How to Explicitly Specialize a Member Function of a Class Template?

Explicit Specialization of a Member Function for a Class Template

Problem Statement:

When defining an explicit specialization of a member function for a class template, one must also explicitly specialize the surrounding class template to avoid compiler errors.

Code Example:

Consider the following code snippet:

template <class C> class X
{
public:
   template <class T> void get_as();
};

template <class C>
void X<C>::get_as<double>()
{
}

int main()
{
   X<int> x;
   x.get_as();
}

This code triggers compiler errors because the surrounding class template X is not explicitly specialized.

Solution:

To resolve this issue, we must explicitly specialize the surrounding class template, as shown below:

template <> template <>
void X<int>::get_as<double>()
{
}

This specialized member function applies only to X and not to other instantiations of X.

Alternative Approach:

An alternative approach is to use function overloads, as seen in the following code:

template <class C> class X
{
   template<typename T> struct type { };

public:
   template <class T> void get_as() { get_as(type<T>()); }

private:
   template<typename T> void get_as(type<T>) {}

   void get_as(type<double>) {}
};

This method allows for explicit specialization without requiring the surrounding class template to be specialized.

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