SFINAE trouble

Okay, so I always fail at applying std::enable_if properly for some reason. So maybe someone here who is better at this can tell why this isn't working?

Code:

---

template<bool B>
struct Test
{
        template<typename T = void>
        std::enable_if_t<B> test1() { std::cout << "test1\n"; }

        template<typename T = void>
        std::enable_if_t<!B> test2() { std::cout << "test2\n"; }
};

int main()
{
        Test<true> a;
        Test<false> b;
}

---

Code:

---

Error        1        error C2039: 'type' : is not a member of 'std::enable_if<false,void>'
Error        2        error C2061: syntax error : identifier 'type'
Error        3        error C2988: unrecognizable template declaration/definition       
Error        4        error C2059: syntax error : '<end Parse>'
Error        5        error C2143: syntax error : missing ';' before '}'
Error        6        error C2238: unexpected token(s) preceding ';'
Error        7        error C2061: syntax error : identifier 'type'
Error        8        error C2988: unrecognizable template declaration/definition
Error        9        error C2059: syntax error : '<end Parse>'
Error        10        error C2334: unexpected token(s) preceding '{'; skipping apparent function body

---

Oh! I recall now. The enable_if construct has to depend on a template parameter. So if we do:

Code:

---

template<bool B>
struct Test
{
    template<bool B2 = B>
    std::enable_if_t<B2> test1() { std::cout << "test1\n"; }
 
    template<bool B2 = B>
    std::enable_if_t<!B2> test2() { std::cout << "test2\n"; }
};
 
int main()
{
    Test<true> a;
    Test<false> b;
}

---

Then it starts working. I should keep this for future reference.

I think you want to do

Code:

---

template<bool B>
struct Test
{
    template<bool B2=B>
    typename std::enable_if_t<B2>::type test1() { std::cout << "test1\n"; }
 
    template<bool B2=B>
    typename std::enable_if_t<!B2>::type test2() { std::cout << "test2\n"; }
};

---

enable_if::type will only be defined if the first template argument is true. You can also pass in a second template parameter, the typename that ::type should be typedefed to, which defaults to void.

std::enable_if_t<B> == typename std::enable_if<B>::type

Ah missed the _t, serves me for reading code on an empty stomach...