如何在 0.3 秒后执行给定次数的动作？

Question

let expecation = expectationWithDescription("do tasks")
for i in 0...40 {

    let afterTiming = 0.3 * Double(i)
    let startTime = CFAbsoluteTimeGetCurrent()

    let delayTime = dispatch_time(DISPATCH_TIME_NOW, Int64(afterTiming * Double(NSEC_PER_SEC)))

    dispatch_after(delayTime, dispatch_get_main_queue()) {
        let timeElapsed = CFAbsoluteTimeGetCurrent() - startTime
        print("\(afterTiming) - \(timeElapsed) : \(i)")
    }
}

waitForExpectationWithTimeout(14)

在 30 执行它几乎一秒后，控制台开始表现怪异，同时显示两行和两行打印

9.0 - 9.88806998729706 : 30
9.3 - 9.88832598924637 : 31

有什么方法可以让 XCTest 更接近实际执行请求 "on correct time"？就像得到应该在 9 秒后完成的请求在 9.88 秒后没有完成..

Answer 1

使用CADisplayLink，高精度定时器

（不是很好的代码，只是快速组合在一起）

var displayLink: CADisplayLink?
var startTime: CFAbsoluteTime = 0
var nextTime: CFAbsoluteTime = 0
var index: Int = 0

func testIncrement() {
    self.startTime = CFAbsoluteTimeGetCurrent()
    self.nextTime = self.startTime

    displayLink = CADisplayLink(target: self, selector: #selector(execute))
    displayLink?.addToRunLoop(NSRunLoop.mainRunLoop(), forMode: NSRunLoopCommonModes)

    let expectation = expectationWithDescription("test")

    self.waitForExpectationsWithTimeout(20.0, handler: nil)
}

func execute() {
    let currentTime = CFAbsoluteTimeGetCurrent()

    if (currentTime - nextTime < 0) {
        return
    }

    let timeElapsed = currentTime - startTime

    print("\(timeElapsed) : \(index)")

    index += 1
    nextTime = startTime + 0.3 * CFAbsoluteTime(index)

    if (index > 30) {
        displayLink?.removeFromRunLoop(NSRunLoop.mainRunLoop(), forMode: NSRunLoopCommonModes)
    }
}

注意该方法实际上执行了多次，您必须在内部检查是否经过了足够的时间。

Answer 2

不确定您是否准备使用分派，但 NSTimer 在我的测试中表现得更加准确。这是一些示例代码运行一个动作 numOfTimes 次

var iterationNumber = 0
var startTime = CFAbsoluteTimeGetCurrent()
let numOfTimes = 30

// Start a repeating timer that calls 'action' every 0.3 seconds 
var timer = NSTimer.scheduledTimerWithTimeInterval(0.3, target: self, selector: #selector(ViewController.action), userInfo: nil, repeats: true)

func action() {
    // Print relevant information
    let timeElapsed = CFAbsoluteTimeGetCurrent() - startTime
    print("\(Double(iterationNumber+1) * 0.3) - \(timeElapsed) : \(CFAbsoluteTimeGetCurrent())")

    // Increment iteration number and check if we've hit the limit
    iterationNumber += 1
    if iterationNumber > numOfTimes {
        timer.invalidate()
    }
}

程序的部分输出：

7.8 - 7.80285203456879 : 25
8.1 - 8.10285001993179 : 26
8.4 - 8.40283703804016 : 27
8.7 - 8.70284104347229 : 28
9.0 - 9.00275802612305 : 29
9.3 - 9.3028250336647 : 30

它保持在预期时间的几毫秒内（我假设这是调用 CFAbsoluteTimeGetCurrent 所需的时间）并且不会漂移或将它们聚集在一起。

这种方法的一个缺点是它不会像在您的代码中那样预先安排每个操作，尽管我不确定这对您是否重要。

如何在 0.3 秒后执行给定次数的动作？

How to perform an action exactly after 0.3 seconds for a given number of times?

swift

grand-central-dispatch

core-foundation