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Partial Specialization using a qualified name


By : Jack He
Date : September 16 2020, 02:00 AM
like below fixes the issue I'm trying to partially specialize a template for a metafunction and ran into a problem. , Using a template template parameter may work out:
code :
template <typename A, typename B>
struct Foo;

template <typename TA, template<class> class TBar,  typename B1>
struct Foo<TA, TBar<B1>> {};
struct A
{
    template<class T>
    struct Bar {};
};
Foo<A, A::Bar<int>> x;


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Multiple Partial specialization and full specialization required <> after type definition


By : user2295726
Date : March 29 2020, 07:55 AM
I hope this helps you . If you want to call a function foo with no arguments you have to write foo() not just foo. That is exactly the same for template types. event is not a type, it is a "function" that takes types and return a type. If you really want to avoid writing the brackets, you can still do something like:
code :
typedef event<> event_;

Specialization of template method in partial class specialization


By : Vikas Ahuja
Date : March 29 2020, 07:55 AM
Hope this helps Don't ask me for exact quotes but according to David Vandevoorde this can't be done. The rationale is that this is essentially a partial specialization of a function template and there is no such thing. One reason to disallow this is that it is unclear what should happen with this specialization if there is a further specialization of the class: is the function specialization inherited or is it not inherited?
Obviously, stating that this can't be done isn't really a lot of help. What can be done instead? A simple approach would be to delegate the implementation to an overloaded private implementation which takes e.g. a pointer argument to disambiguate between the various types. I realize that the whole point probably is to avoid the use of an extra argument, even if it is only used internally. Another possible approach could be the use of fully specialized base class: member function templates of this can be explicitly specialized.

Nested class template full specialization versus partial specialization


By : HexxCube
Date : March 29 2020, 07:55 AM
it helps some times This is an XY problem. You have to simply move it outwards.
code :
template < >
struct Apply::Inheritance< >
{

};

Partial Template Specialization: Why does some of the variables in the partial specialization list different from the pr


By : WilsonGmn
Date : March 29 2020, 07:55 AM
I wish did fix the issue. In a template specialization, the name you choose for your template arguments are silent-variables. All of the following is exactly equivalent:
code :
template<class T, class U, int N> class A<T, U, N> { /*...*/ };
template<class U, class T, int N> class A<U, T, N> { /*...*/ };
template<class F, class G, int Z> class A<F, G, Z> { /*...*/ };
template<class OnceUpon, class ATime, int ThereWere> class A<OnceUpon, ATime, ThereWere> { /*...*/ };
template<class T1, class T2>
struct X
{};
X<int, int>;
template<class T>
struct X<T, T>
{ using type = T; };

undefined reference to full template specialization class member function, but not partial specialization


By : Komal
Date : March 29 2020, 07:55 AM
This might help you You have to declare partial and explicit specializations in every translation unit that uses them (before any use that would implicitly instantiate that specialization). Here, that would look like
code :
template<class T> class A<T,int>;
template<> class A<int,int>;
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