visibility outside namespace propagated by usage of other namespace elements

Question

Take a look at this simple snippet:

namespace Test
{
    struct A {};
    void foo( A _a ) {}
}


int main( int, char** )
{
    foo( Test::A() ); // 1. why foo doesn't require Test:: ?
    Test::foo( A() ); // 2. why A() requires Test:: considering above line?

    return 0;
}

As mentioned in the source:

foo( Test::A() ); why foo doesn't require Test:: here?
Test::foo( A() ); why A() requires Test:: considering above line?

(Visual Studio 2008 and gcc4.8 give the same result, so I suppose this is standard behavior, but I'm wondering which part of standard defines it?)

I do not understand your 2nd point: " why A() requires Test:: considering above line?" ... in your second line, there is no `Test::` in front of `A()` .... — 463035818_is_not_an_ai, Jul 22 '15 at 10:01
@tobi303 Yes, and because of that it doesn't compile. You need to add Test:: to compile it. — Bogdan, Jul 22 '15 at 10:18

score 2 · Accepted Answer · edited May 23 '17 at 11:58

This is Argument-dependent name lookup, a.k.a. Koenig lookup (although Andrew Koenig did not invent it). It is defined in the chapter of the same name in the standard.

It enables things like:

std::string str( "Hello world!" );
std::cout << str;

The overload operator<<( std::ostream &, std::string ) is in the std:: namespace -- and without ADL, it would not be found.

It works only for functions looked for in the namespace of their arguments, not the other way around (your example 2), which would further weaken namespace barriers with no comparable benefit to case 1.

More in-depth discussion of the subject here.

I knew I read about it but I couldn't to find the right topic. Thanks! — Bogdan, Jul 22 '15 at 10:23

visibility outside namespace propagated by usage of other namespace elements

1 Answers1