问题描述
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)
})
}
什么都不输出!
subscribeOn(Schedulers.computation())
和subscribeOn(Schedulers.newThread())
都不起作用
为什么?
解决方法
在代码有时间执行之前,程序正在退出。订阅后请尝试睡眠或输入打印语句
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...
与在不同线程中执行“工作”的其他调度程序(例如.io()
和.computation()
)将具有以下输出:
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()
调度程序视为无界线程池,将computation()
视为有界线程池。