Out-of-line method definition in template specialization by samaursa

The following code is, as far as I know, standard C++. It compiles on gcc but Visual Studio 2008 and 2010 fail to compile it.

    namespace cool
    {
     template <bool, typename = void> struct enable_if {};
     template <typename T> struct enable_if<true, T> { typedef T type; };
    
     template <typename T0, typename T1> struct is_same { enum { value = false }; };
     template <typename T> struct is_same<T, T> { enum { value = true }; };
    }
    
    struct BasePolicy {};
    struct BasePolicy2 {};
    struct Faz {};
    
    template <typename Policy,
    typename = typename cool::enable_if<cool::is_same<BasePolicy, Policy>::value || cool::is_same<BasePolicy2, Policy>::value>::type >
    struct Foo;
    
    template <typename Policy>
    struct Foo<Policy> {
     Foo();
    };
    
    template <typename Policy>
    Foo<Policy>::Foo() {
    }
    
    int main()
    {
     Foo<BasePolicy2> fb;
    }

Moving the constructor BACK to the class itself (i.e. the definition is now in-line so that the declaration is the definition as well) fixes the issue:

namespace cool
{
template <bool, typename = void> struct enable_if {};
template <typename T> struct enable_if<true, T> { typedef T type; };

template <typename T0, typename T1> struct is_same { enum { value = false }; };
template <typename T> struct is_same<T, T> { enum { value = true }; };
}

struct BasePolicy {};
struct BasePolicy2 {};
struct Faz {};

template <typename Policy,
typename = typename cool::enable_if<cool::is_same<BasePolicy, Policy>::value || cool::is_same<BasePolicy2, Policy>::value>::type >
struct Foo;

template <typename Policy>
struct Foo<Policy> {
Foo() {}
};

int main()
{
Foo<BasePolicy2> fb;
}