什么是C ++中默认初始化的全局强类型枚举？ [英] What are global strongly-typed enums initialized to by default in C++?

问题描述

我正在尝试确定初始化全局强类型枚举的默认值。以下代码当然不能编译。

I'm trying to determine the default value to which global strongly-type enums are initialized. The following code of course does not compile.

#include <iostream>
using namespace std;

enum class A{a=10, b=20};

// Global strongly-typed enum, uninitialized
A k;

int main() {
    if(k==A::a)
        cout<<"Equal to a"<<endl;
    else if(k==A::b)
        cout<<"Equal to b"<<endl;
    else if(k==0)
        cout<<"Equal to zero"<<endl;

    return 0;
}

什么是k初始化？

推荐答案

k 具有静态存储持续时间和静态对象为零初始化，我们可以看到 C ++标准草案部分 3.6.2 初始化非局部变量段落 2 ：

k has static storage duration and static objects are zero initialized, we can see this by going to the draft C++ standard section 3.6.2 Initialization of non-local variables paragraph 2:

在任何其他
初始化发生之前，具有静态存储持续时间（3.7.1）或线程存储
持续时间（3.7.2）的变量应为零初始化（8.5）。 [...]

Variables with static storage duration (3.7.1) or thread storage duration (3.7.2) shall be zero-initialized (8.5) before any other initialization takes place. [...]

用于标量类型，意味着初始化为零，其中包含部分 8.5 段落 6 其中说：

for scalar types that means initialization to zero, which is covered section 8.5 paragraph 6 which says:

要初始化一个对象或类型T的对象意思是：

To zero-initialize an object or reference of type T means:

，并包含以下项目符号：

and includes the following bullet:

如果T是标量类型（3.9），则将对象初始化为通过将整数文字0（零）转换为T而获得的值
; 105

if T is a scalar type (3.9), the object is initialized to the value obtained by converting the integer literal 0 (zero) to T;105

我们知道一个枚举是从 3.9 类型段落 9 它说：

we know an enum is a scalar type from section 3.9 Types paragraph 9 which says:

算术类型（3.9.1），枚举类型，指针类型，指向
成员类型的指针3.9.2），std :: nullptr_t和cv-qualified版本的
这些类型（3.9.3）被统称为标量类型[...]

Arithmetic types (3.9.1), enumeration types, pointer types, pointer to member types (3.9.2), std::nullptr_- t, and cv-qualified versions of these types (3.9.3) are collectively called scalar types.[...]

零是一个有效的值，因为un枚举类型可以包含其值和部分 7.2 枚举声明段落 8 表示枚举可以取值未定义通过枚举器：

zero is a valid value since the underlying type can contain its value and section 7.2 Enumeration declarations paragraph 8 says that an enumeration can take a value not defined by its enumerators:

[...]可以定义一个枚举值不是由
定义的任何其枚举器。[...]

[...]It is possible to define an enumeration that has values not defined by any of its enumerators.[...]

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