将虚函数放入一个族中

Question

给定

class A {
    public:
        virtual int foo (int) const = 0;
        virtual void bar (char, double) const = 0;
};

class B : public A {
    virtual int foo (int) const {std::cout << "B::foo() called.\n";  return 3;}
    virtual void bar () const {std::cout << "B::bar() called.\n";}
};

class C : public B {
    virtual int foo (int) const {std::cout << "C::foo() called.\n";  return 8;}
    virtual void bar (char, double) const {std::cout << "C::bar() called.\n";}
};

我想将 foo 和 bar（以及 A 的其他虚函数）放入函数模板族中。这是我到目前为止的想法：

#include <iostream>

enum Enum {Foo, Bar};

template <Enum> struct EnumTraits;

template <> struct EnumTraits<Foo> { using return_type = int; };
template <> struct EnumTraits<Bar> { using return_type = void; };

class A {
    template <Enum> class Execute;
public:
    virtual int foo (int) const = 0;
    virtual void bar (char, double) const = 0;
    template <Enum E, typename... Args>
    typename EnumTraits<E>::return_type execute(Args&&... args) const {
        return Execute<E>(this)(std::forward<Args>(args)...);
    }
};

template <>
class A::Execute<Foo> {
    const A* a;
public:
    Execute (const A* a_) : a(a_) {}
    template <typename... Args>
    int operator()(Args&&... args) const {return a->foo(std::forward<Args>(args)...);}
};

template <>
class A::Execute<Bar> {
    const A* a;
public:
    Execute (const A* a_) : a(a_) {}
    template <typename... Args>
    void operator()(Args&&... args) const {a->bar(std::forward<Args>(args)...);}
};

class B : public A {
    virtual int foo (int) const {std::cout << "B::foo() called.\n";  return 3;}
    virtual void bar () const {std::cout << "B::bar() called.\n";}
};

class C : public B {
    virtual int foo (int) const {std::cout << "C::foo() called.\n";  return 8;}
    virtual void bar (char, double) const {std::cout << "C::bar() called.\n";}
};

int main() {
    A* c = new C;

    int n = c->foo(5);  // C::foo() called.
    c->bar(3, 'c');  // C::bar() called.

    n = c->execute<Foo>(5);  // C::foo() called.
    c->execute<Bar>(3, 'c');  // C::bar() called.
}

但是特化 A::Execute<Foo> 和 A::Execute<Bar> 看起来几乎相同，理想情况下应该不特化（特别是如果除了 foo 和 bar 还有许多其他虚函数).写成这样：

template <Enum N>
class A::Execute {
    const A* a;
public:
    Execute (const A* a_) : a(a_) {}
    template <typename... Args>
    int operator()(Args&&... args) const {return a->???(std::forward<Args>(args)...);}

};

怎么填写？？？部分？理想情况下，我希望使用已经存在的 EnumTraits class。

Answer 1

您不能将继承和多态性与模板元编程相结合。您要做的是使用编译时机制来调用正确的函数。虽然这是可能的，但您不能取消引用指针来调用正确的函数并期望它在编译时变为运行。

本质上，您所做的是在继承层次结构中包装部分模板专业化的层次结构。这有点令人困惑。如果您取消引用具有继承层次结构的指针，则调用将通过在虚拟 table 中查找来解析。

您也可以提取模板编译时分派，但这会有点不同。您需要使用 SFINAE 创建界面并从那里开始工作

Answer 2

这是我的尝试。我已经将 enum 替换为用作标签的 struct，将 EnumTraits 替换为 TagTraits。我更喜欢 struct 方法，因为它允许在不影响现有标签的情况下添加新标签。

#include <iostream>
#include <functional>

template <typename T> struct TagTraits;

// Generic implementation of A based on TagTraits.
class A {

   template <typename Tag, typename... Args>
      class Execute {
         const A* a;
         public:
         Execute (const A* a_) : a(a_) {}
         typename TagTraits<Tag>::return_type operator()(Args&&... args) const
         {
            return (a->*(TagTraits<Tag>::get_funtion_ptr()))(std::forward<Args>(args)...);
         }
      };

   public:

   virtual int foo (int) const = 0;
   virtual void bar (char, double) const = 0;

   template <typename Tag, typename... Args>
      typename TagTraits<Tag>::return_type execute(Args&&... args) const
      {
         return Execute<Tag, Args...>(this)(std::forward<Args>(args)...);
      }
};

// tag for foo and the corresponding TagTraits
struct foo_tag {};

template <> struct TagTraits<foo_tag>
{ 
   using return_type = int;
   static decltype(&A::foo) get_funtion_ptr(){ return &A::foo;}
};

// tag for bar and the corresponding TagTraits
struct bar_tag {};

template <> struct TagTraits<bar_tag>
{
   using return_type = void;
   static decltype(&A::bar) get_funtion_ptr(){ return &A::bar;}
};

// Derived classes of A.

class B : public A {
   virtual int foo (int) const {std::cout << "B::foo() called.\n";  return 3;}
   virtual void bar (char, double) const {std::cout << "B::bar() called.\n";}
};

class C : public B {
   virtual int foo (int) const {std::cout << "C::foo() called.\n";  return 8;}
   virtual void bar (char, double) const {std::cout << "C::bar() called.\n";}
};

// Test B
void test_B()
{
   A* aPtr = new B;

   int n = aPtr->foo(5);  // B::foo() called.
   aPtr->bar(3, 'c');  // B::bar() called.

   n = aPtr->execute<foo_tag>(5);  // B::foo() called.
   aPtr->execute<bar_tag>(3, 'c');  // B::bar() called.
}

// Test C
void test_C()
{
   A* aPtr = new C;

   int n = aPtr->foo(5);  // C::foo() called.
   aPtr->bar(3, 'c');  // C::bar() called.

   n = aPtr->execute<foo_tag>(5);  // C::foo() called.
   aPtr->execute<bar_tag>(3, 'c');  // C::bar() called.
}

int main()
{
   test_B();
   test_C();
}

输出：

B::foo() called.
B::bar() called.
B::foo() called.
B::bar() called.
C::foo() called.
C::bar() called.
C::foo() called.
C::bar() called.

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