运营商枚举 [英] operator for enums

问题描述

只是出于好奇，问这

像下面

的表达有一个

  A =（条件）？ X：Y; //两个输出
  

为什么我们就不能对枚举操作？结果
说，

  myvalue的= F ??? fnApple（）：fnMango（）：fnOrange（）; //没有。在枚举定义
指定输出 

而不是switch语句（尽管重构是可能的）

---

 枚举水果
 {
苹果，
芒果，
橙色
}; 
 
水果F = Fruit.apple; 
  

---

或者是一些无用的运营商<？ / p>

解决方案

我不能说我曾经想这样的运营商 - 这将是令人难以置信的脆弱依靠的顺序枚举值。您可以轻松地使用交换机：

 开关（F）
 {
情况下Fruit.Apple：myvalue的= fnApple（）;打破; 
情况下Fruit.Mango：myvalue的= fnMango（）;打破; 
情况下Fruit.Orange：myvalue的= fnOrange（）;打破; 
默认：抛出新ArgumentOutOfRangeException（F）; 
} 
  

另外，创建一个映射：

 静态只读字典<水果，Func键<富>> FruitFunctions = 
新字典<水果，Func键<富>> {
 {Fruit.Apple，fnApple} 
 {Fruit.Mango，fnMango} 
 {Fruit.Orange，fnOrange} 
}; 
 ... 
 
 myvalue的= FruitFunctions [F]（）; 
  

我已经用在各种情况下这两种技术，并远远甩宁愿建议的运营商，我只怕。

Just out of curiosity, asking this

Like the expression one below

a = (condition) ? x : y; // two outputs

why can't we have an operator for enums?
say,

myValue = f ??? fnApple() : fnMango() : fnOrange(); // no. of outputs specified in the enum definition

instead of switch statements (even though refactoring is possible)

---

enum Fruit
{
    apple,
    mango,
    orange      
};

Fruit f = Fruit.apple;

---

Or is it some kind of useless operator?

解决方案

I can't say I've ever wanted such an operator - which would be incredibly brittle by relying on the ordering of the enum values. You can easily use a switch:

switch (f)
{
    case Fruit.Apple: myValue = fnApple(); break;
    case Fruit.Mango: myValue = fnMango(); break;
    case Fruit.Orange: myValue = fnOrange(); break;
    default: throw new ArgumentOutOfRangeException("f");
}

Alternatively, create a map:

static readonly Dictionary<Fruit, Func<Foo>> FruitFunctions = 
    new Dictionary<Fruit, Func<Foo>> {
    { Fruit.Apple, fnApple },
    { Fruit.Mango, fnMango },
    { Fruit.Orange, fnOrange }
};
...

myValue = FruitFunctions[f]();

I've used both techniques in various situations, and far prefer them to the suggested operator, I'm afraid.

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