Why does endl(std::cout) compile

Surprisingly the below code compiles and runs without error on a variety of compilers and versions.

#include <iostream>

int main() {
    endl(std::cout);
    return 0;
}

Ideone link

How does it compile? I am sure there is no endl in global scope because a code like

std::cout << endl;

would fail unless using is used or else you need std::endl.

Solution

This behavior is called argument dependent lookup or Koenig lookup. This algorithm tells the compiler to not just look at local scope, but also the namespaces that contain the argument's type while looking for unqualified function call.

For ex:

namespace foo {
  struct bar{
    int a;
  };

  void baz(struct bar) {
    ...
  }
};

int main() {
  foo::bar b = {42};
  baz(b);  // Also look in foo namespace (foo::baz)
  // because type of argument(b) is in namespace foo
}

About the piece of code referred in question text:

endl or std::endl is declared in std namespace as following:

template< class CharT, class Traits >
std::basic_ostream<charT,traits>&     endl( std::basic_ostream<CharT, Traits>& os );

std::ostream& endl (std::ostream& os);

And cout or std::cout is declared as

extern std::ostream cout;

So calling std::endl(std::cout); is perfectly fine.

Now when one calls just endl(std::cout);, because the type of argument cout is from std namespace, unqualified supposedly a function endl is searched in std namespace and it is found succesfully and confirmed to be a function and thus a call to qualified function std::endl is made.