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How to Choose a Function Based on Template Parameters Using enable_if: Why does the \'no type named \'type\' in 'struct std::enable_if'\' Error Occur and How Can It Be Resolved?

Susan Sarandon
Release: 2024-10-26 12:00:05
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How to Choose a Function Based on Template Parameters Using enable_if: Why does the

How to Choose a Function Based on Template Parameters Using Enable_if

Problem:

Determining which version of a class member function should be called based on the template parameter of the class can be challenging. Using enable_if can be a solution, but you may encounter errors such as "no type named 'type' in ‘struct std::enable_if’".

Solution:

The issue lies in the substitution of template arguments. enable_if works by eliminating overloads that result in errors during template argument substitution. In the provided code, no substitution occurs because T is already known at the time of member function instantiation.

To resolve this, create a dummy template argument defaulted to T, and use that for SFINAE (Substitution Failure Is Not An Error). Replace the problematic code with the following:

<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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Note:

To prevent users from explicitly specifying template arguments and potentially getting incorrect results, you can add a static assertion to the member functions, as seen in the following example:

<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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