Waiting until the task finishes

Question

How could I make my code wait until the task in DispatchQueue finishes? Does it need any completion handler 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 1

If you need to hide the asynchronous nature of myFunction from the caller, use DispatchGroups to achieve this. Otherwise, use a completion block. Find samples for both below.

---

DispatchGroup Sample

You can either get notified when the group's enter() and leave() calls are balanced:

func myFunction() {
    var a = 0

    let group = DispatchGroup()
    group.enter()

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

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

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

or you can wait:

func myFunction() {
    var a = 0

    let group = DispatchGroup()
    group.enter()

    // avoid deadlocks by not using .main queue here
    DispatchQueue.global(qos: .default).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.

---

Completion Block Sample

func myFunction(completion: @escaping (Int)->()) {
    var a = 0

    DispatchQueue.main.async {
        let b: Int = 1
        a = b
        completion(a) // call completion after you have the result
    }
}

---

// on caller side:
myFunction { result in
    print("result: \(result)")
}