Waiting until the task finishes - swift

Question

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 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)")
}

Answer 2

In Swift 3, there is no need for completion handler when DispatchQueue finishes one task.
Furthermore you can achieve your goal in different ways
One way is this:
var a: Int?
let queue = DispatchQueue(label: "com.app.queue")
queue.sync {
for i in 0..<10 {
print("Ⓜ️" , i)
a = i
}
}
print("After Queue \(a)")
It will wait until the loop finishes but in this case your main thread will block.
You can also do the same thing like this:
let myGroup = DispatchGroup()
myGroup.enter()
//// Do your task
myGroup.leave() //// When your task completes
myGroup.notify(queue: DispatchQueue.main) {
////// do your remaining work
}
One last thing: If you want to use completionHandler when your task completes using DispatchQueue, you can use DispatchWorkItem.
Here is an example how to use DispatchWorkItem:
let workItem = DispatchWorkItem {
// Do something
}
let queue = DispatchQueue.global()
queue.async {
workItem.perform()
}
workItem.notify(queue: DispatchQueue.main) {
// Here you can notify you Main thread
}

Answer 3

Swift 5 version of the solution
func myCriticalFunction() {
var value1: String?
var value2: String?
let group = DispatchGroup()
group.enter()
//async operation 1
DispatchQueue.global(qos: .default).async {
// Network calls or some other async task
value1 = //out of async task
group.leave()
}
group.enter()
//async operation 2
DispatchQueue.global(qos: .default).async {
// Network calls or some other async task
value2 = //out of async task
group.leave()
}
group.wait()
print("Value1 \(value1) , Value2 \(value2)")
}

Answer 4

Use dispatch group
dispatchGroup.enter()
FirstOperation(completion: { _ in
dispatchGroup.leave()
})
dispatchGroup.enter()
SecondOperation(completion: { _ in
dispatchGroup.leave()
})
dispatchGroup.wait() // Waits here on this thread until the two operations complete executing.

Answer 5

In Swift 5.5+ you can take advantage of Swift Concurrency which allows to return a value from a closure dispatched to the main thread
func myFunction() async {
var a : Int?
a = await MainActor.run {
let b = 3
return b
}
print(a)
}
Task {
await myFunction()
}

Answer 6

Swift 4
You can use Async Function for these situations. When you use DispatchGroup(),Sometimes deadlock may be occures.
var a: Int?
#objc func myFunction(completion:#escaping (Bool) -> () ) {
DispatchQueue.main.async {
let b: Int = 3
a = b
completion(true)
}
}
override func viewDidLoad() {
super.viewDidLoad()
myFunction { (status) in
if status {
print(self.a!)
}
}
}

Answer 7

Somehow the dispatchGroup enter() and leave() commands above didn't work for my case.
Using sleep(5) in a while loop on the background thread worked for me though. Leaving here in case it helps someone else and it didn't interfere with my other threads.