c++ - C++中构造函数的默认访问是什么

Question

什么在c++中构造函数的默认访问是什么？为什么？

公共(public)的、私有(private)的还是 protected ？

我如何通过代码检查它？

Answer 1

如果您不自己声明构造函数，编译器将始终为您生成一个public 普通构造函数。它们还将隐式创建公共(public)复制构造函数和复制赋值运算符。

来自 C++ 标准 12.1.5:

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:

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.

最后是 12.8.18、12.8.20、12.8.22:

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.

如果您使用的是 c++11，则不会始终生成移动构造函数。有关详细信息，请参阅第 12.8.20 节。