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C ++枚举副本构造函数和赋值运算符

[英]C++ enumeration copy constructor and assignment operator

原文 2012-08-02 15:22:07 5 2 c++/ enumeration/ copy-constructor/ assignment-operator

I just found out that enumerations have default constructors and assignment operators in C++. 我刚刚发现，枚举在C ++中具有默认的构造函数和赋值运算符。 Does anyone have an example of an enumeration with non-defaults copy constructor and assignment operator ? 有没有人有非默认副本构造函数和赋值运算符的枚举示例？

2 个解决方案

enum are just int s, that's why they will always have "assignment operator". enum只是int ，这就是为什么它们将始终具有“赋值运算符”的原因。

By the standard by default , the first element of an enum always has 0 as value, and all other elements, after the first one, are "previous_value + 1". 默认情况下 ， 默认情况下 ， enum的第一个元素的值始终为0 ，而在第一个元素之后的所有其他元素均为“ previous_value + 1”。

_{You can change the value of the first element, of course.} _{当然，您可以更改第一个元素的值。} _{Actually, you can give values for each "member" of the enum.} _{实际上，您可以为枚举的每个“成员”提供值。} _{Thanks to @Konrad Rudolph for the comment.} _{感谢@Konrad Rudolph的评论。} _{I just didn't mention "by default" at the beginning, as we were talking about "default construction".} _{一开始我只是没有提到“默认”，因为我们在谈论“默认构造”。}

So no, there are no enum s without "assignment operator" and "default/copy constructor". 所以不，没有“赋值运算符”和“默认/复制构造函数”的enum就没有了。

Enumerators behave much like integral types except for fewer implicit conversions. 枚举器的行为与整数类型非常相似，除了隐式转换较少。 I can't think of a case where there would be non-default copy constructor or assignment operators for an enumeration (unless you wrapped it in a class). 我想不出会有非默认的复制构造函数或赋值运算符进行枚举的情况（除非您将其包装在类中）。