Question

It is possible to specialize some class member function outside the template definition:

template<class A>
struct B {
   void f();   
};

template<>
void B<int>::f() { ... }

template<>
void B<bool>::f() { ... }

and in this case I can even omit definition of function f for a general type A.

But how to put this specializations inside the class? Like this:

template<class A>
struct B {
   void f();   

   void f<int>() { ... }
   void f<bool>() { ... }
};

What syntax should I use in this case?

EDIT:
For now the solution with fewest lines of code is to add a fake template function f definition and explicitly call it from original function f:

template<class A>
struct B {
   void f() { f<A>(); }

   template<class B> 
   void f();

   template<> 
   void f<int>() { ... }

   template<> 
   void f<bool>() { ... }
};

Answer 1

You can make B::f a template function within your struct:

struct B {
    template <typename T>
    void f();

    template<>
    void f<int>() { ... }

    template<>
    void f<bool>() { ... }
};

Edit:

According to your comment this may help you, but I’ve not tested if it works:

template <typename A>
struct B {
    template <typename T = A>
    void f() { ... }

    template<>
    void f<int>() { ... }

    template<>
    void f<bool>() { ... }
};