我想为两个不同的源发出两个不同的信号,但是我崩溃了第29行的Thread 1: EXC_BAD_ACCESS (code=1, address=0x0)
中的RxSwift/Event.swift
:return "next(\(value))"
这是我的简化代码:
let scheduler = TestScheduler(initialClock: 0)
locationsFactory = TestableLocationsFactory()
locationsFactory.didReceiveRegion = scheduler.createColdObservable([
.next(100, regionEvents[0]),
.next(200, regionEvents[1])
]).asObservable()
locationsFactory.location = scheduler.createColdObservable([
.next(120, locations[0]),
.next(220, locations[1])
]).asObservable()
let result = scheduler.createObserver(LocationChange.self)
let dispatcher = BestAccuracyLocationsDispatcher(persistenceService: persistenceService, apiClient: api, locationManager: locationsFactory)
subscription = dispatcher.dispatcher.subscribe(result)
scheduler.start()
let events = result.events
XCTAssertEqual(events, [
.next(120, LocationChange(location: locations[0], trigger: .updateLocations)),
.next(220, LocationChange(location: locations[1], trigger: .updateLocations)),
])
当我删除locationsFactory.didReceiveRegion
或locationsFactory.location
时,它起作用。
我可以创建两个不同的调度程序的可观察对象吗?
我使用了区域let region = CLRegion()
的模拟物,这是一个抽象类。当我将其更改为:
let region = CLCircularRegion(center: CLLocationCoordinate2D(latitude: 5.555, longitude: 6.666), radius: Double(50), identifier: "xxx")