Printing function pointer passed as a parameter re

问题:

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

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

int alwaysReturns6()
{ return 6; }

int main()
{
    foo(alwaysReturns6);
    return 0;
}

The above code prints the number '1' on the screen.

I know I should access the function pointer like this: p() (and then 6 gets printed), but I still don't get what the plain p or *p means when used in the foo function.

回答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.

回答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;