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.