英文:

Does rx-java IO Scheduler maintains order?

问题

我想知道 `Schedulers.io()` 是否会按照调用者的顺序执行任务。

```java
public class SaveTest {

    @Inject
    private MyRepository myRepository;

    public void save()  {
        Observable.range(0, 20)
                .map(l -> new MyModel(l))
                .observeOn(Schedulers.io())
                .subscribe(myRepository::save);
    }
}

在插入新数据时，即使不遵守顺序，也不会有问题。但是在更新时，例如：

    BankAccount account = new BankAccount();
    account.deposit(500);
    account.withdraw(50);

并且每个操作在反应流中保存，顺序非常重要。

我的问题是：IO 调度程序是否保持顺序？

<details>
<summary>英文:</summary>

I&#39;m wondering if the `Schedulers.io()` will execute tasks in the order of the caller.


```java
public class SaveTest {

    @Inject
    private MyRepository myRepository;

    public void save()  {
        Observable.range(0, 20)
                .map(l -&gt; new MyModel(l))
                .observeOn(Schedulers.io())
                .subscribe(myRepository::save);
    }
}

When inserting new data, even if the order is not respected there is no problem. But when updating ex :

    BankAccount account = new BankAccount();
    account.deposit(500);
    account.withdraw(50);

and each operation results in a save in a reactive stream the order is very important.

My question is : Does IO Scheduler maintains order ?

答案1

得分: 4

是的，响应式流会维护事件的顺序，即使您将事件交给不同的调度程序处理。

在内部，这些事件被放置在队列中，由调度程序选择的单线程工作器（如此处文档所述）按顺序从队列中获取它们。

英文:

Yes a reactive stream maintains the order of events, even if you hand off the events to a different scheduler like this.

Internally the events are put in a queue, and a single threaded worker selected by the scheduler (as documented here) picks them up from the queue, in order.