[Boost-users] Re: enable_if for function templates

From: Jonathan Turkanis (technews_at_[hidden])
Date: 2005-01-04 19:24:20

Agoston Bejo wrote:
> Hi all,
> I am trying to enable only two version of a function. The caller may
> choose one
>
> of the two versions by specifying a template parameter. I've got two
> versions

> #include <iostream>
> #include <boost/utility/enable_if.hpp>
> #include <boost/type_traits.hpp>
>
> using namespace std;
> using namespace boost;
>
>
> template<unsigned int N, typename T>
> enable_if_c<N==1>
> f(T t)
> {
>  cout << 1 << endl;
> }
>
> template<unsigned int N, typename T>
> enable_if_c<N==2>
> f(T t)
> {
>  cout << 2 << endl;
> }
>
> int _tmain(int argc, _TCHAR* argv[])

 _tmain, _TCHAR ... what are these funny symbols? ;-)

> {
>  f<1>(10); // ERROR
>  f<2>(5.5); // ERROR
>  f<3>(7); // ERROR
>  return 0;
> }

I'd say enable_if is not the right tool for this situation. Try:

    template<int N>
    struct f_impl {
        template<typename T>
        static void execute(T t)
        {
            cout << "f_impl::execute()" << "\n";
        }
    };

    template<int N, typename T>
    void f(T t)
    {
        f_impl<N>::execute(t);
    }

    template<>
    struct f_impl<1> {
        template<typename T>
        static void execute(T t)
        {
            cout << "1" << "\n";
        }
    };

    template<>
    struct f_impl<2> {
        template<typename T>
        static void execute(T t)
        {
            cout << "2" << "\n";
        }
    };

    int main()
    {
        f<1>(10);
        f<2>(5.5);
        f<3>(7);
        return 0;
    }

Jonathan