使用/ typedef的嵌套模板

由 aet发布于 2020-04-09 07:14:23

说我有这个设置：

template<typename T1>
struct A {
  template<typename T2>
  struct B {
    using type = int;
  };
};

我想能够形成一个typedef /使用：

template<typename T1,typename T2>
using type2 = A<T1>::B<T2>::type;
//... and use like
type2<int,char> foo;

GCC complains that I need typename A<T1>::B<T2>::type instead, and afterwards complains that it expects ";" before "<" after B (i.e. typename A<T1>::B)

没有办法对嵌套模板使用“使用”？

负能量

2020-04-09 07:14:24

Note that B is a templated class and type is enclosed in a templated class, hence use the following

#include <iostream>
template<typename T1>
struct A {
    template<typename T2>
    struct B {
        using type = int;
    };
};

template<typename T1,typename T2>
using type2 = typename A<T1>::template B<T2>::type;

int main()
{
    type2<int,char> foo =2;
    std::cout << foo;

}

Guy

2020-04-09 07:14:24

从切换

using type2 = A<T1>::B<T2>::type;

至

using type2 = typename A<T1>::template B<T2>::type;

aet

这家伙很懒，什么都没留下

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