c++ pointer to function not changed

Question

I have defined some functions and I print their address like this:

#include<iostream>
#include <string>

using std::cout;

std::string func()
{
    return "hello world\n";

}

int func2(int n)
{
    if (n==0)
    {
        cout << func2 << std::endl;
        return 1;
    }

    cout << func2 << std::endl;

    return n + func2(n - 1);
}

//================================================
int main()
{
    int (*fun)(int) = func2;

    cout << fun;

    cout << std::endl << func2(3);
}

When I print the function's name (address) they all print 1 on my compiler (Mingw gcc 4.8 ).

Is it OK or it should differ?

Answer 1

It doesn't exist an overload of operator<< for std::ostream that takes a function pointer. Thus the operator<<(std::ostream&, bool) overload is prefered. The address of a function is always evaluated to true when converted to bool. Thus, 1 is printed.

Alternatevely, if a function pointer is not larger than the size of a data pointer, you could cast your function pointer to a void* via reinterpret_cast and evoke the operator<<(std::ostream&, void*) overload and thus get the actual address of the function printed.

int (*fun)(int) = func2;
std::cout << reinterpret_cast<void*>(fun) << std::endl;

Live Demo

However, as correctly Neil and M.M mentioned in the comments there's no standard conversion from a function pointer to a data pointer, and this could evoke undefined behaviour.

Alternatively, and in my humble opinion properly, you could format your function pointer as a char array buffer and convert its address to a string in the following way:

unsigned char *p = reinterpret_cast<unsigned char*>(&func2);
std::stringstream ss;
ss << std::hex << std::setfill('0');
for(int i(sizeof(func2) - 1); i >= 0; --i) ss << std::setw(2) 
                                              << static_cast<unsigned int>(p[i]);
std::cout << ss.str() << std::endl;

Live Demo