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How to Select Member Functions Based on Class Template Parameters Using enable_if in C ?

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2024-10-26 05:50:30208browse

How to Select Member Functions Based on Class Template Parameters Using enable_if in C  ?

Selecting a Member Function Using Different enable_if Conditions

In C , the enable_if metafunction can be used to selectively enable or disable code based on compile-time conditions. This technique is often employed to provide overloads of member functions that handle different types of arguments or perform different actions depending on template parameters.

When attempting to use enable_if to choose between member functions based on class template parameters, it's important to understand that enable_if works by causing substitutions of template arguments that result in errors to be dropped from the overload resolution set.

Problem:

In the provided code, the enable_if conditions are applied directly to the member functions, but this approach doesn't achieve the desired effect because the template argument T is already known when instantiating these functions.

<code class="cpp">template<typename T>
struct Point
{
  void MyFunction(typename std::enable_if<std::is_same<T, int>::value, T &>::type* = 0)
  {
    std::cout << "T is int." << std::endl;
  }

  void MyFunction(typename std::enable_if<!std::is_same<T, int>::value, float &>::type* = 0)
  {
    std::cout << "T is not int." << std::endl;
  }
};</code>

Solution:

To remedy this issue, create a dummy template argument that defaults to T and use that to perform SFINAE (Substitution Failure Is Not An Error) checks. This approach allows the compiler to determine which member function to invoke based on the type of the dummy argument.

<code class="cpp">template<typename T>
struct Point
{
  template<typename U = T>
  typename std::enable_if<std::is_same<U, int>::value>::type
    MyFunction()
  {
    std::cout << "T is int." << std::endl;
  }

  template<typename U = T>
  typename std::enable_if<std::is_same<U, float>::value>::type
    MyFunction()
  {
    std::cout << "T is not int." << std::endl;
  }
};</code>

Preventing Explicit Template Argument Specialization:

As noted by HostileFork, it's possible for users to explicitly specify template arguments for member functions, which could lead to incorrect results. To prevent this, use a static_assert to check that no explicit arguments are provided:

<code class="cpp">template<typename T>
struct Point
{
  template<typename... Dummy, typename U = T>
  typename std::enable_if<std::is_same<U, int>::value>::type
    MyFunction()
  {
    static_assert(sizeof...(Dummy)==0, "Do not specify template arguments!");
    std::cout << "T is int." << std::endl;
  }

  template<typename... Dummy, typename U = T>
  typename std::enable_if<std::is_same<U, float>::value>::type
    MyFunction()
  {
    static_assert(sizeof...(Dummy)==0, "Do not specify template arguments!");
    std::cout << "T is not int." << std::endl;
  }
};</code>

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