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无效的协变返回类型,嵌套类c ++

[英]invalid covariant return type, nested class c++

I have a first class A that contain an iterator nested class with virtual methods: 我有一个第一个类A,它包含一个带有虚方法的迭代器嵌套类:

template <typename T >
class A {
  public:
      class iterator {
      public:
          virtual ~iterator() {};
          virtual T& operator++();
       };

      virtual iterator begin() const = 0;
};

I have a second class B, that override virtuals methods: 我有第二个B类,它覆盖虚拟方法:

template <typename T >
class B : public A<T> {
   public:
      class iterator : A<T>::iterator {
          T& operator++() override { 
              iterator p(*this);
              return p; //for exemple
          }
      };

   iterator begin() const override {
       return iterator(this);// for exemple 
   } 
};

But when i use B class : 但是当我使用B类时:

B<int> test;

I have something like this, compilation error: 我有这样的东西,编译错误:

error: invalid covariant return type for 'B<T>::iterator B<T>::begin() const [with T = int]'
error:  overriding 'B<T>::iterator V<T>::begin() const [with T = int]'

How to implements iterator in B class ? 如何在B类中实现迭代器?

Co-variant return types have a couple of constraints they need to satisfy according to [class.virtual]/8 . 根据[class.virtual] / 8,共变量返回类型有一些它们需要满足的约束。

The return type of an overriding function shall be either identical to the return type of the overridden function or covariant with the classes of the functions. 重写函数的返回类型应与重写函数的返回类型相同,或者与函数类的协变相同。 If a function D​::​f overrides a function B​::​f, the return types of the functions are covariant if they satisfy the following criteria: 如果函数D :: f覆盖函数B :: f,则函数的返回类型如果满足以下条件则是协变的:

  • both are pointers to classes, both are lvalue references to classes, or both are rvalue references to classes 两者都是类的指针,都是对类的左值引用,或者两者都是对类的右值引用
  • the class in the return type of B​::​f is the same class as the class in the return type of D​::​f, or is an unambiguous and accessible direct or indirect base class of the class in the return type of D​::​f 返回类型为B :: f的类与返回类型为D :: f的类相同,或者是返回类型中类的明确且可访问的直接或间接基类D :: f

  • [...] [...]

Yours doesn't inherit publicly, so the base isn't accessible. 您的公开内容不会继承,因此无法访问该基础。 And you aren't returning a pointer or a reference. 而且你没有返回指针或引用。

Returning a type with value semantics is good! 返回具有值语义的类型是好的! You should not give up on that. 你不应该放弃这一点。 You can substitute the attempt at a co-variant return type with the pimpl idiom. 您可以使用pimpl习语替换co-variant返回类型的尝试。 Have iterator manage a polymorphic "iterator implementation" class via pointer. iterator器通过指针管理多态“迭代器实现”类。

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