Template class methods definition with enable_if as template parameter

Question

I asked this question earlier where a solution was presented. The solution is great as far as the question is concerned, but now I am confused on how I would define the methods

Answer 1

Here's how SFINAE can actually work with partial specialization:

template
struct Foo {
    /* catch-all primary template */
    /* or e.g. leave undefined if you don't need it */
};

template
struct Foo::value>::type> {
    /* matches types derived from BasePolicy */
    Foo();
};

The definition for that constructor can then be awkwardly introduced with:

template
Foo::value>::type>::Foo()
{
    /* Phew, we're there */
}

If your compiler supports template aliases (it's a C++11 feature) that then you can cut a lot of the verbosity:

template
using EnableIfPolicy = typename std::enable_if::value>::type;

// Somewhat nicer:

template
struct Foo> {
    Foo();
};

template
Foo>::Foo() {}

Note: your original answer referred to utilies from Boost, like boost::enable_if_c and boost::is_base_of. If you're using that instead of std::enable_if and std::is_base_of (which are from C++11), then usage looks like

typename boost::enable_if >::type

which has the advantage of getting rid of one ::value.