subscribeOn(Schedulers.io())不起作用，但subscribeOn(Schedulers.bra

Question

fun main() {
Single.just("faefw")
.subscribeOn(Schedulers.trampoline())
.map {
println("map in thread ${Thread.currentThread().name}")
}
.flatMap {
return@flatMap Single.just(231).map {
println("map in thread ${Thread.currentThread().name}")
return@map it
}
}
.subscribe({
println(it)
}, {
println(it)
})
}

打印输出：

map in thread main
map in thread main
231

但是

fun main() {
Single.just("faefw")
.subscribeOn(Schedulers.io())
.map {
println("map in thread ${Thread.currentThread().name}")
}
.flatMap {
return@flatMap Single.just(231).map {
println("map in thread ${Thread.currentThread().name}")
return@map it
}
}
.subscribe({
println(it)
}, {
println(it)
})
}

不输出任何内容！

subscribeOn(Schedulers.computation())和subscribeOn(Schedulers.newThread())都不起作用

为什么？

Answer 1

程序在代码执行之前就退出了。尝试睡眠或在订阅后放入打印声明

fun main() {
Single.just("faefw")
.subscribeOn(Schedulers.io())
.map {
println("map in thread ${Thread.currentThread().name}")
}
.flatMap {
return@flatMap Single.just(231).map {
println("map in thread ${Thread.currentThread().name}")
return@map it
}
}
.subscribe({
println(it)
}, {
println(it)
})
println("just subscribed in last statement!")
Thread.sleep(500)
println("exiting main thread...")
}

蹦床的输出是

map in thread main
map in thread main
231
just subscribed in last statement!
exiting main thread...

与其他调度器一起；工作；在不同的线程中(如.io()和.computation()(将有如下输出：

just subscribed in last statement!
map in thread RxComputationThreadPool-1
map in thread RxComputationThreadPool-1
231
exiting main thread...

PS。您可以将io()调度程序视为无边界线程池，将computation()视为有边界线程池。