具有實現接口的類的泛型類型的接口？

Question

擁有一個實現具有“This”類泛型類型的接口的類

有沒有辦法在沒有演員的情況下做到這一點？

簡單代碼：

interface Triggerable<This: Triggerable<This>> {
    var trigger: (This) -> Unit
    fun triggerNow() = trigger(this as This)
}
class Test : Triggerable<Test>{
    override var trigger: (Test) -> Unit = { /*...*/ }
}

同樣稍微復雜一點：

interface TriggerInterface<T> {
    val trigger: (T) -> Unit
    fun triggerNow()
}
interface Triggerable<T: Triggerable<T>>: TriggerInterface<T> {
    override fun triggerNow() = trigger(this as T)
}
interface Signalable<T>: TriggerInterface<T> {
    var value: T
    override fun triggerNow() = trigger(value)
}
class Test : Triggerable<Test>{
    override val trigger: (Test) -> Unit = { /*...*/ }
}

Answer 1

應該可以這樣

 interface TriggerInterface<T: TriggerInterface<T>> {
    val trigger: (T) -> Unit
    fun triggerNow()

    fun getThis(): T
 }

 interface Triggerable<T: TriggerInterface<T>>: TriggerInterface<T> {
     override fun triggerNow() = trigger(getThis())
 }

 class Test : Triggerable<Test>{
     override fun getThis(): Test = this

     override val trigger: (Test) -> Unit = { /*...*/ }
 }

檢查http://www.angelikalanger.com/GenericsFAQ/FAQSections/ProgrammingIdioms.html#FAQ206

我個人還建議您重新考慮是否真的需要 TriggerInterface 和 Triggerable （其中一個繼承自另一個）。

Answer 2

調用override fun triggerNow() = trigger(this as T)您正試圖將Triggerable<T>為T這就是編譯器警告您未經檢查的強制轉換的原因

有

    val trigger: (TriggerInterface<T>) -> Unit

將允許您在不進行轉換的情況下調用觸發器

    override fun triggerNow() = trigger(this)

實際執行

override val trigger: (Test) -> Unit = { /*...*/ }

您需要一個Test實例來傳遞觸發器內部，該實例尚未在代碼中的任何位置聲明