Template specialization for template function members combined with class template

Question

Given the following:

class MyClass
{
   public:
      template <typename T, typename U = T>
      void doSomething(T) {}
   private:
      double parser;
};

template<>
inline void MyClass::doSomething<bool>(bool) {}
template<>
inline void MyClass::doSomething<int, double>(int) {}

int main(int argc, char* argv[])
{
   MyClass foo;
   foo.doSomething<bool>(true);
   foo.doSomething<int, double>(10);
   return 0;
}

This works fine.

However now I want to introduce a class template in order to use a template parameter instead of ParserClass, like so:

template<typename P>
class MyClass
{
   public:
      template <typename T, typename U = T>
      void doSomething(T) {}
   private:
      P parser;
};

template<>
inline void MyClass::doSomething<bool>(bool) {}
template<>
inline void MyClass::doSomething<int, double>(int) {}

int main(int argc, char* argv[])
{
   MyClass<double> foo;
   foo.doSomething<bool>(true);
   foo.doSomething<int, double>(10);
   return 0;
}

Which, however, does no longer work:

test.cpp:12:13: error: 'MyClass' is not a class, namespace, or enumeration
inline void MyClass::doSomething<bool>(bool) {}
            ^
test.cpp:2:7: note: 'MyClass' declared here
class MyClass
      ^
1 error generated.

Is it even possible? How do I combine the member function template parameters and the class template parameters in the above example?

Answer 1

To specialize a function that is a member of a templated class, you need two template<> declarations. Furthermore, you need to specify the typename for MyClass:

template<> template<>
inline void MyClass<some_type>::doSomething<bool>(bool) { ... }
template<> template<>
inline void MyClass<some_type>::doSomething<int, double>(int) { ... }