C++ - statement cannot resolve address for overloaded function

Question

When I types the following as a stand-alone line:

std::endl;

I got the following error:

statement cannot resolve address for overloaded function

Why is that? Cannot I write std::endl; as a stand-alone line?

Thanks.

Answer 1

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!

Answer 2

The most likely reason I can think of is that it's declaration is:

ostream& endl ( ostream& os ); 

In other words, without being part of a << operation, there's no os that can be inferred. I'm pretty certain this is the case since the line:

std::endl (std::cout); 

compiles just fine.

My question to you is: why would you want to do this?

I know for a fact that 7; is a perfectly valid statement in C but you don't see that kind of rubbish polluting my code :-)