Why Does SFINAE Fail for Class Template Member Functions?

SFINAE Fails for Class Template Member Functions

The Substitute Failure, Is Not An Error (SFINAE) mechanism, commonly used in template metaprogramming, appears to exhibit peculiar behavior when applied to class template member functions.

The Problem

Consider the following code snippet:

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

struct A{};
struct B{};

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

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

This code attempts to define two overloads of bar() in the Foo class template, with SFINAE used to conditionally enable each overload based on the value of T. However, the code fails to compile with the following error:

<code class="cpp">error: 'typename std::enable_if<std::is_same<T, B>::value>::type Foo<T>::bar()' cannot be overloaded</code>

The Explanation

SFINAE is typically used to enable or disable template specializations based on template arguments. However, SFINAE only applies to deduced template arguments, meaning arguments that are automatically deduced during overload resolution. In the case of member functions, template arguments are not deduced but rather explicitly specified when instantiating the class. Therefore, SFINAE is not applicable to member functions.

The Solution

There are two main ways to address this issue:

• Use function templates: Define separate function templates for each overload, as shown below:

<code class="cpp">template <typename T>
void bar(Foo<T><- A) {}

template <typename T>
void bar(Foo<T><- B) {}

• Use explicit class template specialization: Define separate class templates for each overload, as shown below:

<code class="cpp">template <typename> struct Foo;

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

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