Question

I have a class

template <typename T>
struct Trait { typedef std::false_type IsGood; };

template <>
struct Trait<int> { typedef std::true_type IsGood; };

A call like this fails to compile on MSVC 2010

template <typename T, typename Enable = void> class Foo;

template <typename T>
class Foo <T, std::enable_if<typename Trait<T>::IsGood::value>::type>
{};

// This fails as well
template <typename T>
class Foo <T, typename std::enable_if<Trait<T>::IsGood::value>::type>
{};

// And this fails horribly
template <typename T>
class Foo <T, typename std::enable_if<typename Trait<T>::IsGood::value>::type>
{};

while

template <typename T>
class Foo <T, typename std::enable_if<std::is_same<std::true_type, 
    typename Trait<T>::IsGood>::value>::type>
{};

works — why?

The error message is:

main.cpp(12): error C2039: 'type' : is not a member of 'std::tr1::enable_if<_Test>'
          with
          [
              _Test=false
          ]
main.cpp(12): error C2146: syntax error : missing ',' before identifier 'type'
main.cpp(12): error C2065: 'type' : undeclared identifier
main.cpp(13): error C2976: 'Foo' : too few template arguments

Answer 1

You’re using typename at wrong place. This is correct:

template <typename T>
class Foo <T, typename std::enable_if<Trait<T>::IsGood::value>::type>
{};        // ^^^^^^^ here should be typename

Now it compiles fine: http://ideone.com/0SwO9

But you’re using typename as:

template <typename T>
class Foo <T, std::enable_if<typename  Trait<T>::IsGood::value>::type>
{};                        //^^^^^^^ wrong place

Trait<T>::IsGood::value is not a type, So you cannot apply typename on it.

GCC error message is very much clear:

prog.cpp:12:62: error: type/value mismatch at argument 1 in template parameter list for ‘template<bool <anonymous>, class _Tp> struct std::enable_if’

See yourself : http://ideone.com/9ujJv