Printing function pointer passed as a parameter results in '1' printed on the screen

Question

I\'ve been experimenting with function pointers and found the behavior of the following program rather misterious:

void foo(int(*p)())
{ std::cout << p         


        

Answer 1

std::cout << p << std::endl;

here an overload of operator<< which accepts a bool is picked up:

basic_ostream& operator<<( bool value );

As p is not null, then 1 is printed.

If you need to print an actual address, then the cast is necessary, as others mention.

Answer 2

Your function pointer is cast to a bool which is true, or 1 without std::boolalpha.

If you want to see the address you can cast it:

std::cout << static_cast<void*>(p) << std::endl;