虚拟继承和静态继承 - 在C ++中混合

问题描述

如果您有这样的：

#include <iostream>

template<typename T> class A
{
public:
    void func()
    {
        T::func();
    }
};

class B : public A<B>
{
public:
    virtual void func()
    {
        std::cout << "into func";
    }
};

class C : public B
{
};

int main()
{
  C c;
  c.func();

  return 0;
}

你实现A类，如果B有一个虚拟重写，它被动态分派，但是如果B不静态分派？

Is func() dynamically dispatched?
How could you implement class A such that if B has a virtual override, that it is dynamically dispatched, but statically dispatched if B doesn't?

编辑：我的代码没有编译？对不起大家。我现在生病了。我的新代码也不编译，但这是问题的一部分。此外，这个问题对我来说，不是常见问题。

My code didn't compile? Sorry guys. I'm kinda ill right now. My new code also doesn't compile, but that's part of the question. Also, this question is for me, not the faq.

#include <iostream>

template<typename T> class A
{
public:
    void func()
    {
        T::func();
    }
};

class B : public A<B>
{
public:
    virtual void func()
    {
        std::cout << "in B::func()\n";
    }
};

class C : public B
{
public:
    virtual void func() {
        std::cout << "in C::func()\n";
    }
};
class D : public A<D> {
    void func() {
        std::cout << "in D::func()\n";
    }
};
class E : public D {
    void func() {
        std::cout << "in E::func()\n";
    }
};

int main()
{
  C c;
  c.func();
  A<B>& ref = c;
  ref.func(); // Invokes dynamic lookup, as B declared itself virtual
  A<D>* ptr = new E;
  ptr->func(); // Calls D::func statically as D did not declare itself virtual
  std::cin.get();

  return 0;
}

visual studio 2010\projects\temp\temp\main.cpp(8): error C2352: 'B::func' : illegal call of non-static member function
      visual studio 2010\projects\temp\temp\main.cpp(15) : see declaration of 'B::func'
      visual studio 2010\projects\temp\temp\main.cpp(7) : while compiling class template member function 'void A<T>::func(void)'
      with
      [
          T=B
      ]
      visual studio 2010\projects\temp\temp\main.cpp(13) : see reference to class template instantiation 'A<T>' being compiled
      with
      [
          T=B
      ]

推荐答案

我不知道我明白你在问什么，但它似乎你缺少必要的CRTP转型：

I'm not sure I understand what you're asking, but it appears you are missing the essential CRTP cast:

template<class T>
struct A {
  void func() {
    T& self = *static_cast<T*>(this);  // CRTP cast
    self.func();
  }
};

struct V : A<V> {  // B for the case of virtual func
  virtual void func() {
    std::cout << "V::func\n";
  }
};

struct NV : A<NV> {  // B for the case of non-virtual func
  void func() {
    std::cout << "NV::func\n";
  }
};

如果T不声明自己的func，这将是无限递归self.func会找到A< ; T> :: func。即使一个T的派生类（例如下面的DV）声明了它自己的函数，但T不是。

If T does not declare its own func, this will be infinite recursion as self.func will find A<T>::func. This is true even if a derived class of T (e.g. DV below) declares its own func but T does not.

测试与不同的最终超控显示发布工作作为广告：

Test with different final overrider to show dispatch works as advertised:

struct DV : V {
  virtual void func() {
    std::cout << "DV::func\n";
  }
};
struct DNV : NV {
  void func() {
    std::cout << "DNV::func\n";
  }
};

template<class B>
void call(A<B>& a) {
  a.func();  // always calls A<T>::func
}

int main() {
  DV dv;
  call(dv);   // uses virtual dispatch, finds DV::func
  DNV dnv;
  call(dnv);  // no virtual dispatch, finds NV::func

  return 0;
}

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