C++ – statement cannot resolve address for overloaded function

std::endl is a function template. Normally, it’s used as an argument to the insertion operator <<. In that case, the operator<< of the stream in question will be defined as e.g. ostream& operator<< ( ostream& (*f)( ostream& ) ). The type of the argument of f is defined, so the compiler will then know the exact overload of the function.

It’s comparable to this:

void f( int ){}
void f( double ) {}
void g( int ) {}
template<typename T> void ft(T){}

int main(){
  f; // ambiguous
  g; // unambiguous
  ft; // function template of unknown type...
}

But you can resolve the ambiguity by some type hints:

void takes_f_int( void (*f)(int) ){}

takes_f_int( f ); // will resolve to f(int) because of `takes_f_int` signature
(void (*)(int)) f; // selects the right f explicitly 
(void (*)(int)) ft; // selects the right ft explicitly 

That’s what happens normally with std::endl when supplied as an argument to operator <<: there is a definition of the function

 typedef (ostream& (*f)( ostream& ) ostream_function;
 ostream& operator<<( ostream&, ostream_function )

And this will enable the compiler the choose the right overload of std::endl when supplied to e.g. std::cout << std::endl;.

Nice question!