Exact rules for matching variadic template template parameters in partial template specialization

Question

While creating this answer for another question I came around the following issue. Consider this program (godbolt):

#include <variant>
#include <iostream>

template <typename T>
struct TypeChecker {
    void operator()() {
        std::cout << "I am other type\n";
    }
};

template<typename... Ts, template<typename...> typename V>
requires std::same_as<V<Ts...>, std::variant<Ts...>>
struct TypeChecker<V<Ts...>>
{
    void operator()()
    {
        std::cout << "I am std::variant\n";
    }
};

int main()
{
    TypeChecker<std::variant<int, float>>{}();
    TypeChecker<int>{}();
}

The output (which is also expected) is the following (with clang 14.0.0 as well as with gcc 12.1):

I am std::variant
I am other type

If however the three dots in the parameter list of the template template are removed, like this (whole program live on godbolt):

template<typename... Ts, template<typename> typename V>

,then the output is different for clang and gcc. The clang 14.0.0 compiled program outputs

I am other type
I am other type

whereas the gcc 12.1 compiled program outputs

I am std::variant
I am other type

It seems that using the non-variadic template template exhibits different matching rules in clang and gcc. So my question is, which behavior is correct if it is even well defined, and why?

Answer 1

Since defect report resolution P0522R0 was adopted, exact matching of template parameter lists for template template parameter match is no longer needed, and correct output according to the standard is:

I am std::variant
I am other type

In the current draft (which also contains changes related to C++20 concepts) relevant standard excerpts are temp.arg.template#3-4 (bold emphasis mine):

3. A template-argument matches a template template-parameter P when P is at least as specialized as the template-argument A. In this comparison, if P is unconstrained, the constraints on A are not considered. If P contains a template parameter pack, then A also matches P if each of A's template parameters matches the corresponding template parameter in the template-head of P. Two template parameters match if they are of the same kind (type, non-type, template), for non-type template-parameters, their types are equivalent ([temp.over.link]), and for template template-parameters, each of their corresponding template-parameters matches, recursively. When P's template-head contains a template parameter pack ([temp.variadic]), the template parameter pack will match zero or more template parameters or template parameter packs in the template-head of A with the same type and form as the template parameter pack in P (ignoring whether those template parameters are template parameter packs).
4. A template template-parameter P is at least as specialized as a template template-argument A if, given the following rewrite to two function templates, the function template corresponding to P is at least as specialized as the function template corresponding to A according to the partial ordering rules for function templates. Given an invented class template X with the template-head of A (including default arguments and requires-clause, if any):
   • (4.1) Each of the two function templates has the same template parameters and requires-clause (if any), respectively, as P or A.
   • (4.2) Each function template has a single function parameter whose type is a specialization of X with template arguments corresponding to the template parameters from the respective function template where, for each template parameter PP in the template-head of the function template, a corresponding template argument AA is formed. If PP declares a template parameter pack, then AA is the pack expansion PP... ([temp.variadic]); otherwise, AA is the id-expression PP. If the rewrite produces an invalid type, then P is not at least as specialized as A.

So, as we see, exact (up to special rules for parameter pack) parameter matching is now considered only in the case parameter list of template template parameter contains a pack (like your original example), otherwise only the new at least as specialized as relation is used to test matching, which defines for both template template parameter and argument respective function templates and tests whether parameter-induced function is at least as specialized as argument-induced function according to partial ordering rules for template functions.

In particular, matching A=std::variant to P=template<typename> typename V we get these corresponding function templates:

template<typename...> class X;
template<typename T> void f(X<T>); // for P
template<typename... Ts> void f(X<Ts...>); // for A

So, to prove A matches P, we need to prove f(X<T>) is at least as specialized as f(X<Ts...>). temp.func.order#2-4 says:

2. Partial ordering selects which of two function templates is more specialized than the other by transforming each template in turn (see next paragraph) and performing template argument deduction using the function type. The deduction process determines whether one of the templates is more specialized than the other. If so, the more specialized template is the one chosen by the partial ordering process. If both deductions succeed, the partial ordering selects the more constrained template (if one exists) as determined below.
3. To produce the transformed template, for each type, non-type, or template template parameter (including template parameter packs thereof) synthesize a unique type, value, or class template respectively and substitute it for each occurrence of that parameter in the function type of the template. ...
4. Using the transformed function template's function type, perform type deduction against the other template as described in [temp.deduct.partial].

Transformed template of f(X<T>) is f(X<U1>) and of f(X<Ts...>) is f(X<U2FromPack>), where U1 and U2FromPack are two synthesized unique types. Now, temp.deduct.partial#2-4,8,10 says:

2. Two sets of types are used to determine the partial ordering. For each of the templates involved there is the original function type and the transformed function type. [Note 1: The creation of the transformed type is described in [temp.func.order]. — end note] The deduction process uses the transformed type as the argument template and the original type of the other template as the parameter template. This process is done twice for each type involved in the partial ordering comparison: once using the transformed template-1 as the argument template and template-2 as the parameter template and again using the transformed template-2 as the argument template and template-1 as the parameter template.
3. The types used to determine the ordering depend on the context in which the partial ordering is done:
   • (3.1) In the context of a function call, the types used are those function parameter types for which the function call has arguments.130
   • (3.2) In the context of a call to a conversion function, the return types of the conversion function templates are used.
   • (3.3) In other contexts the function template's function type is used.
4. Each type nominated above from the parameter template and the corresponding type from the argument template are used as the types of P and A.

8. Using the resulting types P and A, the deduction is then done as described in [temp.deduct.type]. ... If deduction succeeds for a given type, the type from the argument template is considered to be at least as specialized as the type from the parameter template.

10. Function template F is at least as specialized as function template G if, for each pair of types used to determine the ordering, the type from F is at least as specialized as the type from G. F is more specialized than G if F is at least as specialized as G and G is not at least as specialized as F.

Now, in our case, per 3.3, the function type itself is the only one considered among types to determine ordering. So, per 2, 8 and 10, to know whether f(X<T>) is at least as specialized as f(X<Ts...>) we need to see whether void(X<T>) is at least as specialized as void(X<Ts...>), or, equivalently, whether deduction from type for P=void(X<Ts...>) from A=void(X<U1>) succeeds. It does according to temp.deduct.type#9-10:

9. If P has a form that contains <T> or <i>, then each argument P_i of the respective template argument list of P is compared with the corresponding argument A_i of the corresponding template argument list of A. If the template argument list of P contains a pack expansion that is not the last template argument, the entire template argument list is a non-deduced context. If P_i is a pack expansion, then the pattern of P_i is compared with each remaining argument in the template argument list of A. Each comparison deduces template arguments for subsequent positions in the template parameter packs expanded by P_i. ...
10. Similarly, if P has a form that contains (T), then each parameter type P_i of the respective parameter-type-list ([dcl.fct]) of P is compared with the corresponding parameter type A_i of the corresponding parameter-type-list of A. ...

Here, per 10, comparison of functions types results in a single comparison of X<Ts...> and X<U1>, which, according to 9, succeeds.

Thus, deduction is succesful, so f(X<T>) is indeed at least as specialized as f(X<Ts...>), so std::variant matches template<typename> typename V. Intuitively, we gave 'more general' template template argument which should work nicely for intended usage of a more specific template template parameter.

In practice, different compilers enable P0522R0 changes under different circumstances, and cppreference template parameters page (section Template template arguments) contains links and information on GCC, Cland and MSVC. In particular, GCC in C++17+ mode enables it by default (and for previous standards with compiler flag fnew-ttp-matching), but Clang doesn't in any mode unless -frelaxed-template-template-args flag is provided, thus you got the difference in output for them. With the flag, Clang also produces correct behaviour (godbolt).