generics - 如何处理从 Java 迁移到 Kotlin 的泛型边界？

Question

我的目标是让接口(interface)方法接受实现的类类型。这是我到目前为止编写的代码:

internal interface Diff<Diff> {   //in Java I was using <? extends Diff>
    fun lessThan(other: Diff): Boolean
}

private class ADiff(private val value: Int) : Diff<ADiff> {

    override fun lessThan(other: ADiff): Boolean {
        return value < other.value
    }

}

//B can now accept even int types which is not desired
private class BDiff(private val value: Int) : Diff<Int> {

    override fun lessThan(other: Int): Boolean {
        return value < other
    }

}

Answer 1

这在 Java 中“有效”的原因是因为 <T extends Diff>使用 raw type Diff .不要这样做!

您可以获得的最接近的是使用递归类型绑定(bind):

interface Diff<T : Diff<T>> {
    fun lessThan(other: T): Boolean
}

问题是，您可以替换 Diff 的任何其他子类型。 .

但是，当使用 Diff , 使用泛型类型约束 T : Diff<T> :

fun <T : Diff<T>> diffUser(diff1: T, diff2: T) {
    println(diff1.lessThan(diff2))
}

和任何 Diff没有实现 Diff<SameType>将不被接受。

例子:

class CDiff(private val value: Int) : Diff<DDiff> { // <-- Wrong type!
    override fun lessThan(other: DDiff) = value < other.value
}

class DDiff(val value: Int) : Diff<DDiff> {
    override fun lessThan(other: DDiff) = value < other.value
}

fun test() {
    diffUser(CDiff(3), CDiff(4)) // <-- Doesn't compile due to the generic constraint
    diffUser(DDiff(3), DDiff(4))
}

This same approach被 Comparable 使用类(class)。

虽然这可行，但你真正想要的是“自我类型” and this is not supported, although it was on the roadmap at some point .我相信 JetBrains 拒绝了这个请求，尽管我找不到错误报告。

This answer详细介绍了使用 the CRT pattern 的 Java 解决方法，尽管它不一定是类型安全的。