I'm wondering why jest.useFakeTimers
is working with setTimeout
but not with the delay operator of RxJs:
jest.useFakeTimers();
import {Observable} from 'rxjs/Observable';
import 'rxjs';
describe('timers', () => {
it('should resolve setTimeout synchronously', () => {
const spy = jest.fn();
setTimeout(spy, 20);
expect(spy).not.toHaveBeenCalled();
jest.runTimersToTime(20);
expect(spy).toHaveBeenCalledTimes(1);
});
it('should resolve setInterval synchronously', () => {
const spy = jest.fn();
setInterval(spy, 20);
expect(spy).not.toHaveBeenCalled();
jest.runTimersToTime(20);
expect(spy).toHaveBeenCalledTimes(1);
jest.runTimersToTime(20);
expect(spy).toHaveBeenCalledTimes(2);
});
it('should work with observables', () => {
const delay$ = Observable.of(true).delay(20);
const spy = jest.fn();
delay$.subscribe(spy);
expect(spy).not.toHaveBeenCalled();
jest.runTimersToTime(2000);
expect(spy).toHaveBeenCalledTimes(1);
});
});
FYI: using 20 or 2000 as an argument for jest.runTimersToTime
makes no difference.
Using jest.runAllTimers()
makes the test past
The delay
operator does not work with Jest's fake timers because the delay
operator's implementation uses its scheduler's concept of time - which is unrelated to Jest's concept of fake time.
The source is here:
while (queue.length > 0 && (queue[0].time - scheduler.now()) <= 0) {
queue.shift().notification.observe(destination);
}
The current (non-fake) time is assigned to notifications when they are created and the current time is what the scheduler's now
method returns. When Jest's fake timers are used, an insufficient amount of actual (non-fake) time will have elapsed and the notifications will remain in the queue.
To write RxJS tests using fake or virtual time, you can use the VirtualTimeScheduler
. See this answer. Or you can use the TestScheduler
and marble tests.
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