[temp.spec.partial.order] (original) (raw)

13 Templates [temp]

13.7 Template declarations [temp.decls]

13.7.6 Partial specialization [temp.spec.partial]

13.7.6.3 Partial ordering of partial specializations [temp.spec.partial.order]

For two 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 and associated constraintsas 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-listof the simple-template-idof the partial specialization.

[Example 1: template<int I, int J, class T> class X { };template<int I, int J> class X<I, J, int> { }; template<int I> class X<I, I, int> { }; template<int I0, int J0> void f(X<I0, J0, int>); template<int I0> void f(X<I0, I0, int>); template <auto v> class Y { };template <auto* p> class Y<p> { }; template <auto** pp> class Y<pp> { }; template <auto* p0> void g(Y<p0>); template <auto** pp0> void g(Y<pp0>);

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_]

[Example 2: template<typename T> concept C = requires (T t) { t.f(); };template<typename T> concept D = C<T> && requires (T t) { t.f(); };template<typename T> class S { };template<C T> class S<T> { }; template<D T> class S<T> { }; template<C T> void f(S<T>); template<D T> void f(S<T>);

The partial specialization #2 is more specialized than #1 because B is more specialized than A.

— _end example_]