For two class template partial specializations, the first is more specialized than the second if, given the following rewrite to two function templates, the first function template is more specialized than the second according to the ordering rules for function templates:
Each of the two function templates has the same template parameters as the corresponding partial specialization.
Each function template has a single function parameter whose type is a class template specialization where the template arguments are the corresponding template parameters from the function template for each template argument in the template-argument-list of the simple-template-id of the partial specialization.
[ Example:
template<int I, int J, class T> class X { };
template<int I, int J> class X<I, J, int> { }; // #1
template<int I> class X<I, I, int> { }; // #2
template<int I0, int J0> void f(X<I0, J0, int>); // A
template<int I0> void f(X<I0, I0, int>); // B
template <auto v> class Y { };
template <auto* p> class Y<p> { }; // #3
template <auto** pp> class Y<pp> { }; // #4
template <auto* p0> void g(Y<p0>); // C
template <auto** pp0> void g(Y<pp0>); // D
According to the ordering rules for function templates, the function template B is more specialized than the function template A and the function template D is more specialized than the function template C. Therefore, the partial specialization #2 is more specialized than the partial specialization #1 and the partial specialization #4 is more specialized than the partial specialization #3. — end example ]