Explicit operator<< selects 'wrong' overload

Question

I'm playing around with variadic templates, and wrote this, based on this answer :

template <size_t... I>
void print(seq<I...>)
{
    decltype(std::cout) * dummy[sizeof...(I)] = { &(std::cout << I << ' ')... };
}

Because std::cout::operator<< has a return type, it can be stored, so there's no need for the ( ,0) comma trick.

Now, to shut up the "unused variable 'dummy'" warning, and to print a newline, I tried the following statements, but they didn't do what I wanted:

dummy[0]->operator <<('\n'); // prints 10

(apparently called operator<<(int) instead of operator<<(char)

dummy[0]->operator <<("\n"); // prints a pointer

(apparently called operator<<(const void*) instead of operator<<(const char*)

In the end, I had to write

*dummy[0] << '\n';             // prints a newline as desired

My question is, why did the "wrong" overloads get chosen?

What does `std::cout << std::is_same::value<<'\n';` print? – Yakk - Adam Nevraumont Oct 19 '15 at 09:29 — Yakk - Adam Nevraumont, Oct 19 '15 at 09:29

score 9 · Accepted Answer · edited Oct 19 '15 at 09:42

The "wrong" overloads are chosen because only some overloads are members of std::ostream class. The overloads for char and const char* are not members of std::ostream, but free functions, so, in

*dummy[0] << '\n';

argument-dependent lookup will find operator<<(std::ostream&, char), but in

dummy[0]->operator <<('\n');

only member functions will be considered, resulting in std::ostream::operator<<(int) being called.

Explicit operator<< selects 'wrong' overload

1 Answers1