Are Swift4 variables atomic?

Question

I was wondering if Swift 4 variables are atomic or not. So I did the following test.

Below is my test code.

class Test {
    var count = 0
    let lock = NSLock()

    func testA() {
            count = 0

            let queueA = DispatchQueue(label: "Q1")
            let queueB = DispatchQueue(label: "Q2")
            let queueC = DispatchQueue(label: "Q3")

            queueA.async {
                for _ in 1...1000 {
                    self.increase()
                }
            }
            queueB.async {
                for _ in 1...1000 {
                    self.increase()
                }
            }
            queueC.async {
                for _ in 1...1000 {
                    self.increase()
                }

            }
    }

    ///The increase() method:
    func increase() {
//        lock.lock()
        self.count += 1
        print(count)
//        lock.unlock()
    }
}

The conclusion is as follows: lock.lock()and lock.unlock() comment .

3
3
3
4
5
...
2999
3000

The conclusion is as follows: lock.lock()and lock.unlock without commenting .

1
2
3
4
5
...
2999
3000

My problem:
If the variable is countnon-atomic, queueA, queueB, and queueC must invoke asynchronously increase(), resulting in random access and printing count.

So, in my opinion, there is a moment, for example, queueA and queueB got countequal to 15, and both of them increase countby 1 ( count += 1), so the counter should be equal to 16, although two increments are performed.

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1. increase() , .

func increase() {
    lock.lock()
    self.count += 1
    array.append(self.count)
    lock.unlock()
}

2. :

    @IBAction func tapped(_ sender: Any) {
        let testObjc = Test()
        testObj.testA()

        DispatchQueue.main.asyncAfter(deadline: DispatchTime.now()+3) {
            print(self.testObj.array)
        }
    }

NSLock: NSLock:
[1,2,3,...,2999,3000]

Answer 1

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: print count , (2 1000 ):

let t = Test()
t.testA()
DispatchQueue.main.asyncAfter(deadline: .now() + 2.0) {
    // you're likely to get different results each run
    print(t.count)
}
RunLoop.current.run()

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