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

Template Class with Explicitly Specialized Member Function

When defining an explicit specialization for a member function of a class template, a common pitfall occurs when the surrounding class template is also a template. To understand the issue, consider the following example:

#include <iostream>
#include <cmath>

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

template <class C>
void X<C>::get_as<double>() {}  // Explicit specialization for double

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

This code attempts to explicitly specialize the get_as member function for double for the X class template. However, this approach leads to compiler errors because the surrounding class template is not explicitly specialized.

Incorrect Approach:

Attempting to explicitly specialize the member function without specializing the class template, as shown below, is incorrect:

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

Correct Solution:

To fix the issue, both the class template and the member function must be explicitly specialized. For example, to specialize the get_as function only for X, use the following approach:

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

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

Alternative Option:

Alternatively, to keep the surrounding class template unspecialized, an overload of the get_as function can be used, as demonstrated 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>) {}
};

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