协变虚函数的 C++ 规则

Question

我正在阅读 Covariant virtual function .上面写着

假设 B::f 覆盖了虚函数 A::f。如果满足以下所有条件，A::f 和 B::f 的返回类型可能不同:

1) The const or volatile qualification of the pointer or reference returned by B::f has the same or less const or volatile qualification of the pointer or reference returned by A::f.

2) A::f returns an lvalue reference if and only if B::f returns an lvalue reference.

3) The function B::f returns a pointer or a reference to a class of type T, and A::f returns a pointer or a reference to an unambiguous direct or indirect base class of T.

4) The return type of B::f must be complete at the point of declaration of B::f, or it can be of type B

有人会通过给出合适的例子来解释以上两条规则吗？这两条规则到底是什么意思？第二条规则是否适用于 C++11？以下示例是否满足我在这里所说的第一条规则？

#include <iostream>
class Base {
    public:
        virtual const Base& fun() const
        {
            std::cout<<"fun() in Base\n";
            return *this;
        }
        virtual ~Base()
        { }
    private:
        int a=3;
};
class Derived : public Base
{
        const Derived& fun() const
        {
            std::cout<<"fun() in Derived\n";
            return *this;
        }
};
int main(){
    Base* p=new Derived();
    p->fun();
    delete p;
    return 0;
}

如果我哪里错了，请纠正我。我对前两条规则感到困惑。

谢谢

非常感谢您的帮助。

Answer 1

第一条规则意味着如果 版本不会:

struct A
{
    virtual A* foo() { return new A{}; }    
};

struct B : A
{
    B* foo() override {return new B{}; }          //valid
    const B* foo() override {return new B{}; }    //invalid
    volatile B* foo() override {return new B{}; } //invalid
};

如果考虑调用站点，这是有道理的:

A* my_b = new B{};
A* new_b = my_b->foo(); //discards the const qualifier if B::foo() returns const B*

第二条规则意味着您不能将不同的引用或指针类型作为协变返回类型。使用与上面相同的示例:

struct A
{
    virtual A* foo() { return new A{}; }    
};

struct B : A
{
    B* foo() override {return new B{}; }   //valid
    B& foo() override {return new B{}; }   //invalid
    B&& foo() override {return new B{}; }  //invalid
};

再次考虑调用站点:

A* my_b = new B{};
A* new_b = my_b->foo(); //if B::foo() returns a reference, this line is syntactically ill-formed

您的示例满足这两个规则，因为两个返回类型具有相同的 cv 限定并且都是左值引用。