主页 / 问题 / 1163610
Accepted
Ksenia
Asked:2020-08-08 20:36:05 +0800 CST2020-08-08 20:36:05 +0800 CST

Flowable 与 flatMap 的神秘行为

  • 772

不知道为什么，万一我showcaseFlatMap用以下代码调用方法：

fun showcaseFlatMap() { 
    val colors: Flowable<String> = Flowable.just("orange", "red", "green")
        .flatMap<String> { colorName -> simulateRemoteOperation(colorName) }
    colors.subscribe { value -> println("Subscriber received: {$value}") }
}

fun simulateRemoteOperation(color: String): Flowable<String> {
    return Flowable.intervalRange(1, color.length.toLong(), 0, 200, TimeUnit.MILLISECONDS)
            .map { iteration -> color + iteration }
}

我得到以下结果：

Subscriber received: {orange1}
Subscriber received: {red1}
Subscriber received: {green1}
Subscriber received: {orange2}
Subscriber received: {red2}
Subscriber received: {green2}
Subscriber received: {orange3}
Subscriber received: {red3}
Subscriber received: {green3}
Subscriber received: {orange4}
Subscriber received: {green4}
Subscriber received: {orange5}
Subscriber received: {green5}
Subscriber received: {orange6}

但是，如果我更改此方法中的代码：

val colors: Flowable<String> = Flowable.just("orange", "red", "green")
        .flatMap<String> { colorName -> simulateRemoteOperation(colorName) }

编码：

val colors: Flowable<String> = Flowable.just("orange", "red", "green")
colors.flatMap<String> { colorName -> simulateRemoteOperation(colorName) }

结果完全不同：

Subscriber received: {orange}
Subscriber received: {red}
Subscriber received: {green}

造成这种结果变化的原因是什么？我预计输出不会改变。

kotlin

1 个回答

  1. Best Answer

    因为flatMap，实际上，所有运算符都会返回一个Flowable您忽略的新对象（在这种情况下），而不是更改当前对象的状态。事实证明，您只订阅Flowable.just("orange", "red", "green")

    val colors: Flowable<String> = Flowable.just("orange", "red", "green")
val colorsMapped = colors.flatMap<String> { colorName -> simulateRemoteOperation(colorName) }
colorsMapped.subscribe { value -> println("Subscriber received: {$value}") }
    • 2

相关问题