how to get enum value from enum type?

Question

The question in the title may sound trivial, so I better explain with some code what I want to do...

In C++11 I can do this:

#include <iostream>

namespace X {
    enum class  FOO { A,B };
}

template <typename T> void foo(T t) { 
    if (t == T::A) { std::cout << "A"; }
}

int main() {
    foo(X::FOO::A);
}

The important point here is that the template foo does not need to know in what namespace the enum is declared. I could as well call foo(Y::FOO::B) (provided there is a enum class called FOO in namespace Y having members A and B).

Now the question is: How to get the same with plain old enums (and only C++98 stuff)?

This works:

#include <iostream>

namespace X {
    enum FOO { A,B };
}

template <typename T> void foo(T t) { 
    if (t == X::A) { std::cout << "A"; }
}

int main() {
    foo(X::A);
}

but only because foo knows in what namespace the enum is declared. And it wont work for a Y::FOO::B ! (In C++11 it also works if I replace the line with if (t == T::A) ..., even with a plain enum)

Is there a way to get this working in C++98/03 without refering to X in the template explicitly?

For the sake of completeness, in C++98, this

template <typename T> void foo(T t) { 
    if (t == T::A) { std::cout << "A"; }
}

results in

error: ‘A’ is not a member of ‘X::FOO’

PS: I am not allowed to change the enum and the template has to live in a different namespace than the enum.

PPS: a simple if (t == 0) would probably work, but thats something I would like to avoid

Answer 1

Until C++11, there was no way to say "the enumerator name within this enumeration". That's why it was added to C++11, even to unscoped enumerations.

Answer 2

To add to Nicol Bolas' answer, you can kind of hack your way to a solution using ADL:

namespace X {
    enum FOO { A,B };
    bool IsA(FOO t)
    {
        return t == A;
    }
}

template <typename T> void foo(T t) { 
    if (IsA(t)) { std::cout << "A\n"; }
    else{std::cout << "not A\n";}
}

Live Demo