什么是C ++中的构造函数的默认访问权限 [英] What is the default access of constructor in c++

问题描述

什么是c ++中构造函数的默认访问权限，为什么？

What is the default access of constructor in c++ and why ?

公共，私有或受保护？

public, private or protected ?

如何通过 code 检查它？

推荐答案

如果您自己没有声明构造函数，则编译器将始终为您生成一个 public 琐碎的代码。他们还将隐式创建一个公共副本构造函数和副本赋值运算符。

If you do not declare a constructor yourself, the compiler will always generate a public trivial one for you. They will also implicitly create a public copy constuctor and copy assignment operator.

从c ++标准12.1.5开始：

From c++ standard 12.1.5:

如果类X没有用户声明的构造函数，则将不带参数的构造函数隐式声明为默认值。隐式声明的默认构造函数是其类的内联公共成员。

If there is no user-declared constructor for class X, a constructor having no parameters is implicitly declared as defaulted. An implicitly-declared default constructor is an inline public member of its class.

12.8.7和12.8.11：

12.8.7 and 12.8.11:

如果类定义未显式声明一个副本构造函数，则隐式声明一个副本构造函数。 [...]
隐式声明的复制/移动构造函数是其类的内联公共成员。

If the class definition does not explicitly declare a copy constructor, one is declared implicitly. [...] An implicitly-declared copy/move constructor is an inline public member of its class.

最后12.8.18、12.8.20、12.8.22：

Finally 12.8.18, 12.8.20, 12.8.22:

如果类定义未明确声明一个副本赋值运算符，则声明一个隐含地。 [...]
如果类X的定义未明确声明移动赋值运算符，则将隐式声明[...]。
隐式声明的复制/移动赋值运算符是其类的嵌入式公共成员。

If the class definition does not explicitly declare a copy assignment operator, one is declared implicitly. [...] If the definition of a class X does not explicitly declare a move assignment operator, one will be implicitly declared [...]. An implicitly-declared copy/move assignment operator is an inline public member of its class.

如果使用的 c ++ 11 不会总是生成move构造函数。有关更多信息，请参见12.8.20节。

If you are using c++11 the move constructor will not always be generated. See section 12.8.20 for more information.

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