We have a templated class A and derived classes A1 and A2:
template<typename T> class A {
};
template<typename T> class A1: public A<T>{
};
template<typename T> class A2: public A<T>{
};
I need a wrapper that accepts any class of type A*, ie any derived type of A, as a template parameter and modify its template parameter:
template<typename T, typename Atype> class WrapperA {
Atype<pair<T, int>> atypeobj;
};
Hoping to be used as follows:
WrapperA<int, A1<int>> w1;
WrapperB<int, A2<int>> w2;
The WrapperA needs to work only with derived classes of A.
CodePudding user response：
You do not need to explicitly state int as argument. The template and its argument can be dissected from a given instantiation by partial specialization (provided that all derived have same number of arguments). The fact that there is a base class A is actually not that relevant when the derived classes are templates too.
#include <utility>
template <typename T> struct A { };
template <typename T> struct A1 : A<T> {};
// primary template (no definition needed)
template<typename Atype> struct WrapperA;
// specialization when Atype is instantiation of a template D with
// one type argument T
template <template <typename> typename D,typename T> struct WrapperA<D<T>> {
D<std::pair<T,int>> atypeobj;
};
Then use it
WrapperA<A1<int>> w;
CodePudding user response：
You can simply change your typename Atype parameter into a template template parameter, eg:
template<typename T> class A { };
template<typename T> class A1: public A<T>{ };
template<typename T> class A2: public A<T>{ };
template<typename T, template <typename> typename Atype>
class WrapperA {
Atype<pair<T, int>> atypeobj;
};
WrapperA<int, A1> w1;
WrapperB<int, A2> w2;