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

Operator overloading in C++ adds all sorts of nasty complexities. (It lets you do awesome stuff too—but sometimes it's just a headache.)

As explained in the other answers, C++ is doing some automatic type coercion on your function pointers. If you just use the good ol' C-style printf you should get the results you're expecting:

#include 

// ...

printf("func1: %p\nfunc2: %p\n", funcOne, funcTwo);