dispatch_after - GCD in Swift?

Question

I\'ve gone through the iBook from Apple, and couldn\'t find any definition of it:

Can someone explain the structure of dispatch_after?

d         


        

Answer 1

Apple has a dispatch_after snippet for Objective-C:

dispatch_after(dispatch_time(DISPATCH_TIME_NOW, (int64_t)(<#delayInSeconds#> * NSEC_PER_SEC)), dispatch_get_main_queue(), ^{
    <#code to be executed after a specified delay#>
});

Here is the same snippet ported to Swift 3:

DispatchQueue.main.asyncAfter(deadline: DispatchTime.now() + <#delayInSeconds#>) {
  <#code to be executed after a specified delay#>
}

Answer 2

Here is synchronous version of asyncAfter in Swift:

let deadline = DispatchTime.now() + .seconds(3)
let semaphore = DispatchSemaphore.init(value: 0)
DispatchQueue.global().asyncAfter(deadline: deadline) {
    dispatchPrecondition(condition: .onQueue(DispatchQueue.global()))
    semaphore.signal()
}

semaphore.wait()

Along with asynchronous one:

let deadline = DispatchTime.now() + .seconds(3)
DispatchQueue.main.asyncAfter(deadline: deadline) {
    dispatchPrecondition(condition: .onQueue(DispatchQueue.global()))
}

Answer 3

In Swift 4

Use this snippet:

    let delayInSec = 1.0
    DispatchQueue.main.asyncAfter(deadline: .now() + delayInSec) {
       // code here
       print("It works")
    }

Answer 4

use this code to perform some UI related task after 2.0 seconds.

            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 version

Following closure function execute some task after delay on main thread.

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

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

Call this function like:

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

Answer 5

Swift 3+

This is super-easy and elegant in Swift 3+:

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

Older Answer:

To expand on Cezary's answer, which will execute after 1 nanosecond, I had to do the following to execute after 4 and a half seconds.

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)

Edit: I discovered that my original code was slightly wrong. Implicit typing causes a compile error if you don't cast NSEC_PER_SEC to a Double.

If anyone can suggest a more optimal solution I'd be keen to hear it.

Answer 6

Swift 3.0 version

Following closure function execute some task after delay on main thread.

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

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

Call this function like:

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