How to Implement Conditional Member Function Overloads Using enable_if in C ?

Selecting a Member Function with Different enable_if Conditions

In C , enable_if is a tool used to conditionally enable or disable certain code based on whether a template argument meets specific criteria. This can be useful when you want to customize the behavior of a class or function based on its template parameters.

In the given example, the goal is to create a member function MyFunction that behaves differently based on whether the template parameter T is an integer or not. The intended implementation is to use two overloads of MyFunction, one for T = int and one for T != int.

One approach to achieve this is through enable_if, as shown in the code below:

<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>

However, this code will result in compilation errors due to the incorrect use of enable_if. In C , the substitution of template arguments takes place during overload resolution. In this case, there is no substitution occurring because the type of T is known at the time of member function instantiation.

To fix this issue, a dummy template parameter can be introduced and defaulted to T, allowing for SFINAE (Substitution Failure Is Not An Error) to work properly:

<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>

This approach ensures that the correct version of MyFunction is selected based on the value of T.

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