Java switch case - 默认与显式枚举 [英] Java switch case - default vs explicit enumeration

问题描述

我使用的是 Java 6.

I'm using Java 6.

假设我有一个包含 6 个值的枚举，按 A 到 F 的顺序排列.大约 4 个值的处理方式相同.我可以这样写.

Suppose I have an enum with 6 values, ordered A to F. About 4 of the values were handled the same. I could write it like this.

switch (whichType) {
    case A:
    case B:
    case C:
    case D:
        return task();
    case E:
        return someothertask();
    case F:
        return anothersomeothertask();
}

或者像这样.

switch (whichType) {
    case E:
        return someothertask();
    case F:
        return anothersomeothertask();
    default:
        return task();
}

空值永远不会到达这个开关.

Null values will never reach this switch.

在简洁明了方面，第二种方式更好.就明确性而言，我认为第一种方法更好.

In terms of conciseness and clarity, the second approach is better. In terms of being explicit, I think the first approach is better.

每种方法还有其他优点/缺点吗?

Are there any other pros/cons of each approach?

此外，这个简单的问题有重复的风险，但我尝试过，但还没有在任何地方找到它.如果我搜索得不够好，我深表歉意.

Also, this simple question risks being a duplicate, but I tried and couldn't find it asked anywhere yet. I apologize if I haven't searched it good enough.

推荐答案

如果枚举是绝对的，永远固定在六个值，那么两者都很好.否则，请考虑枚举的可能的第七个值.如果 E 和 F 是与此 switch 逻辑相关的仅有的两个可能的异常值，并且任何其他值都将与 A 到 D 落入同一桶中，请继续使用 default.否则，每个值都有一个 case 会更安全.

Both are fine if that enum is absolutely, positively fixed at six values forever. Otherwise, think about what a possible seventh value for the enum might be. If E and F are the only two possible outliers with respect to this switch logic and any other value would fall in the same bucket as A through D, go ahead and use the default. Otherwise, you're safer to have a case per value.

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