How to print function pointers with cout?

Question

I want to print out a function pointer using cout, and found it did not work. But it worked after I converting the function pointer to (void *), so does printf with %p, such

Answer 1

In C++11 one could modify this behavior by defining a variadic template overload of operator<< (whether that is recommendable or not is another topic):

#include
namespace function_display{
template
std::ostream& operator <<(std::ostream& os, Ret(*p)(Args...) ){ // star * is optional
    return os << "funptr " << (void*)p;
}
}

// example code:
void fun_void_void(){};
void fun_void_double(double d){};
double fun_double_double(double d){return d;}

int main(){
    using namespace function_display;
    // ampersands & are optional
    std::cout << "1. " << &fun_void_void << std::endl; // prints "1. funptr 0x40cb58"
    std::cout << "2. " << &fun_void_double << std::endl; // prints "2. funptr 0x40cb5e"
    std::cout << "3. " << &fun_double_double << std::endl; // prints "3. funptr 0x40cb69"
}