g++ "calling" a function without parenthesis (not f() but f; ). Why does it always return 1?

Question

In c++ (GNU GCC g++), my code is "calling" a function without (). The function is not working, but compiles ok.

More surprisingly, the code always returns 1...

Is there any explanation?

I expected the function name to be just a regular pointer, but seems it's a bit different...

Did I get all 1's only by chance?

#include <iostream>
using namespace std;

void pr ()
{
    cout << "sth";
}

int main()
{

pr;
cout << pr;  // output: 1
cout << *pr; // output: 1
cout << &pr; // output: 1

}

Answer 1

You're not actually calling pr in your code, you're passing the function pointer to cout . pr is then being converted to a bool when being passed to cout . If you put cout << boolalpha beforehand you will output true instead of 1 .

EDIT:
With C++11 you can write the following overload:

    template <class RType, class ... ArgTypes>
    std::ostream & operator<<(std::ostream & s, RType(*func)(ArgTypes...))
    {
        return s << "(func_ptr=" << (void*)func << ")(num_args=" 
                 << sizeof...(ArgTypes) << ")";
    }

which means the call cout << pr will print (func_ptr=<address of pr>)(num_args=0) . The function itself can do whatever you want obviously, this is just to demonstrate that with C++11's variadic templates, you can match function pointers of arbitrary arity. This still won't work for overloaded functions and function templates without specifying which overload you want (usually via a cast).

Answer 2

The name of a function, when used without parentheses, can be implicitly cast to a function pointer. In fact, when you dereference or reference it, it remains nothing but a function pointer, or a poointer to a function pointer, etc. These function pointers, when printed, are implicitly cast to bool , which is why they simply output 1. If you want to output the actual memory address of the function, cast it to a void pointer:

cout<<(void*)pr;