我翻遍了苹果的iBook,找不到任何关于它的定义:

有人能解释一下dispatch_after的结构吗?

dispatch_after(<#when: dispatch_time_t#>, <#queue: dispatch_queue_t?#>, <#block: dispatch_block_t?#>)

当前回答

Matt的语法非常好,如果你需要使块失效,你可能想使用这个:

typealias dispatch_cancelable_closure = (cancel : Bool) -> Void

func delay(time:NSTimeInterval, closure:()->Void) ->  dispatch_cancelable_closure? {

    func dispatch_later(clsr:()->Void) {
        dispatch_after(
            dispatch_time(
                DISPATCH_TIME_NOW,
                Int64(time * Double(NSEC_PER_SEC))
            ),
            dispatch_get_main_queue(), clsr)
    }

    var closure:dispatch_block_t? = closure
    var cancelableClosure:dispatch_cancelable_closure?

    let delayedClosure:dispatch_cancelable_closure = { cancel in
        if closure != nil {
            if (cancel == false) {
                dispatch_async(dispatch_get_main_queue(), closure!);
            }
        }
        closure = nil
        cancelableClosure = nil
    }

    cancelableClosure = delayedClosure

    dispatch_later {
        if let delayedClosure = cancelableClosure {
            delayedClosure(cancel: false)
        }
    }

    return cancelableClosure;
}

func cancel_delay(closure:dispatch_cancelable_closure?) {

    if closure != nil {
        closure!(cancel: true)
    }
}

使用方法如下

let retVal = delay(2.0) {
    println("Later")
}
delay(1.0) {
    cancel_delay(retVal)
}

学分

上面的链接似乎坏了。来自Github的原始Objc代码

其他回答

更清晰的结构概念:

dispatch_after(when: dispatch_time_t, queue: dispatch_queue_t, block: dispatch_block_t?)

dispatch_time_t为UInt64。dispatch_queue_t实际上是一个NSObject的类型别名,但是你应该使用你熟悉的GCD方法来获取队列。该区块是一个快速关闭。具体来说,dispatch_block_t定义为()-> Void,相当于()->()。

使用示例:

let delayTime = dispatch_time(DISPATCH_TIME_NOW, Int64(1 * Double(NSEC_PER_SEC)))
dispatch_after(delayTime, dispatch_get_main_queue()) {
    print("test")
}

编辑:

我推荐使用@matt的延时函数。

编辑2:

在Swift 3中,GCD将会有新的包装器。请看这里:https://github.com/apple/swift-evolution/blob/master/proposals/0088-libdispatch-for-swift3.md

原来的例子在Swift 3中是这样写的:

let deadlineTime = DispatchTime.now() + .seconds(1)
DispatchQueue.main.asyncAfter(deadline: deadlineTime) {
    print("test")
}

注意,你可以将deadlineTime声明写为DispatchTime.now() + 1.0,并得到相同的结果,因为+操作符被重写如下(与-类似):

func +(time: DispatchTime, seconds: Double) -> DispatchTime . func +(time: DispatchTime, seconds: Double func +(time: DispatchWalltime, interval: DispatchTimeInterval) -> DispatchWalltime . func +(time: DispatchWalltime, interval: DispatchTimeInterval

这意味着如果您不使用DispatchTimeInterval enum,而只是写一个数字,则假定您使用的是秒。

使用这段代码在2.0秒后执行一些UI相关的任务。

            let delay = 2.0
            let delayInNanoSeconds = dispatch_time(DISPATCH_TIME_NOW, Int64(delay * Double(NSEC_PER_SEC)))
            let mainQueue = dispatch_get_main_queue()

            dispatch_after(delayInNanoSeconds, mainQueue, {

                print("Some UI related task after delay")
            })

Swift 3.0版本

以下闭包函数在主线程上执行一些延迟后的任务。

func performAfterDelay(delay : Double, onCompletion: @escaping() -> Void){

    DispatchQueue.main.asyncAfter(deadline: DispatchTime.now() + delay, execute: {
       onCompletion()
    })
}

像这样调用这个函数:

performAfterDelay(delay: 4.0) {
  print("test")
}

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())
}

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。