Question

为什么这些代码的输出不同：

Observable.fromArray("Red", "Orange", "Blue")
            .doOnNext(color ->{
                Log.i(TAG, "Color "+  color + " pushed through on $ " +Thread.currentThread());})
            .map(String::length)
            .subscribe(length ->
                    Log.i(TAG, "Length $" + length + " being recieved on $" + Thread.currentThread()));

输出：

Color Red pushed through on $ Thread[main,5,main]
Length $3 being recieved on $Thread[main,5,main]
Color Orange pushed through on $ Thread[main,5,main]
Length $6 being recieved on $Thread[main,5,main]
Color Blue pushed through on $ Thread[main,5,main]
Length $4 being recieved on $Thread[main,5,main]

和

Observable.fromArray("Red", "Orange", "Blue")
                .doOnNext(color ->{
                    Log.i(TAG, "Color "+  color + " pushed through on $ " +Thread.currentThread());})
                .map(String::length)
                .observeOn(Schedulers.computation())
                .subscribe(length ->
                        Log.i(TAG, "Length $" + length + " being recieved on $" + Thread.currentThread()));

输出：

Color Red pushed through on $ Thread[main,5,main]
Color Orange pushed through on $ Thread[main,5,main]
Color Blue pushed through on $ Thread[main,5,main]
Length $3 being recieved on $Thread[main,5,main]
Length $6 being recieved on $Thread[main,5,main]
Length $4 being recieved on $Thread[main,5,main]

为什么输出的排列不同？我在observeon方法中使用的每个调度程序，但输出不同什么时候触发doonnext并订阅？

Answer 1

doOnNext在当前线程上完成，订阅在Schedulers.computation线程上完成。因此，由于多线程竞争条件，它们完成的顺序可能是随机的。 subscriptionOn仅适用于直接调用的对象。

rxjava的输出安排不是唯一的

1 个答案: