Multiple Inheritance from two derived classes

I have an abstract base class which acts as an interface.

I have two "sets" of derived classes, which implement half of the abstract class. ( one "set" defines the abstract virtual methods related to initialization, the other "set" defines those related to the actual "work". )

I then have derived classes which use multiple inheritance to construct fully defined classes ( and does not add anything itself ).

So: ( bad pseudocode )

class AbsBase {
  virtual void init() = 0;
  virtual void work() = 0;
}

class AbsInit : public AbsBase {
  void init() { do_this(); }
  // work() still abs
}

class AbsWork : public AbsBase {
  void work() { do_this(); }
  // init() still abs
}

class NotAbsTotal : public AbsInit, public AbsWork {
  // Nothing, both should be defined
}

First of all, can I do this? Can I inherit from two classes which are both derived from the same Base? (I hope so).

Here is the "real problem", though (I lied a bit above to simplify the example).

What I have really gone and done is add non abstract accessors methods to the base class:

class AbsBase {
public:
  void init() { init_impl(); }
  void work() { work_impl(); }

private:
  virtual void init_impl() = 0;
  virtual void work_impl() = 0;
}

Because, a common idiom is to make all virtual methods private.

Unfortunately, now both AbsInit, and AbsWork inherit these methods, and so NotAbsTotal inherits "two of each" ( I realize I may be butchering what is really happening at compile time ).

Anyway, g++ complains that: "request for member init() is ambiguous" when trying to use the class.

I assume that, had I used my AbsBase class as a pure interface, this would have been avoided ( assuming that the top example is valid ).

So: - Am I way off with my implementation? - Is this a limitation of the idiom of making virtual methods private? - How do I refactor my code to do what I want? ( Provide one common interface, but allow a way to swap out implementations for "sets" of member functions )

Edit:

Seems I am not the first one: http://en.wikipedia.org/wiki/Diamond_problem

Seems Virtual Inheritance is the solution here. I have heard of virtual inheritance before, but I have not wrapped my head around it. I am still open to suggestions.

---

It looks like you want to do virtual inheritance. Whether that turns out to actually be a good idea is another question, but here's how you do it:

class AbsBase {...};
class AbsInit: public virtual AbsBase {...};
class AbsWork: public virtual AbsBase {...};
class NotAbsTotal: public AbsInit, public AbsWork {...};

Basically, the default, non-virtual multiple inheritance will include a copy of each base class in the derived class, and includes all their methods. This is why you have two copies of AbsBase -- and the reason your method use is ambiguous is both sets of methods are loaded, so C++ has no way to know which copy to access!

Virtual inheritance condenses all references to a virtual base class into one datastructure. This should make the methods from the base class unambiguous again. However, note: if there is additional data in the two intermediate classes, there may be some small additional runtime overhead, to enable the code to find the shared virtual base class.

---

您需要将继承声明为虚拟：

struct AbsBase {
          virtual void init() = 0;
          virtual void work() = 0;
};

struct AbsInit : virtual public AbsBase {
          void init() {  }
};

struct AbsWork : virtual public AbsBase {
          void work() { }
};

struct NotAbsTotal : virtual public AbsInit, virtual public AbsWork {
};

void f(NotAbsTotal *p)
{
        p->init();
}

NotAbsTotal x;

---

It can be done, although it gives most the shivers.

You need to use "virtual inheritance", the syntax for which is something like

class AbsInit: public virtual AbsBase {...};
class AbsWork: public virtual AbsBase {...};
class NotAbsTotal: public AbsInit, public AbsWork {...};

Then you have to specify which function you want to use:

NotAbsTotal::work()
{
    AbsInit::work_impl();
}

(UPDATED with correct syntax)

链接地址: http://www.djcxy.com/p/54828.html

上一篇: Mixins vs scala中的组合

下一篇: 来自两个派生类的多重继承