Kotlin 'when' 语句与 Java 'switch' [英] Kotlin 'when' statement vs Java 'switch'

问题描述

Kotlin 中的模式匹配很好，并且在 90% 的用例中它不执行下一个模式匹配这一事实很好.

Pattern matching in Kotlin is nice and the fact it does not execute the next pattern match is good in 90% of use cases.

在 Android 中，当数据库更新时，如果我们不中断代码看起来像这样，我们将使用 Java 开关属性继续下一个案例:

In Android, when database is updated, we use Java switch property to go on next case if we do not put a break to have code looking like that:

switch (oldVersion) {
    case 1: upgradeFromV1();
    case 2: upgradeFromV2(); 
    case 3: upgradeFromV3();
}

因此，如果某人有一个使用 DB 版本 1 的应用程序并且错过了使用 DB v2 的应用程序版本，他将执行所有需要的升级代码.

So if someone has an app with version 1 of the DB and missed the app version with DB v2, he will get all the needed upgrade code executed.

转换为 Kotlin 后，我们会变得一团糟:

Converted to Kotlin, we get a mess like:

when (oldVersion) {
    1 -> {
        upgradeFromV1()
        upgradeFromV2()
        upgradeFromV3()
    }
    2 -> {
        upgradeFromV2()
        upgradeFromV3()
    }
    3 -> {
        upgradeFromV3()
    }
}

这里我们只有 3 个版本，想象一下当 DB 达到版本 19 时.

Here we have only 3 versions, imagine when DB reaches version 19.

无论如何，当以同样的方式行事时，然后切换?我试图继续，但运气不佳.

Anyway to makes when acting in the same way then switch? I tried to continue without luck.

推荐答案

简单但冗长的解决方案是:

Simple but wordy solution is:

if (oldVersion <= 1) upgradeFromV1()
if (oldVersion <= 2) upgradeFromV2()
if (oldVersion <= 3) upgradeFromV3()

函数引用的另一种可能解决方案:

Another possible solution with function references:

fun upgradeFromV0() {}
fun upgradeFromV1() {}
fun upgradeFromV2() {}
fun upgradeFromV3() {}

val upgrades = arrayOf(::upgradeFromV0, ::upgradeFromV1, ::upgradeFromV2, ::upgradeFromV3)

fun upgradeFrom(oldVersion: Int) {
    for (i in oldVersion..upgrades.lastIndex) {
        upgrades[i]()
    }
}

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