immutability - Kotlin 中带有 val 的循环引用

Question

在 Kotlin 中，假设我有 data class A (val f: B)和 data class B (val f: A) .我要初始化本地var a: A和 var b: B这样a.f是 b和 b.f是 a . A.f和 B.f必须保持 vals。这种循环实例化可能吗？

data class A(val f: B)

data class B(val f: A)

fun foo() {
    var a: A
    var b: B
    // instantiate a and b here with a.f == b and b.f == a ??
}

Answer 1

如果你明确声明你的自我引用，你可以这样做 val如Lazy :

sealed class MyData {
    data class A(val x: Int) : MyData()
    data class B(val x : Int, val rb: Lazy<MyData>) : MyData() {
        val r: MyData by rb
    }
}

fun <A : Any> rec(body: (Lazy<A>) -> A): A {
    lateinit var a: A
    a = body(lazy { a })
    return a
}

fun MyData.print(gas: Int): String = if (gas <= 0) "..." else
    when(this) {
        is MyData.A -> "A(x=$x)"
        is MyData.B -> {
            val rbString = 
                if (rb.isInitialized()) 
                    r.print(gas - 1)
                else 
                    "<thunk>"
            "B(x=$x, rb=$rbString)" 
        }
    }

fun main() {
    val a = MyData.A(42)
    val b1 = MyData.B(1, lazy { a })
    println(b1.r) // Force value
    println(b1)
    val b2 = rec<MyData.B> { b2 -> MyData.B(1, b2) }
    println(b2.r.print(4))
}

这打印

A(x=42)
B(x=1, rb=A(x=42))
B(x=1, rb=B(x=1, rb=B(x=1, rb=B(x=1, rb=...))))