Why do function pointers all have the same value?

Question

For example:

using namespace std;
#include 

void funcOne() {
}

void funcTwo( int x ) {
}

int main() {

  void (*ptrOne)() = funcOne;
  cout &l         


        

Answer 1

Try this instead:

void (*ptrOne)() = funcOne;
cout << reinterpret_cast(ptrOne) << endl;

void (*ptrTwo)( int x ) = funcTwo;
cout << reinterpret_cast(ptrTwo) << endl;

int (*ptrMain)() = main;
cout << reinterpret_cast(ptrMain) << endl;

I think normally, function overloading rules mean that the version of << being called is operator<<(bool), which means:

cout << ptrOne << endl;

Gets transformed into:

cout << (ptrOne != NULL) << endl;

Which is the same as:

cout << true << endl;