Why function pointer address is printing in bool type in c++?

Question

From the below code snippet I am getting the address of the function is 1. why ?

#include<iostream>
using namespace std;
int  add(int x, int y)
{
  int z;
  z = x+y;
  cout<<"Ans:"<<z<<endl;
}

int main()
{
  int a=10, b= 10;
  int (*func_ptr) (int,int);
  func_ptr  = &add;
  cout<<"The address of function add()is :"<<func_ptr<<endl;
  (*func_ptr) (a,b);
}

Answer 1

Function pointers aren't convertible to data pointers. You'd get a compiler error if you were to try and assign one to a void* variable. But they are implicitly convertible to bool!

That is why the bool overload for operator<< is chosen over the const void* one.

To force the overload you want, you'd need to use a very strong C++ cast, that will almost completely ignore the static type information.

#include<iostream>
using namespace std;
int  add(int x, int y)
{
  int z;
  z = x+y;
  cout<<"Ans:"<<z<<endl;
}

int main()
{
  int a=10, b= 10;
  int (*func_ptr) (int,int);
  func_ptr  = &add;
  cout<<"The address of function add()is :"<< reinterpret_cast<void*>(func_ptr) <<endl;
  (*func_ptr) (a,b);
}

Note that casting and treating function pointers as data pointers is only conditionally supported (from the C++ standard standpoint). Using it for anything other than casting back to the same function pointer will have an implementation specific outcome, that can very greatly among compilers.

Answer 2

The overload used is

ostream& ostream::operator<< (bool val);

Which prints 1 as your function pointer is not null.