std::enable_if string/char type

Question

Possible Duplicate:
“What happened to my SFINAE” redux: conditional template class members?

Why can't I pass other parameters to my template class? I'm trying to enable a specific parameter only if the type passed is a literal type. If it isn't, accept other types but do not enable the Case-Sensitive parameter.

Why does the below not compile?

#include <iostream>
#include <type_traits>

template<typename T>
struct is_literal
{
   enum{value = false};
};

template<>
struct is_literal<char>
{
   enum{value = true};
};

template<>
struct is_literal<char*>
{
   enum{value = true};
};

template<>
struct is_literal<const char*>
{
   enum{value = true};
};

template<typename Char, typename Traits, typename Alloc>
struct is_literal<std::basic_string<Char, Traits, Alloc>>
{
   enum
   {
      value = true
   };
};

template<typename T>
class Test
{
    public:
        bool Contains(T DataType, typename std::enable_if<is_literal<T>::value, bool>::type  CaseSensitive = true);
};

template<typename T>
bool Test<T>::Contains(T DataType, typename std::enable_if<is_literal<T>::value, bool>::type CaseSensitive)
{
    return true;
}


int main()
{
    Test<int> F;    //This line gives errors.. It gives none if I pass char, char*, const char*, std::string.
    F.Contains(1);
}

Answer 1

SFINAE can only occur when substitution is performed during overload resolution. That means that you must have a function template. In this case, you have a non-template function in a class template. That does not work. You can add a default template parameter to fix this.

template <typename U = T>
bool Contains(U DataType, typename std::enable_if<is_literal<U>::value, bool>::type  CaseSensitive = true);