Why doesn't the << operator defined inside a namespace participate in overload resolution?

Question

In the following scenario, the operator<< function participates in overload resolution (when it is in the global namespace):

template <typename T>
std::ostream &operator<<(std::ostream &os, const std::vector<T> &vec) {
    os << '[';
    for (const T &val : vec)
        os << val << ", ";
    return os << "\b\b]";
}

... but below, when defined inside a namespace, it does not:

namespace blah {
    template <typename T>
    std::ostream &operator<<(std::ostream &os, const std::vector<T> &vec) {
        os << '[';
        for (const T &val : vec)
            os << val << ", ";
        return os << "\b\b]";
    }
}

When called from main, as such:

int main() {
    std::vector<int> vec{5, 4, 3, 6, 4, 3, 5, 6, 8, 1};
    std::cout << vec << std::endl;
    return 0;
}

... the compiler is not able to find the operator<< overload when it is defined in blah.

What is going on?

Edit:

When calling the overload with the scope resolution operator, everything compiles fine:

blah::operator<<(std::cout, vec);

Why is this?

How can the operator declared in the user namespace be found without qualifying it? — Vlad from Moscow, Feb 10 '23 at 22:48
Indeed, but what I then do not understand is why an operator overload such as `<<` defined for custom classes can remain within the namespace the classes are defined in, and still be found during lookup? — Gregor Hartl Watters, Feb 10 '23 at 22:50

score 5 · Accepted Answer · edited Feb 10 '23 at 23:07

5

In the first case, there is used unqualified name lookup and/or ADL (argument-dependent lookup) lookup that includes the global namespace where the operator is declared.

In the second case, the operator will not be found because the namespace blah is not considered for unqualified name lookup.

Either you could use qualified the name, like:

blah::operator <<( std::cout, vec ) << std::endl;

Or, you could use a using directive to include the namespace blah in the set of namespaces for unqualified name lookup, as:

using namespace blah;
std::cout << vec << std::endl;

edited Feb 10 '23 at 23:07

Remy Lebeau

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answered Feb 10 '23 at 22:57

Vlad from Moscow

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Indeed, although you sure don't seem like a beginner! :) – Gregor Hartl Watters Feb 10 '23 at 23:06
@GregorHartlWatters In Russia nobody suggests me a job because everyone thinks that I know too little. – Vlad from Moscow Feb 10 '23 at 23:08
well, it seems like you know quite a lot, so I'm sure you will find a job eventually! – Gregor Hartl Watters Feb 10 '23 at 23:16

Why doesn't the << operator defined inside a namespace participate in overload resolution?

1 Answers1