Swift - Prevent function from being called twice

Question

My main code is to long to put here so i'll try with a simpler explanation. Let's say i have a global variable called "count" that gets used by functions. If count was to get called twice for w.e reason as shown in the code below... how can i make it so the output "10 , 10" rather then "10 , 20".

var count = 0

override func viewDidLoad() {

  helloWorld()

}


override func viewWillAppear(_ animated: Bool) {
    helloWorld()
}



func helloWorld(){
    var counter = 0
    for i in 1...10{
        count = count + 1
        counter = counter + 1
    }
    print(count)
}

Possible duplicate of [Whither dispatch\_once in Swift 3?](https://stackoverflow.com/questions/37801407/whither-dispatch-once-in-swift-3) — ielyamani, Sep 27 '18 at 20:56

Oleh Zayats · Answer 1 · 2018-09-27T21:11:46.530

Use DispatchQueue with static token (cached through app lifecycle), here's a useful extension:

extension DispatchQueue {

    private static var tokens: [String] = [] // static, so tokens are cached

    class func once(executionToken: String, _ closure: () -> Void) {
        // objc_sync_enter/objc_sync_exit is a locking mechanism
        objc_sync_enter(self)
        defer { objc_sync_exit(self) }
        // if we previously stored token then do nothing
        if tokens.contains(executionToken) { return }
        tokens.append(executionToken)
        closure()
    }
}

func doSomething() {
    DispatchQueue.once(executionToken: "doSomething") {
        print("This should be executed once")
    }
}

doSomething()
doSomething()
doSomething()

"This should be executed once" will be printed only once.

You can also use the lazy mechanism:

class SomeClass {
    lazy var executeOnce: () -> Void = {
        doSomething()
        return {}
    }()

    func doSomething() {
        print("This should be executed once")
    }
}

let someClass = SomeClass()
someClass.executeOnce()
someClass.executeOnce()
someClass.executeOnce()

It's thread-safe out of the box so you don't need to worry about locking.

I'm trying to implement the lazy option in my main code. It works when the application is initialized, but when i leave the view controller and come back to the page. My executeOnce function doesn't even get called. Could it be because my view controller is being instantiated also with a "Lazy". — iloveStackOverflow, Sep 27 '18 at 22:13

score 0 · Answer 2 · answered Sep 27 '18 at 21:52

Change

func helloWorld(){
var counter = 0
for i in 1...10{
    count = count + 1
    counter = counter + 1
}
print(count)
}

to

func helloWorld(){
var counter = self.count
for i in 1...10{
    count = count + 1
}
print(count)
self.count = counter
}

This way you store into counter the count value before it changes, then make it equal to the previous one again.

NOTE: This way you will always get 10 when you call helloWorld(), no matter how much time you call the function.

You can change the function as shown below:

func helloWorld(_ countValue: Int){
var counter = self.count
for i in 1..<countValue{
    count = count + 1
}
print(count)
self.count = counter

}

So you can choose the upper element of the counter simply by calling:

helloWorld(10)
helloWorld(20)

ADVICE: it's not a good practice to have a function that changes an external value

Swift - Prevent function from being called twice

2 Answers2