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How to use enable_if to enable member functions based on template parameter of class

Asked 2010-11-11T20:00:24.100
14

In code:

template<class T>
struct is_builtin
{
    enum {value = 0};
};

template<>
struct is_builtin<char>
{
    enum {value = 1};
};

template<>
struct is_builtin<int>
{
    enum {value = 1};
};

template<>
struct is_builtin<double>
{
    enum {value = 1};
};

template<class T>
struct My
{
    typename enable_if<is_builtin<T>::value,void>::type f(T arg)
    {
        std::cout << "Built-in as a param.\n";
    }


    typename enable_if<!is_builtin<T>::value,void>::type f(T arg)
    {
        std::cout << "Non - built-in as a param.\n";
    }
};

struct A
{
};

int main()
{
    A a;
    My<int> m;
    My<A> ma;
    m.f(1);
    ma.f(a);
    return 0;
}

I'm getting an error:

error C2039: 'type' : is not a member of 'std::tr1::enable_if<_Test,_Type>'    

Obviously I don't understand how to use enable_if. What I was thinking was that I can enable one or the second one member function from a set of member functions during compilation time but it does not work. Could anyone please explain to me how to do it correctly?
Edited
What I really can't understand is why isn't there typedef in one of those def. Compiler cannot find it and it wont compile it.

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1 Answer

-3

enable_if expects a metafunction. To use a bool you need enable_if_c. I'm surprised you're not getting errors explaining THAT problem.

You can fix your metafunction by declaring a 'type' typedef inside that is simply itself. Then you can use boost::enable_if<is_builtin<T>>::type

answered 2010-11-11T20:35:40.223

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