Question

我有以下代码

enum class State {
    EMPTY, LOADING, DATA, ERROR
}
private var currentState = State.EMPTY

private var timer: Timer? = null

fun startTimer() {
    if (timer == null) {
        timer = fixedRateTimer(period = 1000, action = {

            mainvView.showView(currentState)
            currentState = when (currentState) {
                State.EMPTY -> State.LOADING
                State.LOADING -> State.DATA
                State.DATA -> State.ERROR
                State.ERROR -> State.EMPTY
            }
        })
    }
}

以1s的固定间隔迭代枚举。

我不是特别喜欢这件作品

            currentState = when (currentState) {
                State.EMPTY -> State.LOADING
                State.LOADING -> State.DATA
                State.DATA -> State.ERROR
                State.ERROR -> State.EMPTY
            }

因为我希望更多地采用下面的内容，所以我不需要明确写出所有的状态转换

            if (currentState == State.ERROR) {
                currentState = State.EMPTY
            } else {
                currentState++
            }

显然currentState++是不可能的。还有其他方法吗？

Answer 1

如果您使用像

这样的运算符重载，则可以执行currentState++之类的操作

operator fun State.inc(): State = when(this){
            State.EMPTY -> State.LOADING
            State.LOADING -> State.DATA
            State.DATA -> State.ERROR
            State.ERROR -> State.EMPTY
        }

或者只是想圈出枚举

operator fun State.inc(): State {
             val size = State.values().size
             return State.values()[(this.ordinal + 1) % size]
}

Answer 2

您可以使用通用功能

inline operator fun <reified E : Enum<E>> E.inc() =
    enumValues<E>()[(ordinal + 1) % enumValues<E>().size]

迭代Kotlin中的枚举值

2 个答案: