构造一个更改mixin基类的类型

时间:2016-02-12 14:20:01

标签: c++ templates

我希望能够改变mixin最内层类型的类型。草图如下所示:

struct PointTypeA {
    double x,y;
};

struct PointTypeB {
    double x,y,z;
};

template<class Base>
class MyMixin1 : public Base {
public:
    double someProperty;
};

template<class Base>
class MyMixin2 : public Base {
public:
    double someProperty;
};

// how do we automatically construct MyMixin2<MyMixin1<TNewInside> > from MyMixin2<MyMixin1<PointTypeA> >
// template <typename T, typename TNewInside>
// ChangedType ChangeInsideType(T object) {
//     return ChangedType(object);
// }

int main() {
    typedef MyMixin2<MyMixin1<PointTypeA> > Mixed12A;
    Mixed12A mixed12A;

    //MyMixin2<MyMixin1<PointTypeB> > mixed12B = ChangeInsideType<Mixed12AType, PointTypeB>(mixed12A);
    return 0;
}

这样的事情可能吗?

1 个答案:

答案 0 :(得分:1)

可以使用以下方法替换内部模板参数:

template <class ToReplace, class With>
struct replace_inner {
    using type = With;  
};

template <template <class> class Outer, class Inner, class With>
struct replace_inner<Outer<Inner>, With> {
    using type = Outer<typename replace_inner<Inner, With>::type>;
};

template <class ToReplace, class With>
using replace_inner_t = typename replace_inner<ToReplace,With>::type;

我们这样使用它:

using replaced = replace_inner_t<MyMixin1<MyMixin2<PointTypeA>>, PointTypeB>;
static_assert(
    std::is_same<replaced, MyMixin1<MyMixin2<PointTypeB>>>::value, "wat"
);

然后你只需要在点类型和mixins之间用不同的模板参数编写转换构造函数。