subscribeOn(Schedulers.io()) 不起作用但 subscribeOn(Schedulers.trampoline()) 起作用
subscribeOn(Schedulers.io()) not working but subscribeOn(Schedulers.trampoline()) does
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)
})
}
打印出来:
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())
也不行
为什么?
程序在代码有时间执行之前就退出了。在 subscribe
之后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...")
}
用蹦床输出是
map in thread main
map in thread main
231
just subscribed in last statement!
exiting main thread...
与其他在不同线程中执行“工作”的调度程序(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()
调度程序视为无界线程池,computation()
视为有界线程池。
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)
})
}
打印出来:
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())
也不行
为什么?
程序在代码有时间执行之前就退出了。在 subscribe
之后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...")
}
用蹦床输出是
map in thread main
map in thread main
231
just subscribed in last statement!
exiting main thread...
与其他在不同线程中执行“工作”的调度程序(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()
调度程序视为无界线程池,computation()
视为有界线程池。