rx-java - 一个操作符默认不对任何特定的 Scheduler 进行操作是什么意思

Question

默认情况下运算符不对任何特定调度程序进行操作是什么意思？

示例:CombineLatest .

RxJava implements this operator as combineLatest. It may take between two and nine Observables (as well as the combining function) as parameters, or a single List of Observables (as well as the combining function). It does not by default operate on any particular Scheduler.

Answer 1

这意味着运营商遵循 Scheduler 中规定的准则文档——

By default, an Observable and the chain of operators that you apply to it will do its work, and will notify its observers, on the same thread on which its Subscribe method is called.

它所做的一切都将发生在调用订阅的线程上。大多数运算符(operator)都是这样工作的。有些无法在同一线程上执行工作(例如 sample )，因此需要第二个线程来执行工作。这些运算符通常有一个用于工作的默认 Scheduler，以及一个接受 Scheduler 作为参数的版本，以便您选择他们获取第二个的位置要处理的线程。