Why does 'std::endl' require the namespace qualification when used in the statement 'std::cout << std::endl;", given argument-dependent lookup?

Question

I was looking at the Wikipedia entry on argument-dependent lookup, and (on Jan 04, 2014) the following example was given:

#include<iostream>

int main() 
{
  std::cout << "Hello World, where did operator<<() come from?" << std::endl;
}

... with the following comment:

Note that std::endl is a function but it needs full qualification, since it is used as an argument to operator<< (std::endl is a function pointer, not a function call).

My thought is that the comment is incorrect (or simply unclear). I am considering changing the comment to say, instead

Note that std::endl needs full qualification, because ADL does not apply to the arguments of a function call; it only applies to the function name itself.

Am I correct that the Wikipedia comment is incorrect? Is my proposed change correct? (I.e., is my understanding of ADL correct in this example?)

Answer 1

There's nothing wrong about what Wikipedia says.

std::cout << "Hello World, where did operator<<() come from?" << std::endl

is equivalent to the following (assuming operator<< is implemented as a free function)

operator<<(
    operator<<(std::cout, "Hello World, where did operator<<() come from?"),
    std::endl)

which clearly requires namespace qualification for both cout and endl because this is argument-dependent lookup (of the function), not "argument lookup".
The arguments determine the function to be called, not the way around.