Assuming I have a templated type, e.g.
template<typename A, typename B, typename C>
struct mytype { };
How do I write a concept that checks whether a type is an instantiation of that template?
template<typename T>
concept MyType = requires(T x) { ??? }
I can't figure an obvious way of doing it without resolving to old-style specialised detector types or maybe a marker base type.
The act of creating a new definition of a function, class, or member of a class from a template declaration and one or more template arguments is called template instantiation.
Template instantiation involves generating a concrete class or function (instance) for a particular combination of template arguments. For example, the compiler generates a class for Array<int> and a different class for Array<double>.
When a function template is first called for each type, the compiler creates an instantiation. Each instantiation is a version of the templated function specialized for the type. This instantiation will be called every time the function is used for the type.
In order for any code to appear, a template must be instantiated: the template arguments must be provided so that the compiler can generate an actual class (or function, from a function template).
You can define your own meta-function (type trait) for that purpose:
template <typename T>
struct is_mytype : std::false_type { };
template <typename A, typename B, typename C>
struct is_mytype<mytype<A, B, C>> : std::true_type { };
template <typename T>
concept MyType = is_mytype<T>::value;
But to say the truth, I don't know whether there isn't a way how to defining such a concept directly without the need of a separate metafunction.
Using C++17 class template argument deduction, you should be able to do something like this:
template<typename A, typename B, typename C>
struct mytype { };
template<class T>
concept C1 = requires(T x) {
{ mytype{x} } -> std::same_as<T>;
};
mytype{x}
uses class template argument deduction to deduce A
, B
and C
, so this is valid if you can construct a mytype<A, B, C>
from a T
. In particular, this is valid if mytype
is copy-constructible since you have an implicitly declared copy-deduction guide similar to:
template <typename A, typename B, typename C>
mytype(mytype<A, B, C>) -> mytype<A, B, C>;
Checking that T
is also the constructed mytype
instantiation avoid matching other deduction guides, e.g., this would match for any type without the -> std::same_as<T>
:
template <class A, class B, class C>
struct mytype {
mytype(A);
};
template <class A>
mytype(A) -> mytype<A, A, A>;
The proposed solution does not work for non copy-constructible classes, even though should be possible to make it work for move-only classes.
Tested with clang and gcc: https://godbolt.org/z/ojdcrYqKv
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