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.
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!