ios - 为什么只有第一个任务退出后才调用dispatchGroup.notify？

Question

为什么只有第一个任务退出后才调用dispatchGroup.notify？

下面的代码输出如下:

1) Did the other thing 

**2) Did all the things** 

3) Did one thing 

4) done waiting

我期望:

1) Did the other thing 

2) Did one thing 

3) done waiting

**4) Did all the things** 

<小时/>

DispatchQueue.global().async {

        let dispatchGroup = DispatchGroup()

        dispatchGroup.notify(queue: DispatchQueue.main) {
            print("Did all the things")
        }


        dispatchGroup.enter()
        DispatchQueue.global().asyncAfter(deadline: .now() + 10) {
        print("Did one thing")
            dispatchGroup.leave()

        }


        dispatchGroup.enter()
        DispatchQueue.global().async {
            print("Did the other thing")
            dispatchGroup.leave()
        }

        dispatchGroup.wait()
        print("done waiting")

    }

<小时/>

作为旁注，如果我在主线程上执行此操作，它会按预期工作。

Answer 1

根据最简单的苹果文档:https://developer.apple.com/documentation/dispatch/dispatchgroup

func notify(queue: DispatchQueue, work: DispatchWorkItem) Schedules a work item to be submitted to a queue when a group of previously submitted block objects have completed.

在上面的示例中，我在将 block 提交到队列之前调用了dispatchQueue.notify。通过如下更新代码，我能够获得预期的行为。

DispatchQueue.global().async {

        let dispatchGroup = DispatchGroup()



        dispatchGroup.enter()
        DispatchQueue.global().asyncAfter(deadline: .now() + 10) {
        print("Did one thing")
            dispatchGroup.leave()

        }


        dispatchGroup.enter()
        DispatchQueue.global().async {
            print("Did the other thing")
            dispatchGroup.leave()
        }

        dispatchGroup.notify(queue: DispatchQueue.main) {
            print("Did all the things")
        }

        dispatchGroup.wait()
        print("done waiting")

    }