C ++的默认继承访问说明符？ [英] C++'s default inheritance access specifier?

问题描述

我有一些遗留代码，我必须换行，我遇到了这个声明：

I have some legacy code that I have to wrap, and I have come across this declaration:

class Foo : Bar
{
    // ...
};

这似乎是在GCC下编译的。我知道这是坏的，但我不能改变它。我的问题是，如果没有继承访问说明符，C ++编译器如何处理它？<​​/ p>

This seems to compile under GCC. I know it's bad, but I can't change it. My question is, if no inheritance access specifier is present, how does the C++ compiler handle it?

推荐答案

称为访问修饰符。它被称为访问说明符

BTW, it is not called access modifier. It is called access specifier

$ 11.2 / 2 - 没有
基类的访问限定符，$当派生的
类用class-key
struct定义时，假定b $ b public;当
类用class-key
类定义时，假定private。

$11.2/2 - "In the absence of an access-specifier for a base class, public is assumed when the derived class is defined with the class-key struct and private is assumed when the class is defined with the class-key class."

在您的上下文中，'Bar'是'Foo'的私有基类

In your context, 'Bar' is a private base class of 'Foo'

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