无法访问派生类中的基类的受保护成员 | 无法访问派生类中的基类的受保护成员

本文介绍了无法访问派生类中的基类的受保护成员的处理方法，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

我有以下代码：

struct A {
protected:
    A() {}

    A* a;
};

struct B : A {
protected:
    B() { b.a = &b; }

    A b;
};

奇怪的是不能编译。原因是 ba =& b; 赋值：GCC和clang都抱怨 A（）这不应该是一个问题，因为B继承了A.我进入了哪个黑暗角落的标准？

It strangely doesn't compile. The culprit is the b.a = &b; assignment: both GCC and clang complain that A() is protected, which shouldn't be a problem because B inherits A. Which dark corner of the standard have I come into?

推荐答案

protected 是指派生类型可以访问它自己的基类的成员而不是任何随机对象. In your case, you care trying to modify b's member which is outside of your control (i.e. you can set this->a, but not b.a)

感兴趣，但更好的解决方案是重构代码，而不依赖于hack。例如，您可以在 A 中提供一个构造函数，它以 A * 作为参数（此构造函数应为public ），然后在 B 的初始化器列表中初始化它：

There is a hack to get this to work if you are interested, but a better solution would be to refactor the code and not depend on hacks. You could, for example, provide a constructor in A that takes an A* as argument (this constructor should be public) and then initialize it in the initializer list of B:

A::A( A* p ) : a(p) {}
B::B() : b(&b) {}

protected 您自己类型的任何实例中的基础成员或派生自您自己的类型。

protected grants you access to the base member in any instance of your own type or derived from your own type.

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