blocks|key|2379255|text|我测试的所有编译器都抱怨了几件事，特别是受保护的构造函数，即使删除了赋值语句，它也会成为问题。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2379256|您无法访问从中派生的类型的任何实例的protected成员。这个问题在11.4p1的示例中得到了澄清。|offset|length|style|CODE|2379257|class+B+{
protected:
++int+i;
++static+int+j;
};

class+D1+:+public+B+{
};

class+D2+:+public+B+{
++void+mem(B*,+D1*);
};

void+D2::mem(B*+pb,+D1*+p1)+{
++pb->i+=+1;+//+ill-formed
++p1->i+=+2;+//+ill-formed
++//+...
}|code-block|syntax|javascript|2379258|entityMap^0|0|I|9|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|P|8|@$D|Q|E|R|F|G]]|9|@]|A|$]]|$1|H|3|I|5|J|7|S|8|@]|9|@]|A|$K|L]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

All the compilers that I tested complained about several things, and specifically the protected constructor would be a problem even if the assignment statement were removed.

You don't get to access the <code>protected</code> members of any instance of a type you derive from. This issue is clarified in the examples of 11.4p1.

<pre><code>class B {
protected:
 int i;
 static int j;
};

class D1 : public B {
};

class D2 : public B {
 void mem(B*, D1*);
};

void D2::mem(B* pb, D1* p1) {
 pb-&gt;i = 1; // ill-formed
 p1-&gt;i = 2; // ill-formed
 // ...
}
</code></pre>

blocks|key|4284634|text|实际上这里有两个独立的问题。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4284635|首先，这一行不只是执行赋值，而是尝试初始化基类(它工作得很好)和成员b。为了创建b成员，它需要构造它，并且作为一个成员，它需要对构造函数的public访问，而这是它所没有的。|offset|length|style|CODE|4284636|那么赋值也不能访问b的非公共成员，因为它不是B类型，而是A类型。|4284637|请记住，protected意味着您只能通过B对象(或子对象)访问A的某些部分。|4284638|在这种情况下，告诉我们您真正的问题，我们可以尝试帮助解决它。继承和组合相同的类型是一种设计的味道。|4284639|entityMap^0|0|Y|1|14|1|1X|6|0|9|1|M|1|S|1|0|4|9|L|1|W|1|0|0^^$0|@$1|2|3|4|5|6|7|P|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|Q|8|@$D|R|E|S|F|G]|$D|T|E|U|F|G]|$D|V|E|W|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|X|8|@$D|Y|E|Z|F|G]|$D|10|E|11|F|G]|$D|12|E|13|F|G]]|9|@]|A|$]]|$1|J|3|K|5|6|7|14|8|@$D|15|E|16|F|G]|$D|17|E|18|F|G]|$D|19|E|1A|F|G]]|9|@]|A|$]]|$1|L|3|M|5|6|7|1B|8|@]|9|@]|A|$]]|$1|N|3|-4|5|6|7|1C|8|@]|9|@]|A|$]]]|O|$]]

There are actually two separate problems here.

The first is that the line doesn't just do an assignment, but tries to initialize the base class (which works fine) and the member <code>b</code>. To create the <code>b</code> member it needs to construct it, and as a member it needs <code>public</code> access to a constructor, which it doesn't have.

Then the assignment also is unable to access non-public member of <code>b</code> because again, it's not of type <code>B</code> but type <code>A</code> instead.

Remember that <code>protected</code> means you can access parts of <code>A</code> through a <code>B</code> object (or child) only.

In this case tell us your real problem and we can try to help solve it. Inheriting and composing from the same type is a design smell.

blocks|key|2379297|text|这似乎是C%2B%2B语言的一个很大的限制。你将如何解决这样的问题：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2379298|class+Node
{
public:
+void+Save();
protected:
+virtual+void+SaveState(int+type)+=+0;
};

class+BinaryNode+:+public+Node
{
protected:
+Node+*left;
+Node+*right;

+virtual+void+SaveState(int+type)+override
+{
++++left->SaveState(type);
++++right->SaveState(type);
+}
};|code-block|syntax|javascript|2379299|在本例中，我不想让方法SaveState在Node层次结构之外可见。只有方法Save应为public。|offset|length|style|CODE|2379300|entityMap^0|0|0|B|9|L|4|12|4|18|6|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@$I|R|J|S|K|L]|$I|T|J|U|K|L]|$I|V|J|W|K|L]|$I|X|J|Y|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|N|$]]

It seems like a big limitation of the C++ language. How would you solve problem like this:

<pre><code>class Node
{
public:
 void Save();
protected:
 virtual void SaveState(int type) = 0;
};

class BinaryNode : public Node
{
protected:
 Node *left;
 Node *right;

 virtual void SaveState(int type) override
 {
 left-&gt;SaveState(type);
 right-&gt;SaveState(type);
 }
};
</code></pre>

In this example I do not want to make method <code>SaveState</code> visible outside <code>Node</code> hierarchy. Only method <code>Save</code> should be <code>public</code>.

I have the following code:

<pre><code>struct A {
protected:
 A() {}

 A* a;
};

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

 A b;
};
</code></pre>

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

Cannot access protected member of base class in derived class

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

文章

问答

视频

学习中心

腾讯云实验室

直播

竞赛

腾讯云代码分析专区

腾讯iOA零信任安全管理系统专区

腾讯云架构师技术同盟交流圈

腾讯云数据库专区

腾讯云顾问专区

腾讯云原生专区

腾讯混元专区

腾讯云TCE专区

腾讯云Lighthouse专区

腾讯云HAI专区

腾讯云Edgeone专区

腾讯云存储专区

腾讯云智能专区

腾讯轻联专区 

腾讯云开发专区

TAPD专区

腾讯轻量云游戏服专区

腾讯云最具价值专家

腾讯云架构师技术同盟

腾讯云创作之星

腾讯云开发者先锋

腾讯云代码助手

云原生构建

TAPD 敏捷项目管理

Cloud Studio

SDK中心

API中心

命令行工具

涵盖代码开发、场景应用、自动测试全流程，助你从零构建专属AI助手

一站式MCP教程库，解锁AI应用新玩法

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，它不能编译。罪魁祸首是b.a = &b;赋值: GCC和clang都抱怨A()是受保护的，这应该不是问题...

问无法访问派生类中的基类的受保护成员
EN

Stack Overflow用户

社区

活动

圈层

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问无法访问派生类中的基类的受保护成员EN