Why does endl(std::cout) compile

Question

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

#include 

int main() {
    endl(std::cout);
         


        

Answer 1

It is the way the stream manipulators work. Manipulators are functions that passed to operator << as arguments. Then within the operator they are simply called.

So you have function declared like

template 
basic_ostream& endl(basic_ostream& os);

and you pass its pointer to operator <<. And inside the operator that declared something like

ostream& ostream::operator << ( ostream& (*op)(ostream&));

the function is called.like

return (*endl )(*this);

Thus when you see record

std::cout << std::endl;

then std::endl is function pointer that is passed to the operator << as argument.

In the record

std::endl( std::cout );

namespace prefix before name endl may be omitted because in this case the compiler will use the Argument Dependent Lookup. Thus this record

endl( std::cout );

will compile successfully.

However if to enclose the function name in parentheses then ADL is not used and the following record

( endl )( std::cout );

will not compiled.