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:

template
struct Point
{
  // Check if T is int and call MyFunction for int
  void MyFunction(typename std::enable_if<:is_same int>::value, T &>::type* = 0)
  {
    std::cout ::value, float &>::type* = 0)
  {
    std::cout 
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:
template
struct Point
{
  template
  typename std::enable_if<:is_same int>::value>::type
    MyFunction()
  {
    std::cout 
  typename std::enable_if<:is_same float>::value>::type
    MyFunction()
  {
    std::cout