Class template specialization priority/ambiguity

Question

While trying to implement a few things relying on variadic templates, I stumbled accross something I cannot explain. I boiled down the problem to the following code snippet:

template <typename ... Args>
struct A {};

template <template <typename...> class Z, typename T>
struct test;

template <template <typename...> class Z, typename T>
struct test<Z, Z<T>> {
    static void foo() {
        std::cout << "I'm more specialized than the variadic spec, hehe!" << std::endl;
    }
};

template <template <typename...> class Z, typename T, typename ... Args>
struct test<Z, Z<T, Args...>> {
    static void foo() {
        std::cout << "I'm variadic!" << std::endl;
    }
};

int main() {
    test<A, A<int>>::foo();
}

Under gcc, it produces an error because it considers both specializations to be equally specialized when trying to instantiate test<A, A<int>>:

main.cpp: In function 'int main()':

main.cpp:25:24: error: ambiguous template instantiation for 'struct test<A, A<int> >'

         test<A, A<int>>::foo();

                        ^~

main.cpp:11:12: note: candidates are: template<template<class ...> class Z, class T> struct test<Z, Z<T> > [with Z = A; T = int]

     struct test<Z, Z<T>> {

            ^~~~~~~~~~~~~

main.cpp:18:12: note:                 template<template<class ...> class Z, class T, class ... Args> struct test<Z, Z<T, Args ...> > [with Z = A; T = int; Args = {}]

     struct test<Z, Z<T, Args...>> {

However, clang deems the first specialization "more specialized" (through partial ordering: see next section) as it compiles fine and prints:

I'm more specialized than the variadic spec, hehe!

A live demo can be found on Coliru. I also tried using gcc's HEAD version and got the same errors.

My question here is: since these two well-known compilers behave differently, which one is right and is this piece of code correct C++?

---

Standard interpretation (C++14 current draft)

From the sections §14.5.5.1 and $14.5.5.2 of the C++14 standard draft, partial ordering is triggered to determine which specialization should be chosen:

(1.2) — If more than one matching specialization is found, the partial order rules (14.5.5.2) are used to determine whether one of the specializations is more specialized than the others. If none of the specializations is more specialized than all of the other matching specializations, then the use of the class template is ambiguous and the program is ill-formed.

Now according to §14.5.5.2, the class template specializations are transformed into function templates through this procedure:

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 (14.5.6.2):

(1.1) — the first function template has the same template parameters as the first partial specialization and has a single function parameter whose type is a class template specialization with the template arguments of the first partial specialization, and

(1.2) — the second function template has the same template parameters as the second partial specialization and has a single function parameter whose type is a class template specialization with the template arguments of the second partial specialization.

Therefore, I tried to reproduce the issue with the function template overloads that the transformation described above should generate:

template <typename T>
void foo(T const&) {
    std::cout << "Generic template\n";
}

template <template <typename ...> class Z, typename T>
void foo(Z<T> const&) {
    std::cout << "Z<T>: most specialized overload for foo\n";
}

template <template <typename ...> class Z, typename T, typename ... Args>
void foo(Z<T, Args...> const&) {
    std::cout << "Z<T, Args...>: variadic overload\n";
}

Now trying to use it like this:

template <typename ... Args>
struct A {};

int main() {
    A<int> a;
    foo(a);
}

yields a compilation error [ambiguous call] in both clang and gcc: live demo. I expected clang would at least have a behavior consistent with the class template case.

Then, this is my interpretation of the standard (which I seem to share with @Danh), so at this point we need a language-lawyer to confirm this.

Note: I browsed a bit LLVM's bug tracker and could not find a ticket for the behavior observed on function templates overloads in this question.

Answer 1

From temp.class.order:

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 ([temp.func.order]):

• 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.

The order of:

template <template <typename...> class Z, typename T>
struct test<Z, Z<T>> {
    static void foo() {
        std::cout << "I'm more specialized than the variadic spec, hehe!" << std::endl;
    }
};

template <template <typename...> class Z, typename T, typename ... Args>
struct test<Z, Z<T, Args...>> {
    static void foo() {
        std::cout << "I'm variadic!" << std::endl;
    }
};

depends on the order of:

template <template <typename...> class Z, typename T>
void bar(test<Z, Z<T>>); // #1
template <template <typename...> class Z, typename T, typename ... Args>
void bar(test<Z, Z<T, Args...>>); // #2

From [temp.func.order]:

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.

To produce the transformed template, for each type, non-type, or template template parameter (including template parameter packs ([temp.variadic]) 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.

Using the transformed function template's function type, perform type deduction against the other template as described in [temp.deduct.partial].

By those paragraph, for any function transformed from any synthesized template Z0 and type T0, which can form #1, we can do type deduction with #2. But functions transformed from #2 with fictitious template Z2 with any type T2 and any non-empty set of Args2 can't be deduced from #1. #1 is obviously more specialized than #2.

clang++ is right in this case.

---

Actually, this one and this one are failed to compile (because of ambiguous) in both g++ and clang. It seems like both compilers have hard time with template template parameters. (The latter one is clearly ordered because its order is the same of no function call).