Waiting until the task finishes

Bartosz Woźniak picture Bartosz Woźniak · Feb 27, 2017 · Viewed 107.1k times · Source

How could I make my code wait until the task in DispatchQueue finishes? Does it need any CompletionHandler or something?

func myFunction() {
    var a: Int?

    DispatchQueue.main.async {
        var b: Int = 3
        a = b
    }

    // wait until the task finishes, then print 

    print(a) // - this will contain nil, of course, because it
             // will execute before the code above

}

I'm using Xcode 8.2 and writing in Swift 3.

Answer

shallowThought picture shallowThought · Feb 27, 2017

Use DispatchGroups to achieve this. You can either get notified when the group's enter() and leave() calls are balanced:

func myFunction() {
    var a: Int?

    let group = DispatchGroup()
    group.enter()

    DispatchQueue.main.async {
        a = 1
        group.leave()
    }

    // does not wait. But the code in notify() gets run 
    // after enter() and leave() calls are balanced

    group.notify(queue: .main) {
        print(a)
    }
}

or you can wait:

func myFunction() {
    var a: Int?

    let group = DispatchGroup()
    group.enter()

    // avoid deadlocks by not using .main queue here
    DispatchQueue.global(attributes: .qosDefault).async {
        a = 1
        group.leave()
    }

    // wait ...
    group.wait()

    print(a) // you could also `return a` here
}

Note: group.wait() blocks the current queue (probably the main queue in your case), so you have to dispatch.async on another queue (like in the above sample code) to avoid a deadlock.