
  cout<<"calling inside pow \n";  
  cout<<pow(2,*fun())<<endl;
  return 0;
 }

Output-

simple call 
value of a = 1 
calling inside sizeof 
8 
calling inside pow 
value of a = 2
4

My doubt is how sizeof operator calculate size of return type of function. I assume function is first called then it return address to it's static variable and sizeof operator is operated on that value. But here, it's directly printing sizeof pointer. And if I use same fun inside pow() it is first called and then further calculation is done.

Answer 1

No, sizeof never evaluates its operands (except for C variable length arrays but, since this is a C++ question, that's irrelevant). From C++17 [expr.sizeof]:

The sizeof operator yields the number of bytes in the object representation of its operand. The operand is either an expression, which is an unevaluated operand (Clause 8), or a parenthesized type-id.

In your case, since fun() returns int * (thew compiler knows this because you told it so), it's as if you had asked for sizeof(int *).