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foobar(以及A的其他虚拟函数)放入函数的模板系列中。到目前为止,这是我想到的:
#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>A::Execute<Bar>看起来几乎相同,并且理想情况下应保留非特化(特别是如果存在foobar之外的许多其他虚函数)。写类似:
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)...);}

};

如何填写???部分?理想情况下,我希望使用已经存在的EnumTraits类。

最佳答案

这是我的尝试。我已将enum替换为用作标签的struct,并将EnumTraits替换为TagTraits。我喜欢struct方法,因为它允许添加新标签而不影响现有标签。

#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.

关于c++ - 将虚拟功能融入家庭,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/32240503/

10-11 23:11
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