Why Doesn't SFINAE Work with Member Functions of Class Templates?

Why doesn't SFINAE (enable_if) work for member functions of a class template?

In C , SFINAE (Substitution Failure Is Not An Error) allows you to enable or disable code depending on the type of a template argument. However, when dealing with member functions of a class template, SFINAE often doesn't work as expected.

Here's an example that demonstrates the issue:

<code class="cpp">#include <type_traits>

struct A {};
struct B {};

template <typename T>
struct Foo
{
    typename std::enable_if<std::is_same<T, A>::value>::type bar()
    {}

    typename std::enable_if<std::is_same<T, B>::value>::type bar()
    {}
};</code>

In this example, Foo defines two overloaded member functions bar(). The first overload is enabled when T is A, and the second is enabled when T is B. However, if you try to compile this code, you'll get an error message indicating that the overloads cannot be resolved.

The reason for this error is that SFINAE only works for deduced template arguments. In the case of member functions of a class template, the template argument is not deduced but rather specified explicitly. To fix the issue, you can use one of the following techniques:

• Use explicit template arguments:

  <code class="cpp">struct Foo
  {
    void bar(A) {}
    void bar(B) {}
  };</code>

• Use std::enable_if inside the member functions:

  <code class="cpp">template <typename T>
  struct Foo
  {
    template<typename U = T>
    typename std::enable_if<std::is_same<U, A>::value>::type bar() {}
  
    template<typename U = T>
    typename std::enable_if<std::is_same<U, B>::value>::type bar() {}
  };</code>

• Use explicit class template specialization:

  <code class="cpp">template <> struct Foo<A> { void bar() {} };
  template <> struct Foo<B> { void bar() {} };</code>

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