具有相同名称的C ++虚拟覆盖函数

I have something like that (simplified)

我有类似的东西(简化)

class A
{
  public:
    virtual void Function () = 0;
};

class B
{
  public:
    virtual void Function () = 0;
};

class Impl : public A , public B
{
  public:
        ????
};

How can I implement the Function () for A and the Function() for B ? Visual C++ lets you only define the specific function inline (i.e. not in the cpp file), but I suppose it's an extension. GCC complains about this. Is there a standard C++ way to tell the compiler which function I want to override?

如何为A实现Function()和为B实现Function()? Visual C ++允许您只定义内联的特定函数(即不在cpp文件中),但我认为它是一个扩展。海湾合作委员会抱怨这一点。是否有标准的C ++方式告诉编译器我要覆盖哪个函数?

(visual c++ 2008)

(visual c ++ 2008)

class Impl : public A , public B
{
  public:
     void A::Function () {  cout << "A::Function" << endl; }
     void B::Function () {  cout << "B::Function" << endl; }
};

Thank you!

4 个解决方案

#1

You cannot use qualified names there. I you write void Function() { ... } you are overriding both functions. Herb Sutter shows how it can be solved.

你不能在那里使用合格的名字。我写了void Function(){...}你正在覆盖这两个函数。 Herb Sutter展示了如何解决它。

Another option is to rename those functions, because apparently they do something different (otherwise i don't see the problem of overriding both with identical behavior).

另一个选择是重命名这些函数,因为显然它们做了不同的事情(否则我没有看到以相同的行为覆盖两者的问题)。

#2

As a workaround, try

作为一种解决方法,请尝试

struct Impl_A : A
{ 
     void Function () {  cout << "A::Function" << endl; } 
}; 


struct Impl_B : B
{
    void Function () { cout << "B::function" << endl; }
};

struct Impl : Impl_A, Impl_B {};

#3

I can suggest another way to resolve this issue. You can add wrapper Typed which changes Function signature by adding dummy parameter. Thus you can distinguish methods in your implementation.

我可以建议另一种方法来解决这个问题。您可以添加包装类型,通过添加虚拟参数来更改功能签名。因此,您可以区分实现中的方法。

class A {
public:
  virtual void Function() = 0;
  virtual ~A() = default;
};

class B {
public:
  virtual void Function() = 0;
  virtual ~B() = default;
};

template<typename T>
class Typed : public T {
public:
  virtual void Function(T* dummy) = 0;
  void Function() override {
    Function(nullptr);
  }
};

class Impl : public Typed<A>, public Typed<B> {
public:
  void Function(A* dummy) override {
    std::cerr << "implements A::Function()" << std::endl;
  }
  void Function(B* dummy) override {
    std::cerr << "implements B::Function()" << std::endl;
  }
};

The benefit of such solution is that all implementation are placed in one class.

这种解决方案的好处是所有实现都放在一个类中。

#4

-1

If A and B are interfaces, then I would use virtual derivation to "join" them (make them overlap). If you need different implementations for your Function if called through a pointer to A or to B then I would strongly recommend to choose another design. That will hurt otherwise.

如果A和B是接口,那么我将使用虚拟推导来“连接”它们(使它们重叠)。如果通过指向A或B的指针调用函数需要不同的实现,那么我强烈建议您选择其他设计。否则会受到伤害。

Impl "derives from" A and B means Impl "is a" A and B. I suppose you do not mean it.

Impl“源自”A和B意味着Impl“是”A和B.我想你不是这个意思。

Impl "implements interface" A and B means Impl "behaves like" A and B. then same interface should mean the same behavior.

Impl“实现接口”A和B意味着Impl“表现得像”A和B.然后相同的接口应该表示相同的行为。

In both cases having a different behavior according to the type of pointer used would be "schizophrenic" and is for sure a situation to avoid.

在两种情况下,根据所使用的指针类型具有不同的行为将是“精神分裂症”并且肯定是要避免的情况。

#1