Parameter pack must be at the end of the parameter

Question

I don't get the reason for which a parameter pack must be at the end of the parameter list if the latter is bound to a class, while the constraint is relaxed if the parameter list is part of a member method declaration.

In other terms, this one compiles:

class C {
    template<typename T, typename... Args, typename S>
    void fn() { }
};

The following one does not:

template<typename T, typename... Args, typename S>
class C { };

Why is the first case considered right and the second one is not?
I mean, if it's legal syntax, shouldn't it be in both the cases?

To be clear, the real problem is that I was defining a class similar to the following one:

template<typename T, typename... Args, typename Allocator>
class C { };

Having the allocator type as the last type would be appreciated, but I can work around it somehow (anyway, if you have a suggestion it's appreciated, maybe yours are far more elegant than mine!!).
That said, I got the error:

parameter pack 'Args' must be at the end of the template parameter list

So, I was just curious to fully understand why it's accepted in some cases, but it is not in some others.

Here is a similar question, but it simply explains how to solve the problem and that was quite clear to me.

Answer 1

It is valid for function templates but only when argument deduction can help the compiler resolve the template parameters, as it stands your function template example is virtually useless because

template<typename T, typename... Args, typename S> void fn() { }
int main() { fn<int, int, int>(); }

test.cpp: In function 'int main()':
test.cpp:2:32: error: no matching function for call to 'fn()'
 int main() { fn<int, int, int>(); }
                                ^
test.cpp:1:57: note: candidate: template<class T, class ... Args, class S> void fn()
 template<typename T, typename... Args, typename S> void fn() { }
                                                         ^
test.cpp:1:57: note:   template argument deduction/substitution failed:
test.cpp:2:32: note:   couldn't deduce template parameter 'S'
 int main() { fn<int, int, int>(); }

the compiler has no way of determining which template parameters belong to the parameter pack, and which to S. In fact as @T.C. points out it should actually be a syntax error because a function template defined in this manner cannot ever be instantiated.

A more useful function template would be something like

template<typename T, typename... Args, typename S> void fn(S s) { }

as now the compiler is able to unambiguously match the function parameter s with the template type S, with the side effect that S will always be deduced - all explicit template parameters after the first will belong to Args.

None of this works for (primary) class templates, parameters aren't deduced and it's expressly forbidden:

From draft n4567

http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2015/n4567.pdf

[temp.param] / 11

[...]If a template-parameter of a primary class template or alias template is a template parameter pack, it shall be the last template-parameter.[...]

(if they were deduced it would be ambiguous as in the function template example).

Answer 2

The first one is not right. The compiler is just buggy and failed to diagnose it. [temp.param]/11:

A template parameter pack of a function template shall not be followed by another template parameter unless that template parameter can be deduced from the parameter-type-list of the function template or has a default argument (14.8.2).

---

If the function type T(Args...) is meaningful to the end-user, one way to fix this would be to use a partial specialization instead:

template<class F, class Alloc> class C; //undefined
template<class T, class... Args, class Alloc>
class C<T(Args...), Alloc> {
    // implementation
};

Depending on the actual requirements, type-erasing the allocator might also be worth considering.