c ++中的隐藏规则是什么? [英] What is the hiding rule in c++?

问题描述

我对名称隐藏"和信息隐藏"一词感到困惑.最重要的是，c ++中的隐藏规则是什么?有人可以给我一个定义吗?

I am so confused with the term name hiding and information hiding. Most importantly, What is the hiding rule in c++? Can someone give me a definition?

推荐答案

隐藏名称会在您覆盖某个类时发生:

Name hiding happens when you override a class:

struct A
{
    int x;
    int y;

    void foo();
    void bar();
};

struct B : A
{
    int y;
    void bar();
};

在类 B 中，名称 x 和 foo 是明确的，并且引用基类中的名称.但是，名称 y 和 bar 隐藏是基本名称.请考虑以下内容:

In class B, the names x and foo are unambiguous and refer to the names in the base class. However, the names y and bar hide the base names. Consider the following:

B b;
b.bar();

函数名称指的是函数 B :: bar 的名称，因为基名是隐藏的，没有歧义.如果需要基本功能，则必须明确地说出它: b.A :: bar().另外，您可以在类定义中添加 using A :: bar; 来取消隐藏名称，但随后必须处理歧义(如果存在明显的超载).

The function name refers to the name of the function B::bar without ambiguity, since the base name is hidden. You have to say it explicitly if you want the base function: b.A::bar(). Alternatively, you can add using A::bar; to your class definition to unhide the name, but then you have to deal with the ambiguity (this makes sense if there are distinct overloads, though).

最常见的名称隐藏示例之一是 operator = 的示例，它存在于每个类中，并且隐藏任何基类运算符；因此，为每个类都提供了一个隐式的运算符定义，而该定义没有提供显式的定义，如果您已经在基类中定义了一个赋值运算符，这可能会令人惊讶.

Perhaps one of the most frequently confusing examples of name hiding is that of the operator=, which exists in every class and hides any base class operator; consequently, an implicit definition of the operator is produced for each class which doesn't provide an explicit one, which may come as a surprise if you already defined an assignment operator in a base class.

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