Why does printing a function name returns a value?

Question

I accidentally printed a function name without the parenthesis and it printed a value. I am just curious about how this happens? Output is same irrespective of the function name or definition, and for every time I run it.

EDIT: The answers cleared my doubt, to anyone else who is reading this - Explicitly converting the function name to int works, i.e int k=(int)foo;

This test code will make things more clear:

#include <iostream>
#include <stdio.h>
#include <conio.h>
using namespace std;

void foo(){cout<<'Á';}      //FUNCTION IS NEVER CALLED

int main()
{
    while(_kbhit())         //JUST TO MAKE SURE BUFFER IS CLEARED
    {   getch();}           //SAME RESULT WITHOUT THESE TWO STATEMENTS

    cout<<foo;              //OUTPUT 1
    printf("\n%u", foo);    //OUTPUT 4199232
    /*int k=foo;            //CANNOT CONVERT VOID(*)() TO 'INT'*/
    return 0;
}

Answer 1

A mention of a function name without parentheses is interpreted as a function pointer. The pointer is interpreted as an unsigned int by the %u format specifier, which is undefined behaviour but happens to work on most systems.

The reason int k = foo doesn't work is that a function pointer normally needs a cast to be converted to int. However, printf is much more lenient because it uses varargs to parse its argument string; the type of the arguments is assumed to match the type requested in the format string.

Answer 2

This statement printf("\n%u", foo); passes the address of function foo() to printf. So the value that gets printed is the address of this function in the memory of the running program.

Answer 3

std::cout << foo;

Outputs 1 because foo is a function pointer, it will be converted to bool with std::cout.

To print its address, you need explicitly cast it:

std::cout << reinterpret_cast<void*>(foo) << std::endl;

---

With printf("\n%u", foo);, %u expects unsigned int, what you saw is the value of the function pointer converted to unsgigned int.