枚举除VS减法和铸造 [英] Enum addition vs subtraction and casting

问题描述

为什么除了需要强制，但减法作品，未经强制转换？请参见下面的code明白我在问

Why does addition require a cast but subtraction works without a cast? See the code below to understand what I am asking

public enum Stuff
{
    A = 1,
    B = 2,
    C = 3
}

var resultSub = Stuff.A - Stuff.B; // Compiles
var resultAdd = Stuff.A + Stuff.B; // Does not compile
var resultAdd2 = (int)Stuff.A + Stuff.B; // Compiles     

请注意：对于这两个加减没关系结果是否超出范围（枚举），或者未在上述三个例子

note: For both addition and subtraction it does not matter whether result is out of range (of the enum) or not in all three examples above.

推荐答案

好问题 - 我感到惊讶的是第一和第三线的工作

Good question - I was surprised that the first and third lines worked.

不过，他们的是的在C＃语言规范的支持 - 在第7.8.4，它谈论枚举另外：

However, they are supported in the C# language specification - in section 7.8.4, it talks about enumeration addition:

每个枚举类型都隐式提供下列pre定义运算符，其中E为枚举类型，U是底层的E型：

Every enumeration type implicitly provides the following pre-defined operators, where E is the enum type and U is the underlying type of E:

E operator +(E x, U y)
E operator +(U x, E y)

     

在运行时，这些运营商完全ealuated为（E）（（U）x +（U）Y）

At runtime, these operators are ealuated exactly as (E)((U)x + (U)y)

和第7.8.5：

每个枚举类型都隐式提供下列predefined运营商，其中E为枚举类型，U是底层的E型：

Every enumeration type implicitly provides the following predefined operator, where E is the enum type and U is the underlying type of E:

U operator -(E x, E y)

该运营商也正是评为（U）（（U）x - （U）Y））。换言之，运算符计算 X 和是，以及结果的类型的序数值之间的差是枚举的基础类型。

This operator is evaluated exactly as (U)((U)x - (U)y)). In other words, the operator computes the difference between the ordinal values of x and y, and the type of the result is the underlying type of the enumeration.

E operator -(E x, U y);

该运营商也正是评为（E）（（U）x - Y）。换句话说，操作者减去从枚举的基础类型的值，得到枚举的值。

This operator is evaluated exactly as (E)((U)x - y). In other words, the operator subtracts a value from the underlying type of the enumeration, yielding a value of the enumeration.

所以这就是为什么编译器的行为就像是 - 因为这就是C＃的规范说的事：）

So that's why the compiler behaves like that - because it's what the C# spec says to do :)

我不知道的任意的这些运营商的存在，我从来没有故意见过他们使用。我怀疑的原因，他们的存在是介于语言设计会议埋指出，埃里克利珀偶尔潜入 - 但我也不会感到惊讶，如果他们感到遗憾的是增加功能没有什么好处。再说，也许他们的真正的在某些情况下非常有用：）

I wasn't aware that any of these operators exist, and I've never knowingly seen them used. I suspect the reasons for their existence are buried somewhere in the language design meeting notes that Eric Lippert occasionally dives into - but I also wouldn't be surprised if they were regretted as adding features for little benefit. Then again, maybe they're really useful in some situations :)

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