How to Use enable_if to Select Member Functions Based on Template Arguments?

Selecting a Member Function Using Different enable_if Conditions

The enable_if metafunction is utilized to specify template function parameters and select appropriate member functions based on them. Consider the following code:

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

  // Check if T is not int and call MyFunction for non-int
  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 may cause compiler errors indicating that "no type named ‘type’ in ‘struct std::enable_if’".

Understanding enable_if

enable_if ensures that only viable function overloads are considered during overload resolution. If a template argument substitution fails, that overload is removed from the candidate set.

In the example above, the template argument T is already known when instantiating the member functions. To implement the desired behavior, we can create a dummy template argument defaulted to T and perform SFINAE using it:

<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>

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