How to program a delay in Swift 3

In earlier versions of Swift, one could create a delay with the following code:

let time = dispatch_time(dispatch_time_t(DISPATCH_TIME_NOW), 4 * Int64(NSEC_PER_SEC))
dispatch_after(time, dispatch_get_main_queue()) {
    //put your code which should be executed with a delay here
}

But now, in Swift 3, Xcode automatically changes 6 different things but then the following error appears: "Cannot convert DispatchTime.now to expected value dispatch_time_t aka UInt64."

How can one create a delay before running a sequence of code in Swift 3?

Solution 1:

After a lot of research, I finally figured this one out.

DispatchQueue.main.asyncAfter(deadline: .now() + 2.0) { // Change `2.0` to the desired number of seconds.
   // Code you want to be delayed
}

This creates the desired "wait" effect in Swift 3 and Swift 4.

Inspired by a part of this answer.

Solution 2:

I like one-line notation for GCD, it's more elegant:

    DispatchQueue.main.asyncAfter(deadline: .now() + 42.0) {
        // do stuff 42 seconds later
    }

Also, in iOS 10 we have new Timer methods, e.g. block initializer:

(so delayed action may be canceled)

    let timer = Timer.scheduledTimer(withTimeInterval: 42.0, repeats: false) { (timer) in
        // do stuff 42 seconds later
    }

Btw, keep in mind: by default, timer is added to the default run loop mode. It means timer may be frozen when the user is interacting with the UI of your app (for example, when scrolling a UIScrollView) You can solve this issue by adding the timer to the specific run loop mode:

RunLoop.current.add(timer, forMode: .common)

At this blog post you can find more details.