std::enable_if to conditionally compile a member function

Question

I am trying to get a simple example to work to understand how to use std::enable_if. After I read this answer, I thought it shouldn't be too hard to come up with a simple example. I want to use std::enable_if to choose between two member-functions and allow only one of them to be used.

Unfortunately, the following doesn't compile with gcc 4.7 and after hours and hours of trying I am asking you guys what my mistake is.

#include <utility>
#include <iostream>

template< class T >
class Y {

    public:
        template < typename = typename std::enable_if< true >::type >
        T foo() {
            return 10;
        }
        template < typename = typename std::enable_if< false >::type >
        T foo() {
            return 10;
        }

};


int main() {
    Y< double > y;

    std::cout << y.foo() << std::endl;
}

gcc reports the following problems:

% LANG=C make CXXFLAGS="-std=c++0x" enable_if
g++ -std=c++0x    enable_if.cpp   -o enable_if
enable_if.cpp:12:65: error: `type' in `struct std::enable_if<false>' does not name a type
enable_if.cpp:13:15: error: `template<class T> template<class> T Y::foo()' cannot be overloaded
enable_if.cpp:9:15: error: with `template<class T> template<class> T Y::foo()'

Why doesn't g++ delete the wrong instantiation for the second member function? According to the standard, std::enable_if< bool, T = void >::type only exists when the boolean template parameter is true. But why doesn't g++ consider this as SFINAE? I think that the overloading error message comes from the problem that g++ doesn't delete the second member function and believes that this should be an overload.

Answer 1

SFINAE only works if substitution in argument deduction of a template argument makes the construct ill-formed. There is no such substitution.

I thought of that too and tried to use std::is_same< T, int >::value and ! std::is_same< T, int >::value which gives the same result.

That's because when the class template is instantiated (which happens when you create an object of type Y<int> among other cases), it instantiates all its member declarations (not necessarily their definitions/bodies!). Among them are also its member templates. Note that T is known then, and !std::is_same< T, int >::value yields false. So it will create a class Y<int> which contains

class Y<int> {
    public:
        /* instantiated from
        template < typename = typename std::enable_if< 
          std::is_same< T, int >::value >::type >
        T foo() {
            return 10;
        }
        */

        template < typename = typename std::enable_if< true >::type >
        int foo();

        /* instantiated from

        template < typename = typename std::enable_if< 
          ! std::is_same< T, int >::value >::type >
        T foo() {
            return 10;
        }
        */

        template < typename = typename std::enable_if< false >::type >
        int foo();
};

The std::enable_if<false>::type accesses a non-existing type, so that declaration is ill-formed. And thus your program is invalid.

You need to make the member templates' enable_if depend on a parameter of the member template itself. Then the declarations are valid, because the whole type is still dependent. When you try to call one of them, argument deduction for their template arguments happen and SFINAE happens as expected. See this question and the corresponding answer on how to do that.