Template friend function instantiation

Question

Why do I get a linker error for the following?

template<typename T, typename U>
class A {
public:
    class B;
};

template<typename T, typename U>
class A<T, U>::B {
    friend bool operator==(const B& b1, const B& b2);
};

template<typename T, typename U>
bool operator==(const typename A<T, U>::B& b1, const typename A<T, U>::B& b2) {
    return true;
}

int main() {
    A<char, int>::B b1;
    A<char, int>::B b2;

    if (b1 == b2) {
        return 0;
    } else {
        return 1;
    }
}

The error I get is:

Undefined symbols for architecture x86_64:
  "operator==(A<char, int>::B const&, A<char, int>::B const&)", referenced from:
      _main in test-qSCyyF.o

jrok · Accepted Answer · 2013-08-01T19:57:20.630

2

Because you're calling it and it has no definition!

This

template<typename T, typename U>
class A<T, U>::B {
    friend bool operator==(const B& b1, const B& b2);
};

is a declaration of a non-template friend function of A<T,U>::B class.

It matches the call (b1 == b2) just as well as the template operator you define further down but it is preferred because it is a non-template.

GCC even gives a warning with -Wnon-template-friend

warning: friend declaration 'bool operator==(const A::B&, const A::B&)' declares a non-template function [-Wnon-template-friend]
note: (if this is not what you intended, make sure the function template has already been declared and add <> after the function name here)

To fix, it provide the definition

template<typename T, typename U>
class A<T, U>::B {
    friend bool operator==(const B& b1, const B& b2) {
        return true;
    }
};

and get rid of the templated operator OR keep only the template one.

edited Aug 01 '13 at 19:57

answered Aug 01 '13 at 19:45

jrok

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So… how do I fix it. =P – fumoboy007 Aug 01 '13 at 19:53
So I can't have a templated friend function that's defined outside of the class?? – fumoboy007 Aug 01 '13 at 20:53
@fumoboy007 You usualy can, but in your case, with a nested non-template class? I don't think so, but I'm not sure :) I'd say you don't need it, but I don't know exactly what you're trying to achieve. – jrok Aug 01 '13 at 21:03
I modified my hierarchy so that I don't have that nested class anymore. Thanks for your help! =) – fumoboy007 Aug 02 '13 at 03:36

Template friend function instantiation

1 Answers1