隐式复制构造函数 [英] Implicit copy constructor

问题描述

n3337的报价12.3.1 / 3

Quote from n3337 12.3.1/3

非显式复制/移动构造函数（12.8） b构造函数。隐式声明的复制/移动构造函数不是一个
显式构造函数;

A non-explicit copy/move constructor (12.8) is a converting constructor. An implicitly-declared copy/move constructor is not an explicit constructor; it may be called for implicit type conversions.

来自ANSI ISO IEC 14882 2003的报价

Quote from ANSI ISO IEC 14882 2003

非显式复制构造函数（12.8）是转换构造函数。
隐式声明的拷贝构造函数不是显式构造函数;
它可能被调用隐式类型转换。

A non-explicit copy-constructor (12.8) is a converting constructor. An implicitly-declared copy constructor is not an explicit constructor; it may be called for implicit type conversions.

我没有想法，构造函数可用于 implicit 类型转换。如果是在标准的错误印刷/错误，为什么它不会更正，因为C ++ 03标准？任何链接和示例（如果我们可以使用类型转换）非常感谢。

I have no ideas, how copy-constructor can be used for implicit type conversions. And if it's misprint/error in standard, why it's not corrected since C++03 standard? Any links and examples (if we can use it for type conversions) are really appreciated.

推荐答案

复制构造函数可以通过对派生类型的对象进行切片来转换：

A copy constructor can convert from an object of a derived type by slicing it:

struct A {};
struct B : A {};

B b;
A a = b; // uses A::A(A const&) to convert B to A

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