Kotlin 挂起函数递归调用

突然发现递归调用挂起函数比调用相同的函数但没有调用要花更多的时间suspend修饰符，因此请考虑下面的代码片段（基本斐波那契数列计算）：

suspend fun asyncFibonacci(n: Int): Long = when {
    n <= -2 -> asyncFibonacci(n + 2) - asyncFibonacci(n + 1)
    n == -1 -> 1
    n == 0 -> 0
    n == 1 -> 1
    n >= 2 -> asyncFibonacci(n - 1) + asyncFibonacci(n - 2)
    else -> throw IllegalArgumentException()
}

如果我调用此函数并使用以下代码测量其执行时间：

fun main(args: Array<String>) {
    val totalElapsedTime = measureTimeMillis {
        val nFibonacci = 40

        val deferredFirstResult: Deferred<Long> = async {
            asyncProfile("fibonacci") { asyncFibonacci(nFibonacci) } as Long
        }
        val deferredSecondResult: Deferred<Long> = async {
            asyncProfile("fibonacci") { asyncFibonacci(nFibonacci) } as Long
        }

        val firstResult: Long = runBlocking { deferredFirstResult.await() }
        val secondResult: Long = runBlocking { deferredSecondResult.await() }
        val superSum = secondResult + firstResult
        println("${thread()} - Sum of two $nFibonacci'th fibonacci numbers: $superSum")
    }
    println("${thread()} - Total elapsed time: $totalElapsedTime millis")
}

我观察到进一步的结果：

commonPool-worker-2:fibonacci - Start calculation...
commonPool-worker-1:fibonacci - Start calculation...
commonPool-worker-2:fibonacci - Finish calculation...
commonPool-worker-2:fibonacci - Elapsed time: 7704 millis
commonPool-worker-1:fibonacci - Finish calculation...
commonPool-worker-1:fibonacci - Elapsed time: 7741 millis
main - Sum of two 40'th fibonacci numbers: 204668310
main - Total elapsed time: 7816 millis

但如果我删除suspend修饰符来自asyncFibonacci函数，我会得到这个结果：

commonPool-worker-2:fibonacci - Start calculation...
commonPool-worker-1:fibonacci - Start calculation...
commonPool-worker-1:fibonacci - Finish calculation...
commonPool-worker-1:fibonacci - Elapsed time: 1179 millis
commonPool-worker-2:fibonacci - Finish calculation...
commonPool-worker-2:fibonacci - Elapsed time: 1201 millis
main - Sum of two 40'th fibonacci numbers: 204668310
main - Total elapsed time: 1250 millis

我知道最好重写这样的函数tailrec它将增加其执行时间 apx。几乎 100 次，但无论如何，这是什么suspend关键字这是否会将执行速度从 1 秒降低到 8 秒？

用 来标记递归函数是不是完全愚蠢的想法suspend?

作为介绍性评论，您的测试代码设置太复杂。这个更简单的代码在压力方面实现了相同的效果suspend fun递归：

fun main(args: Array<String>) {
    launch(Unconfined) {
        val nFibonacci = 37
        var sum = 0L
        (1..1_000).forEach {
            val took = measureTimeMillis {
                sum += suspendFibonacci(nFibonacci)
            }
            println("Sum is $sum, took $took ms")
        }
    }
}

suspend fun suspendFibonacci(n: Int): Long {
    return when {
        n >= 2 -> suspendFibonacci(n - 1) + suspendFibonacci(n - 2)
        n == 0 -> 0
        n == 1 -> 1
        else -> throw IllegalArgumentException()
    }
}

我试图通过编写一个简单的函数来重现它的性能，该函数近似于suspend函数必须执行以下操作才能实现可挂起性：

val COROUTINE_SUSPENDED = Any()

fun fakeSuspendFibonacci(n: Int, inCont: Continuation<Unit>): Any? {
    val cont = if (inCont is MyCont && inCont.label and Integer.MIN_VALUE != 0) {
        inCont.label -= Integer.MIN_VALUE
        inCont
    } else MyCont(inCont)
    val suspended = COROUTINE_SUSPENDED
    loop@ while (true) {
        when (cont.label) {
            0 -> {
                when {
                    n >= 2 -> {
                        cont.n = n
                        cont.label = 1
                        val f1 = fakeSuspendFibonacci(n - 1, cont)!!
                        if (f1 === suspended) {
                            return f1
                        }
                        cont.data = f1
                        continue@loop
                    }
                    n == 1 || n == 0 -> return n.toLong()
                    else -> throw IllegalArgumentException("Negative input not allowed")
                }
            }
            1 -> {
                cont.label = 2
                cont.f1 = cont.data as Long
                val f2 = fakeSuspendFibonacci(cont.n - 2, cont)!!
                if (f2 === suspended) {
                    return f2
                }
                cont.data = f2
                continue@loop
            }
            2 -> {
                val f2 = cont.data as Long
                return cont.f1 + f2
            }
            else -> throw AssertionError("Invalid continuation label ${cont.label}")
        }
    }
}

class MyCont(val completion: Continuation<Unit>) : Continuation<Unit> {
    var label = 0
    var data: Any? = null
    var n: Int = 0
    var f1: Long = 0

    override val context: CoroutineContext get() = TODO("not implemented")
    override fun resumeWithException(exception: Throwable) = TODO("not implemented")
    override fun resume(value: Unit) = TODO("not implemented")
}

你必须调用这个

sum += fakeSuspendFibonacci(nFibonacci, InitialCont()) as Long

where InitialCont is

class InitialCont : Continuation<Unit> {
    override val context: CoroutineContext get() = TODO("not implemented")
    override fun resumeWithException(exception: Throwable) = TODO("not implemented")
    override fun resume(value: Unit) = TODO("not implemented")
}

基本上，要编译一个suspend fun编译器必须将其主体转变为状态机。每次调用还必须创建一个对象来保存机器的状态。当您恢复时，状态对象会告诉您要转到哪个状态处理程序。以上还不是全部，真正的代码更加复杂。

在解释模式下（java -Xint），我得到了与实际几乎相同的性能suspend fun，并且比启用 JIT 的真实速度快不到两倍。相比之下，“直接”函数实现的速度大约快 10 倍。这意味着所示的代码解释了可挂起性开销的很大一部分。