ADL using std types fails to find operator

Question

The following code fails to compile

namespace A {
using C = std::vector<std::string>;
std::ostream& operator << (std::ostream& lhs, const C& rhs) {
    lhs << 5;
    return lhs;
}
}
int main()
{
    A::C f;
    std::cout << f;
    return 0;
}

with the error

Error   C2679   binary '<<': no operator found which takes a right-hand operand of type 'A::C' (or there is no acceptable conversion)   

Obviously it cant find the << operator presumably due to considering C to be a class from the std namespace. Is there some way to ensure the compiler finds this operator or otherwise work around the problem?

Answer 1

A::C is just a type alias, and aliases are transparent. They don't "remember" where they came from. When we do argument-dependent lookup and figure out what the associated namespaces are, we only consider the associated namespaces of the types - not the alias that got us there. You can't just add associated namespaces to existing types. The specific associated namespace of f (which is of type std::vector<std::string>) is std, which doesn't have an operator<< associated with it. Since there's no operator<< found using ordinary lookup, nor is there one found using ADL, the call fails.

Now, I said you can't just add associated namespaces to existing types. But you can of course just create new types:

namespace A {
    struct C : std::vector<std::string> { };
}

or:

namespace A {
    // template parameters are also considered for associated namespaces
    struct S : std::string { };
    using C = std::vector<S>;
}