本文介绍了规则三为内部类,外部类的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

假设我的类定义为

#ifndef A_H
#define A_H

#include <vector>

class A
{
    int n;
    std::vector<A::B> elements;
    public:
        A();
        A(int);
        class B
        {
            int m;
            A* a;
            public:
                B();
                B(int);
                B(int, A*);
        };

};

#endif

A.cpp / p>

A.cpp

#include "A.h"

A::A()
 : n(0)
{

}

A::A(int x)
 : n(x), elements(std::vector<A::B>(n))
{
    for (int j = 0; j < this->n; j++)
    {
        B newElement(j, this);
        this->elements[j] = newElement;
    }
}

A::B::B()
 : m(0), a(0)
{

}

A::B::B(int j)
 : m(j), a(0)
{

}

A::B::B(int j, A* aPtr)
 : m(j), a(aPtr)
{

}

如何定义复制构造函数,重载赋值运算符和析构函数,递归删除,我避免,如果可能,必须使用 new ? A :: B 需要指向 A 的指针来定义程序员可能决定添加的操作符。此外,程序员可以决定编写扩展A的 C C > < $ / p>

How should I define copy constructors, overloaded assignment operators, and destructors such that I avoid infinite recursion on deletion, and that I avoid, if possible, having to use new? A::B needs pointer to A to define operators that programmer may decide to add. Also, programmer may decide to write class C that extends A and C::D that extends A::B

推荐答案

A :: elements的所有权由您的声明决定。 〜A 将销毁owned元素成员中的任何 B 。这意味着 A :: B :: a 可能必须被视为未知的指针,并且在〜B 运行。如果在某些情况下 A :: B :: a 由 B 的实例拥有,则 B 可能需要添加一个成员来跟踪此所有权,并且只在这些情况下释放。或者, A :: B :: a 可以被声明为 std :: shared_ptr< A> (或类似)。但是,请记住,如果所有权是循环的, std :: shared_ptr 仍然可能会泄露。

Ownership of A::elements is determined by your declaration. ~A is going to destroy any B's in the owned elements member. This means that A::B::a likely has to be treated as an unowned pointer and not deleted when ~B is run. If, under some circumstances, A::B::a is owned by an instance of B, then B likely needs to add a member to track this ownership and free a only under these circumstances. Alternatively A::B::a could be declared as std::shared_ptr<A> (or similar). However, remember that std::shared_ptr can still potentially leak if ownership is circular.

这种模糊性和复杂性如果有可能确定什么拥有什么,并得出一个比一切都拥有一切更不透明的关系。

This ambiguity and complexity would be avoided if it's possible to determine 'what owns what' and come up with a relationship that is less opaque than 'everything owns everything else'.

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08-20 20:21