为什么在第一个任务退出后调用 dispatchGroup.notify ？

Question

为什么在第一个任务退出后调用 dispatchGroup.notify ？

在以下代码中，输出如下：

1) Did the other thing 

**2) Did all the things** 

3) Did one thing 

4) done waiting

我希望：

1) Did the other thing 

2) Did one thing 

3) done waiting

**4) Did all the things** 

---

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")

    }

---

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

根据非常小的 Apple 文档： https : //developer.apple.com/documentation/dispatch/dispatchgroup

func notify(queue: DispatchQueue, work: DispatchWorkItem) 当一组先前提交的块对象完成时，安排一个工作项提交到队列。

在上面的示例中，我在将块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")

    }