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>
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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