如何在Kotlin中实现KFunctionN?

问题描述

如果我定义

class MyF : KFunction0<Int> {
    override val name: String = "f"
    override fun invoke() = 42

    override val annotations: List<Annotation>
        get() = TODO()
    ....
}

我会写

val f0: KFunction0<Int> = MyF()
assertEquals("f", f0.name)

但如果我尝试

assertEquals(42, f0())

我得到java.lang.ClassCastException: MyF cannot be cast to kotlin.jvm.functions.Function0

如何定义自己的KFunction0实现?

我不能使用() -> Int，因为我需要name属性.

I can't work with () -> Int because I need the name property.

我正在使用Kotlin 1.3.21.

I'm using Kotlin 1.3.21.

另外-看来我可以跑步

val f02 = MyF()
assertEquals(42, f02())

我的实际用例是

推荐答案

从Kotlin JVM显式实现内部类绝对不是一个好主意.由于某种原因，您在IntelliJ或Android Studio中没有该类的代码完成

It is definitely not a good idea to explicitly implement an internal class from Kotlin JVM. You do not have code completion for that classes in IntelliJ or Android Studio for a reason

您可以改用可调用的引用，以使Kotlin编译器为您生成所有必需的类. https://kotlinlang.org/docs/reference/reflection.html#callable-引用

You may use callable references instead, to make Kotlin compiler generate all necessary classes for you.https://kotlinlang.org/docs/reference/reflection.html#callable-references

    fun myfun() = 42
    val kFunction = ::myfun


    println(kFunction.name)
    println(kFunction())

好处-将来的Kotlin版本不太可能破坏该代码(并可能破坏您的继承者类)

The benefit - future version of Kotlin will unlikely to break that code (and may break you inheritor class)

如果您需要一个更长的名称，可以这样声明

Should you need a longer name for the function, you may declare it like that

fun `my function with long name`() = 43

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