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<:is_same int>::value, T >::type* = 0) {
    std::cout ::value, float >::type* = 0) {
    std::cout <p>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.</p>
<p>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:</p>
<pre class="brush:php;toolbar:false"><code class="cpp">template<typename t>
struct Point {
  template<typename u="T">
  typename std::enable_if<:is_same int>::value>::type MyFunction() {
    std::cout 
  typename std::enable_if<:is_same float>::value>::type MyFunction() {
    std::cout <p>This approach ensures that the correct version of MyFunction is selected based on the value of T.</p></:is_same></:is_same></typename></typename></code>

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