默认继承访问说明符 [英] Default inheritance access specifier

问题描述

如果我有两个类 A 和 B ，使得 B 继承 A 如下：

If I have for example two classes A and B, such that class B inherits A as follows:

B类：public A

class B: public A

在这种情况下，我会执行 public 继承。

In this case, I'm doing public inheritance.

如果我写如下代码：

B类：A

我将在这里做什么类型的继承（即public）？换句话说，什么是默认访问说明符？

What type of inheritance will I be doing here (i.e; public)? In other words, what is the default access specifier?

这里只是一个侧面的问题。我是否调用上一行代码语句？特别是我记得我在 C ++ Without Fear：A Beginner's Guide that Make You Feel Smart 中看到语句是以; 。

Just a side question here. Do I call the previous line of codes statements? Especially that I remember I read in the C++ Without Fear: A Beginner's Guide That Makes You Feel Smart book that statements are that that end with ;. What do you think about that?

感谢。

推荐答案

一个小的除了所有现有的答案：继承的默认类型取决于继承类型（B），而不是正在继承的（A）。例如：

Just a small addition to all the existing answers: the default type of the inheritance depends on the inheriting type (B), not on the one that is being inherited (A). For example:

class A {};
struct B: /* public */ A {};

struct A {};
class B: /* private */ A {};

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