使用可变参数模板的递归继承

Question

考虑以下代码：

#include <iostream>

struct ActionOption {
    virtual void foo(int) const = 0;
};

template <int> struct ActionType;

template <> struct ActionType<0> : ActionOption {
    virtual void foo(int) const override {std::cout << "ActionType<0>::foo(int) called.\n";}
};

template <> struct ActionType<1> : ActionOption {
    virtual void foo(int) const override {std::cout << "ActionType<1>::foo(int) called.\n";} 
};

template <> struct ActionType<2> : ActionOption {
    virtual void foo(int) const override {std::cout << "ActionType<2>::foo(int) called.\n";}
};

template <> struct ActionType<3> : ActionOption {
    virtual void foo(int) const override {std::cout << "ActionType<3>::foo(int) called.\n";}
};

template <> struct ActionType<4> : ActionOption {
    virtual void foo(int) const override {std::cout << "ActionType<4>::foo(int) called.\n";}
};

template <int...> struct PossibleActions;

template <> struct PossibleActions<> { void operator()(int) const {} };

template <int First, int... Rest>
struct PossibleActions<First, Rest...> : ActionType<First>, PossibleActions<Rest...> {
    void operator()(int a) const {
        ActionType<First>::foo(a);
        PossibleActions<Rest...>::operator()(a);
    }
};

// Anything that can call ActionType<2>::foo(int) can also call ActionType<3>::foo(int).
struct Object : PossibleActions<1,  2,3,  4> {
    void foo(int a) {PossibleActions<1,2,3,4>()(a);}
};

struct Blob : PossibleActions<0,  2,3,  4> {
    void foo(int a) {PossibleActions<0,2,3,4>()(a);}
};

int main() {
    Object object;
    object.foo(12);  // ActionType<1>::foo(int) called  ActionType<2>::foo(int) called   ActionType<3>::foo(int) called  ActionType<4>::foo(int) called
    std::cout << std::endl;

    Blob blob;
    blob.foo(12);  // ActionType<0>::foo(int) called  ActionType<2>::foo(int) called   ActionType<3>::foo(int) called  ActionType<4>::foo(int) called
    std::cout << std::endl;
}

它运行除了这里是问题所在：任何可以调用 ActionType<2>::foo(int) 的东西也可以调用 ActionType<3>::foo(int)。因此，每次我定义一个新类时，如果我使用 2 或 3，我必须在 PossibleActions<I...> 中同时使用它们。这对于维护当然是有问题的（比如我决定将来使用 2 也必须使用 3、7 和 20）。解决方案如下：

using TwoAndThree = PossibleActions<2,3>;
struct Object : PossibleActions<1,4>, TwoAndThree {
    void foo(int a) {PossibleActions<1,4>()(a);  TwoAndThree()(a);}
};

struct Blob : PossibleActions<0,4>, TwoAndThree {
    void foo(int a) {PossibleActions<0,4>()(a);  TwoAndThree()(a);}
};

不可接受，因为我需要 ActionType<N>::foo(int) 按 数字顺序 调用。拆分 PossibleActions<1,4>()(a); 是一个非常也是糟糕的解决方案，因为它遇到了同样的维护问题（我认为维护更糟）。

template <> struct ActionType<2> : ActionOption { virtual void foo(int) const override {std::cout << "ActionType<2>::foo(int) called.\n";} };
template <> struct ActionType<3> : ActionType<2> { virtual void foo(int) const override {std::cout << "ActionType<3>::foo(int) called.\n";} };

由于歧义而无法编译（并且使用虚拟继承没有帮助），我想不出其他任何东西。这个问题有解决办法吗？

也许用 template <typename... Args> struct PossibleActions; 重新定义 PossibleActions？但是递归丢失了。

还是？

相关问题：有没有一种方法可以使用 Args 执行递归...其中有些类型是 int 但有些不是（对于那些不使用递归的类型）定义这些类型的整数）？例如

PossibleActions<1, TwoAndThree, 4, EightAndTen, 20>()(a);

根据需要遍历 1,2,3,4,8,10,20 因为 TwoAndThree = PossibleActions<2,3> 和 EightAndTen = PossibleActions<8,10>???如果可能的话，那将解决问题。

Answer 1

归功于 Piotr。 S 对于这个解决方案（我希望我能给他加分，但他出于某种原因喜欢隐藏他的惊奇）。尽管他的第二个解决方案也不错，但我更喜欢他的第一个解决方案提供的语法。他的 Sort 结构必须用

概括

template <typename, typename...> struct Sort;

template <typename T, typename A, typename B>
struct Sort<T,A,B> {
    using type = typename Merge<T,A,B>::type;
};

template <typename T, typename First, typename Second, typename... Rest>
struct Sort<T, First, Second, Rest...> {
    using type = typename Sort<T, typename Sort<T, First, Second>::type, Rest...>::type;
};

所以我为他做了那件事。这允许语法

struct Widget : Sort<PossibleActions<0,5>, OneAndFour, TwoAndThree>

我更喜欢哪个。我也在图片中添加了模板模板：

#include <iostream>

namespace Detail {
    template <typename T, typename, typename, T...> struct Merge;

    template <typename T, template <T...> class S, T... Ks>
    struct Merge<T, S<>, S<>, Ks...> {
        using type = S<Ks...>;
    };

    template <typename T, template <T...> class S, T... Is, T... Ks>
    struct Merge<T, S<Is...>, S<>, Ks...> {
        using type = S<Ks..., Is...>;
    };

    template <typename T, template <T...> class S, T... Js, T... Ks>
    struct Merge<T, S<>, S<Js...>, Ks...> {
        using type = S<Ks..., Js...>;
    };

    template <typename T, bool, typename, typename, T...> struct Strip;

    template <typename T, template <T...> class S, T I, T... Is, T J, T... Js, T... Ks>
    struct Strip<T, true, S<I, Is...>, S<J, Js...>, Ks...> {
        using type = Merge<T, S<I, Is...>, S<Js...>, Ks..., J>;
    };

    template <typename T, template <T...> class S, T I, T... Is, T J, T... Js, T... Ks>
    struct Strip<T, false, S<I, Is...>, S<J, Js...>, Ks...> {
        using type = Merge<T, S<Is...>, S<J, Js...>, Ks..., I>;
    };

    template <typename T, template <T...> class S, T I, T... Is, T J, T... Js, T... Ks>
    struct Merge<T, S<I, Is...>, S<J, Js...>, Ks...> : Strip<T, (I > J), S<I, Is...>, S<J, Js...>, Ks...>::type {};

    template <typename, typename...> struct Sort;

    template <typename T, typename A, typename B>
    struct Sort<T,A,B> {
        using type = typename Merge<T,A,B>::type;
    };

    // Piotr S.'s Sort generalized to accept any number of template arguments.
    template <typename T, typename First, typename Second, typename... Rest>
    struct Sort<T, First, Second, Rest...> {
        using type = typename Sort<T, typename Sort<T, First, Second>::type, Rest...>::type;
    };
}

template <typename... P>
using Sort = typename Detail::Sort<int, P...>::type;

struct ActionOption {
    virtual void foo(int) const = 0;
};

template <int> struct ActionType;

template <> struct ActionType<0> : ActionOption {
    virtual void foo(int) const override {std::cout << "ActionType<0>::foo(int) called.\n";}
};

template <> struct ActionType<1> : ActionOption {
    virtual void foo(int) const override {std::cout << "ActionType<1>::foo(int) called.\n";} 
};

template <> struct ActionType<2> : ActionOption {
    virtual void foo(int) const override {std::cout << "ActionType<2>::foo(int) called.\n";}
};

template <> struct ActionType<3> : ActionOption {
    virtual void foo(int) const override {std::cout << "ActionType<3>::foo(int) called.\n";}
};

template <> struct ActionType<4> : ActionOption {
    virtual void foo(int) const override {std::cout << "ActionType<4>::foo(int) called.\n";}
};

template <> struct ActionType<5> : ActionOption {
    virtual void foo(int) const override {std::cout << "ActionType<5>::foo(int) called.\n";}
};

template <int...> struct PossibleActions;

template <> struct PossibleActions<> { void operator()(int) const {} };

template <int First, int... Rest>
struct PossibleActions<First, Rest...> : ActionType<First>, PossibleActions<Rest...> {
    void operator()(int a) const {
        ActionType<First>::foo(a);
        PossibleActions<Rest...>::operator()(a);
    }
};

using OneAndFour = PossibleActions<1,4>;
using TwoAndThree = PossibleActions<2,3>;

struct Thing : PossibleActions<0,1,2,3,4> {
    void foo(int a) {PossibleActions<0,1,2,3,4>::operator()(a);}
};

struct Object : Sort<PossibleActions<1,4>, TwoAndThree> {
    void foo(int a) {Sort<PossibleActions<1,4>, TwoAndThree>()(a);}
};

struct Blob : Sort<PossibleActions<0,4>, TwoAndThree> {
    void foo(int a) {Sort<PossibleActions<0,4>, TwoAndThree>()(a);}
};

struct Widget : Sort<PossibleActions<0,5>, OneAndFour, TwoAndThree> {
    void foo(int a) {Sort<PossibleActions<0,5>, OneAndFour, TwoAndThree>()(a);} 
};

int main() {
    Thing thing;
    thing.foo(12);  // ActionType<0>::foo(int)  ActionType<1>::foo(int) called  ActionType<2>::foo(int) called   ActionType<3>::foo(int) called  ActionType<4>::foo(int) called
    std::cout << std::endl;

    Object object;
    object.foo(12);  // ActionType<1>::foo(int) called  ActionType<2>::foo(int) called   ActionType<3>::foo(int) called   ActionType<4>::foo(int) called
    std::cout << std::endl;

    Blob blob;
    blob.foo(12);  // ActionType<0>::foo(int) called  ActionType<2>::foo(int) called   ActionType<3>::foo(int) called   ActionType<4>::foo(int) called
    std::cout << std::endl;

    Widget widget;
    widget.foo(12);  // ActionType<0>::foo(int) called  ActionType<1>::foo(int) called   ActionType<2>::foo(int) called   ActionType<3>::foo(int) called   ActionType<4>::foo(int) called   ActionType<5>::foo(int) called
}

但是请注意，如果原始包本身未分类，则该解决方案实际上会失败。在执行上述操作之前，这可能需要先在原始包上使用辅助排序器结构。

使用可变参数模板的递归继承

Recursive inheritance with variadic templates

c++

templates

variadic

c++11