Specializing member template for enum type arguments

Question

In the code below, Foo<T>::setValue works well for my purposes, except in cases where where T is a class enum named TYPE e.g. Bar::TYPE and Baz:TYPE.

Therefore, I'd appreciate help specializing Foo<T>::setValue without naming Bar and Baz, because there could be dozens of such classes.

class Bar
{
public:
    enum TYPE{ ONE , TWO };
};

class Baz
{
public:
    enum TYPE{ SIX , TEN };
};

template<typename T>
class Foo
{
public:
    void setValue(){} // Need a different setValue if T is a class enum

private:
    T m_value;
};


int main()
{
    Foo<int> f1;
    Foo<Bar::TYPE> f2;
    Foo<Baz::TYPE> f3; 
    return EXIT_SUCCESS;
}

Answer 1

You can do this with something like:

#include <type_traits>

template<typename T>
class Foo
{
public:
    void setValue() {
      setValueImpl<T>();
    }
private:
    template <class X>
    typename std::enable_if<std::is_enum<X>::value, void>::type
    setValueImpl() { std::cout << "Is enum" << std::endl; }

    template <class X>
    typename std::enable_if<!std::is_enum<X>::value, void>::type
    setValueImpl() { std::cout << "Not enum" << std::endl; } 

    T m_value;
};

Where enable_if picks which version to use based on the is_enum type trait.

The example used C++11 enable_if and is_enum but boost has similar for pre-C++11 also.

Answer 2

Consider this:

#include <iostream>

class Bar
{
public:
    enum TYPE{ ONE , TWO };
};

class Baz
{
public:
    enum TYPE{ SIX , TEN };
};

template<typename T>
class Foo
{
public:
    template<typename C> void setValue(const C &m_value, ...) 
    {
        std::cout << "normal" << std::endl;
    }

    template<typename C> void setValue(const C &m_value, typename C::TYPE fake = C::TYPE()) 
    {
        std::cout << "TYPE'ed" << std::endl;
    }

private:
    T m_value;
};


int main()
{
    Foo<int> f1;
    Foo<Bar> f2;
    Foo<Baz> f3; 

    f1.setValue(1);
    f2.setValue(Bar());
    f3.setValue(Baz());

    return 0;
}