RxJava Observable Single显示当前线程

Question

I'm trying to understand how observable works and I'd like to show the current thread to get a better understanding. The whole app is built with java 8 and make use of lambda expressions. No having much experience with it I'm finding some troubles to show my current thread in an expression like this:

.subscribeOn(Schedulers.io())                          
.observeOn(Schedulers.computation())
.flatMap(..... -> {

I like to put something like this:

log.info("Current Thread:", Thread.currentThread().name)

But I couldn't find a way to do it as subscribeOn(Schedulers.io()) and observeOn(Schedulers.computation()) return and Single and there is no way to put something like this:

.subscribeOn(Schedulers.io())
.log.info("Current Thread:", Thread.currentThread().name)
.observeOn(Schedulers.computation())
.log.info("Current Thread:", Thread.currentThread().name)
.flatMap(..... -> {

Thanks

Answer 1

You are trying just to log the current Thread, correct?

Just use map for log and return the same value:

.subscribeOn(Schedulers.io())
.map(it -> {
    log.info("Current Thread:", Thread.currentThread().name)
    return it;
})
.observeOn(Schedulers.computation())
.map(it -> {
    log.info("Current Thread:", Thread.currentThread().name)
    return it;
})
.flatMap(..... -> {