std::cin 提取到枚举

发布于 2024-10-14 06:32:36 字数 130 浏览 2 评论 0原文

我不明白为什么我不能这样做：

enum MyEnum {X=1, Y};
...
X x;
std::cin >> x;

问题是 C++ 不够聪明，或者我弄错了什么？

原文

I don't understand why I can't to that:

enum MyEnum {X=1, Y};
...
X x;
std::cin >> x;

the problem is that C++ is not smart enougth or I'm mistaking something?

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叶落知秋 2024-10-21 06:32:36

可以，但您需要编写自定义 operator>> 重载：

std::istream& operator>>(std;:istream& is, MyEnum& e) { ... }

无法使用默认 std::istream operator>> 执行此操作的原因是 重载的特点是右侧参数必须与运算符重载的参数完全匹配，因为它是由非常量引用获取的（因为运算符将通过分配给对象来修改对象）。

另一种选择是将流中的整数表示形式提取到 int 中，然后将其转换为枚举类型：

int i;
MyEnum e;

if (!(std::cin >> i)) { /* handle error */ }
e = static_cast<MyEnum>(i);

您可能希望在此处执行一些错误检查，除非您确定提取的值是可以用MyEnum表示。（从技术上讲，您还应该小心提取到 int，因为 int 可能无法表示 MyEnum 的所有值。有一个在另一个问题的答案中解释如何做到这一点，如何扩展词法转换以支持枚举类型？）

You can, but you need to write a custom operator>> overload:

std::istream& operator>>(std;:istream& is, MyEnum& e) { ... }

The reason you cannot do this with the default std::istream operator>> overloads is that the right side argument must exactly match the parameter of the operator overload because it is taken by non-const reference (because the operator is going to modify the object by assigning to it).

Another option would be to extract the integer representation from the stream into an int and then cast it to the enumeration type:

int i;
MyEnum e;

if (!(std::cin >> i)) { /* handle error */ }
e = static_cast<MyEnum>(i);

You probably want to perform some error checking here, unless you are certain that the extracted value is able to be represented by MyEnum. (Technically, you should also be careful with extracting to int, since int may not be able to represent all the values of MyEnum. There's an explanation of how to do this in an answer to another question, How can I extend a lexical cast to support enumerated types?)

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