Reputation: 50540
Consider the following code:
template<typename>
struct S { };
template<typename, typename>
struct B;
template <typename R, typename... Args, template<class> class C>
struct B<R(Args...), C<R>> {
void f() { }
};
int main() {
B<void(), S<void>> b;
b.f();
}
It compiles and has no problem.
Anyway, whenever one decides to use B
, it has to provide two types.
What I'd like to achieve is to default somehow the second parameter (I know, partial specializations do not accept a default for their parameters) and let an user define it's type as B<void()>
instead of B<void(), S<void>>
.
Unfortunately, because of template template, partial specialization and the dependency existent between the parameters, all together they lead to a puzzle against which I'm struggling since a couple of hours.
Is there any clever solution to do that?
So far, I have been able to solve it with intermediate structures, but I don't like it so much...
Upvotes: 1
Views: 84
Reputation: 275385
Here we change B
into a template alias.
B_t
does the default arg work.
B_impl
is the implementation of B
without any default args.
B
is a using
alias that gets the result of B_t
.
template<class> struct S {};
template<class, class>
struct B_impl;
template<class R, class... Args, template<class...> class C>
struct B_impl<R(Args...), C<R>> {
void f() { }
};
template<class, class=void>
struct B_t;
template<class R, class...Args>
struct B_t<R(Args...), void>:
B_t<R(Args...),S<R>>
{};
template<class R, class... Args, template<class...> class C>
struct B_t<R(Args...), C<R>> {
using type=B_impl<R(Args...), C<R>>;
};
template<class Sig, class Z=void>
using B=typename B_t<Sig,Z>::type;
The downside is that pattern-matching on B
won't work well.
Upvotes: 1
Reputation: 303007
Partial specializations don't accept default parameters, but the primary does. You can just add it there:
template<typename Sig, typename X = S<return_type_t<Sig>>>
struct B;
Then all you need to do is implement a return type metafunction for a signature. Something like:
template <class Sig>
struct return_type;
template <class Sig>
using return_type_t = typename return_type<Sig>::type;
template <class R, class... Args>
struct return_type<R(Args...)> {
using type = R;
};
Upvotes: 2