用 C++ 编写类型类

Question

我正在尝试用 C++ 为一组类型编写类似的东西类，但我正在努力解决如何安排模板签名，或者是否有可能做我想做的事情做。

为了将其分解为最小的示例，假设我有这个：

template<typename S, typename T>
struct Homomorphism {
    //Defined in specialization: static const T morph(const S&);
    static constexpr bool is_instance = false;
    using src  = S;
    using dest = T;
};

template<typename S, typename T>
struct Monomorphism : Homomorphism<S, T> {
    //Defined in specialization: static const T morph(const &S);
    static constexpr bool is_instance = false;
    using src  = S;
    using dest = T;
};

我在我的程序中对这些类（和其他态射）的数据类型进行了专门化处理。

我现在想做的是编写一个结构模板，它将采用两个同态或两个单态，并将它们组合起来分别生成一个新的同态或单态结构，即类似于：

template<typename S, typename T, typename U,
         typename HST = Homomorphism<S, T>,
         typename HTU = Homomorphism<T, U>,
         typename HSU = Homomorphism<S, U> >
struct CompositionMorphism : HSU {
    static const U morph(const S &s) {
        return HTU::morph(HST::morph(s));
    }
    static constexpr bool is_instance = true;
    using src  = S;
    using dest = U;
}

这实际上适用于通过以下方式组合特殊的同态实例：

CompositionMorphism<Class1, Class2, Class3>::morph(class1Instance);

当我有：

struct Homomorphism<Class1, Class2> {
    static const Class2 morph(const Class1 &c) {
        ...
    }
};

和 Homomorphism<Class2, Class3> 类似。

不过，现在我想写：

template<typename S, typename T, typename U,
        typename MST = Monomorphism<S, T>,
        typename MTU = Monomorphism<T, U>,
        typename MSU = Monomorphism<S, U> >
struct CompositionMorphism : MSU {
    static const U morph(const S &s) {
        return MTU::morph(MST::morph(s));
    }
    static constexpr bool is_instance = true;
    using src  = S;
    using dest = U;
};

但是编译器毫不奇怪地抱怨 CompositionMorphism.

的重复定义

有没有一种方法可以用 Homomorphism 和 Monomorphism 来编写 CompositionMorphism 及其特化，这样我就可以做一些事情，比如 call:

template<> struct Homomorphism<Class1, Class2> { ... };
template<> struct Homomorphism<Class2, Class3> { ... };
CompositionMorphism<Class1, Class2, Class3>::morph(c1Instance);

或：

template<> struct Monomorphism<Class1, Class2> { ... };
template<> struct Monomorphism<Class2, Class3> { ... };
CompositionMorphism<Class1, Class2, Class3>::morph(c1Instance);

或：

template<> struct Monomorphism<Class1, Class2> { ... };
template<> struct Homomorphism<Class2, Class3> { ... };
CompositionMorphism<Class1, Class2, Class3>::morph(c1Instance);

并让编译器根据我的态射层次选择最接近的 CompositionMorphism 特化？

Answer 1

好吧，有时我需要多思考一下，但这可能就是您要找的：

#include <type_traits>
#include <cstdint>
#include <tuple>

template<typename S, typename T>
struct Homomorphism;

template<typename S, typename T>
struct Monomorphism;

class Class1{};
class Class2{};
class Class3{};

template<> struct Homomorphism<Class1, Class2> 
{ 
     static const Class2 morph(const Class1&); 
     static constexpr bool is_instance = true;#
};

template<> struct Homomorphism<Class2, Class3> 
{
    static const Class3 morph(const Class2&);
    static constexpr bool is_instance = true;
};

template<typename S, typename T>
struct Homomorphism {
    //Defined in specialization: static const T morph(const S&);
    static constexpr bool is_instance = false;
    using src  = S;
    using dest = T;
};

template<typename S, typename T>
struct Monomorphism : Homomorphism<S, T> {
    //Defined in specialization: static const T morph(const &S);
    static constexpr bool is_instance = false;
    using src  = S;
    using dest = T;
};


namespace details {
    template<typename T, typename U, std::enable_if_t<Homomorphism<T,U>::is_instance>* = nullptr>
    U morph (const T& t)
    {return  Homomorphism<T,U>::morph(t);}

    template<typename T, typename U,  std::enable_if_t<Monomorphism<T,U>::is_instance>* = nullptr>
    U morph (const T& t)
    {return  Monomorphism<T,U>::morph(t);}


 }

template <typename S, typename T, typename U>
class CompositionMorphism
{
public:
    static U morph (const S& s)  {return  details::morph<T,U>(details::morph<S,T>(s));}
    static constexpr bool is_instance = true;
};


 int main(int, char**)
{
    Class1 c1Instance;
    CompositionMorphism<Class1, Class2, Class3>::morph(c1Instance);
    std::ignore = d;
}

并且您可能想要手动创建复合 Homo/Mono 态射，如下所示：

template <> class Monomorphism<Class1,Class3> : public CompositionMorphism<Class1, Class2, Class3> {};

然后它们可以被 CompositionMorphism 自动重用。

Answer 2

您可以尝试在 morph 函数上基于 SFINAE 编写 select Homomorphism 或 Monomorphism 的模板。

template <typename S, typename T, typename = void>
struct SelectMorphism {
    using type = Homomorphism<S, T>;
};

template <typename S, typename T>
struct SelectMorphism<S, T, std::enable_if_t<std::is_same_v<decltype(Monomorphism<S, T>::morph(std::declval<S>())), const T>>> {
    using type = Monomorphism<S, T>;
};

这将检查 Monomorphism<S, T>::morph(S) 是否会 return 一个 T，如果是 select 一个 Monomorphism<S, T>。如果不是，SFINAE 将失败并默认为 Homomorphism<S, T>.

然后我们更改 CompositionMorphism 以像这样使用此模板

template<typename S, typename T, typename U,
         typename HST = typename SelectMorphism<S, T>::type,
         typename HTU = typename SelectMorphism<T, U>::type,
         typename HSU = typename SelectMorphism<S, U>::type >
struct CompositionMorphism : HSU {
    static const U morph(const S &s) {
        return HTU::morph(HST::morph(s));
    }
    static constexpr bool is_instance = true;
    using src  = S;
    using dest = U;
};

您可以查看此完整工作示例的 live demo here。它需要 c++17，但也可以写成 c++11（稍微冗长）。

#include <iostream>

template<typename S, typename T>
struct Homomorphism {
    //Defined in specialization: static const T morph(const S&);
    static constexpr bool is_instance = false;
    using src  = S;
    using dest = T;
};

template<typename S, typename T>
struct Monomorphism : Homomorphism<S, T> {
    //Defined in specialization: static const T morph(const &S);
    static constexpr bool is_instance = false;
    using src  = S;
    using dest = T;
};

template <typename S, typename T, typename = void>
struct SelectMorphism {
    using type = Homomorphism<S, T>;
};

template <typename S, typename T>
struct SelectMorphism<S, T, std::enable_if_t<std::is_same_v<decltype(Monomorphism<S, T>::morph(std::declval<S>())), const T>>> {
    using type = Monomorphism<S, T>;
};

struct Class1 {};

struct Class2 {};

struct Class3 {};

template<>
struct Monomorphism<Class1, Class2> : Homomorphism<Class1, Class2> {
    static const Class2 morph(const Class1&) { std::cout << "Morphing in Mono<Class1, Class2>" << std::endl; return Class2{}; }
    static constexpr bool is_instance = false;
    using src  = Class1;
    using dest = Class2;
};

template<>
struct Homomorphism<Class2, Class3> {
    static const Class3 morph(const Class2&) { std::cout << "Morphing in Homo<Class2, Class3>" << std::endl; return Class3{}; }
    static constexpr bool is_instance = false;
    using src  = Class2;
    using dest = Class3;
};

template<typename S, typename T, typename U,
         typename HST = typename SelectMorphism<S, T>::type,
         typename HTU = typename SelectMorphism<T, U>::type,
         typename HSU = typename SelectMorphism<S, U>::type >
struct CompositionMorphism : HSU {
    static const U morph(const S &s) {
        return HTU::morph(HST::morph(s));
    }
    static constexpr bool is_instance = true;
    using src  = S;
    using dest = U;
};

int main ()
{
    CompositionMorphism<Class1, Class2, Class3>::morph(Class1{});
}

Answer 3

根据 Super 的观察，如果您仅传递 T、U 和 V，编译器不知道是否选择 Homomorphism 或 Monomorphism.

所以我想你应该通过Homomorphism<T, U>和Homomorphism<U, V>（Homomorphism<T, V>可以构造）或者Monomorphism<T, U>和Monomorphism<U, V>

如果你想强加两个 Homomorphism 或两个 Monomorphism （我的意思是：如果你想排除一个 Monomorphism 和一个 Homomorphism) 你可以这样写

template <typename, typename>
struct CompositionMorphism;

template <template <typename, typename> class C,
          typename S, typename T, typename U>
struct CompositionMorphism<C<S, T>, C<T, U>>
 {
   using comp = C<S, U>;

   static const U morph (const S & s)
    { return C<T, U>::morph(C<S, T>::morph(s)); }
 };

并如下调用

   Homomorphism<int, long>        h0;
   Homomorphism<long, long long>  h1;
   Monomorphism<int, long>        m0;
   Monomorphism<long, long long>  m1;

   CompositionMorphism<decltype(h0), decltype(h1)>  h2;
   CompositionMorphism<decltype(m0), decltype(m1)>  m2;

   // compiler error
   //CompositionMorphism<decltype(h0), decltype(m1)>  hm;

下面是一个完整的编译示例

#include <array>
#include <iostream>

template <typename S, typename T>
struct Homomorphism
 {
   //Defined in specialization: static const T morph(const S&);
   static constexpr bool is_instance = false;
   using src  = S;
   using dest = T;
 };

template <typename S, typename T>
struct Monomorphism : Homomorphism<S, T>
 {
   //Defined in specialization: static const T morph(const &S);
   static constexpr bool is_instance = false;
   using src  = S;
   using dest = T;
 };

template <typename, typename>
struct CompositionMorphism;

template <template <typename, typename> class C,
          typename S, typename T, typename U>
struct CompositionMorphism<C<S, T>, C<T, U>>
 {
   using comp = C<S, U>;

   static const U morph (const S & s)
    { return C<T, U>::morph(C<S, T>::morph(s)); }
 };


int main ()
 { 
   Homomorphism<int, long>        h0;
   Homomorphism<long, long long>  h1;
   Monomorphism<int, long>        m0;
   Monomorphism<long, long long>  m1;

   CompositionMorphism<decltype(h0), decltype(h1)>  h2;
   CompositionMorphism<decltype(m0), decltype(m1)>  m2;

   // compiler error
   //CompositionMorphism<decltype(h0), decltype(m1)>  hm;

   static_assert( std::is_same<Homomorphism<int, long long>,
                               decltype(h2)::comp>{}, "!" );

   static_assert( std::is_same<Monomorphism<int, long long>,
                               decltype(m2)::comp>{}, "!" );
 }

用 C++ 编写类型类

Writing typeclasses in C++

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metaprogramming

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