这对我很管用。

斯威夫特3:

let time1 = 8.23
let time2 = 3.42

// Delay 2 seconds

DispatchQueue.main.asyncAfter(deadline: .now() + 2.0) {
    print("Sum of times: \(time1 + time2)")
}

objective - c:

CGFloat time1 = 3.49;
CGFloat time2 = 8.13;

// Delay 2 seconds

dispatch_after(dispatch_time(DISPATCH_TIME_NOW, (int64_t)(2.0 * NSEC_PER_SEC)), dispatch_get_main_queue(), ^{
    CGFloat newTime = time1 + time2;
    NSLog(@"New time: %f", newTime);
});

保留当前队列!

除了很好地回答这个问题之外，您还可以考虑保留当前队列以防止不必要的主队列操作(例如，当您试图延迟一些异步操作时)。

func after(_ delay: TimeInterval,
           perform block: @escaping ()->(),
           on queue: DispatchQueue = OperationQueue.current?.underlyingQueue ?? .main) { // So this `queue` preserves the current queue and defaulted to the `main`. Also the caller can pass in the desired queue explicitly
    queue.asyncAfter(deadline: .now() + delay, execute: block)
}

用法:

after(3) {
    // will be executed on the caller's queue
    print(Date())
}

在Swift 5中，在下面使用:

 DispatchQueue.main.asyncAfter(deadline: .now() + 0.2, execute: closure) 

// time gap, specify unit is second
DispatchQueue.main.asyncAfter(deadline: .now() + .seconds(2)) {
            Singleton.shared().printDate()
        }
// default time gap is second, you can reduce it
    DispatchQueue.main.asyncAfter(deadline: .now() + 0.2) {
          // just do it!
    }

我经常使用dispatch_after，所以我写了一个顶级实用函数来简化语法:

func delay(delay:Double, closure:()->()) {
    dispatch_after(
        dispatch_time(
            DISPATCH_TIME_NOW,
            Int64(delay * Double(NSEC_PER_SEC))
        ),
        dispatch_get_main_queue(), closure)
}

现在你可以这样说:

delay(0.4) {
    // do stuff
}

哇，一种你可以改进的语言。还有什么比这更好的呢?

更新Swift 3, Xcode 8种子6

看起来几乎不值得费心，现在他们已经改进了调用语法:

func delay(_ delay:Double, closure:@escaping ()->()) {
    let when = DispatchTime.now() + delay
    DispatchQueue.main.asyncAfter(deadline: when, execute: closure)
}

Swift 3和4:

您可以在DispatchQueue上创建一个扩展，并在内部添加使用DispatchQueue asyncAfter函数的函数delay

extension DispatchQueue {
    static func delay(_ delay: DispatchTimeInterval, closure: @escaping () -> ()) {
        let timeInterval = DispatchTime.now() + delay
        DispatchQueue.main.asyncAfter(deadline: timeInterval, execute: closure)
    }
}

使用:

DispatchQueue.delay(.seconds(1)) {
    print("This is after delay")
}

斯威夫特 3+

这在Swift 3+中是超级简单和优雅的:

DispatchQueue.main.asyncAfter(deadline: .now() + 4.5) {
    // ...
}

年长的回答:

为了扩展Cezary的答案，它将在1纳秒后执行，我必须执行以下操作以在4秒半后执行。

let delay = 4.5 * Double(NSEC_PER_SEC)
let time = dispatch_time(DISPATCH_TIME_NOW, Int64(delay))
dispatch_after(time, dispatch_get_main_queue(), block)

编辑:我发现我原来的代码有一点错误。如果不将NSEC_PER_SEC转换为Double类型，隐式类型将导致编译错误。

如果有人能提出一个更优的解决方案，我很乐意听听。