在我的应用程序中早期调用 DispatchGroup 的通知 block 时出现问题，并制作了这个 playground 示例进行实验。根据输出，有时它甚至在第一个 .leave() 之前被调用。感觉我错过了一些明显的东西，现在我已经看了太久了。

let s = DispatchSemaphore(value: 1)
let dg = DispatchGroup()
func go() -> Void {
    for i in 1...2 {
        doWork(attemptNo: i, who: "Lily", secs: Double.random(in: 1.0...5.0))
        doWork(attemptNo: i, who: "Emmie", secs: Double.random(in: 1.0...10.0))
        doWork(attemptNo: i, who: "Wiley", secs: Double.random(in: 1.0...3.0))
    }
}
func doWork(attemptNo: Int, who: String, secs: TimeInterval) -> Void {
    DispatchQueue.global().async {
        dg.enter()
        print("\(who) wants to work, will wait \(secs) seconds, attempt #\(attemptNo)")
        if s.wait(timeout: .now() + secs) == .timedOut {
            print("\(who) was denied. No soup for me! Task #\(attemptNo) not going to happen.")
            dg.leave()
            return
        }
        let workSecs = UInt32(Int.random(in: 1...3))
        print("\(who) went to work for \(workSecs) seconds on task #\(attemptNo)")

        DispatchQueue.global().asyncAfter(deadline: .now() + TimeInterval(workSecs)) {
            print("\(who) is sliding down the dinosaur tail. Task #\(attemptNo) all done!")
            s.signal()
            dg.leave()
       }
   }
}
go()
dg.notify(queue: .global(), execute: {print("Everyone is done.")})

示例输出:

Emmie wants to work, will wait 4.674405654828654 seconds, attempt #1
Lily wants to work, will wait 1.5898288206500877 seconds, attempt #1
Wiley wants to work, will wait 1.2182416407288 seconds, attempt #1
Lily wants to work, will wait 3.3225083978280647 seconds, attempt #2
Everyone is done.
Wiley wants to work, will wait 2.801577828588925 seconds, attempt #2
Emmie wants to work, will wait 8.9696422949966 seconds, attempt #2
Lily went to work for 3 seconds on task #2
Wiley was denied. No soup for me! Task #1 not going to happen.
Lily was denied. No soup for me! Task #1 not going to happen.
Wiley was denied. No soup for me! Task #2 not going to happen.
Lily is sliding down the dinosaur tail. Task #2 all done!
Emmie went to work for 3 seconds on task #1
Emmie is sliding down the dinosaur tail. Task #1 all done!
Emmie went to work for 2 seconds on task #2
Emmie is sliding down the dinosaur tail. Task #2 all done!

在这种情况下，“每个人都完成了”几乎是立即发生的，因为 .leave 伴随着没有得到信号量或在“工作”完成之后，这没有意义。请帮忙。

最佳答案

您的主要问题是您在错误的地方调用了 dg.enter()。您总是希望在异步调用之前调用 enter()，并希望在异步调用完成时调用 leave()。

在编写代码时，for 循环在第一次调用 enter 之前完成，这就是触发 notify 的原因立即。

func doWork(attemptNo: Int, who: String, secs: TimeInterval) -> Void {
    dg.enter()
    DispatchQueue.global().async {
        print("\(who) wants to work, will wait \(secs) seconds, attempt #\(attemptNo)")
        if s.wait(timeout: .now() + secs) == .timedOut {
            print("\(who) was denied. No soup for me! Task #\(attemptNo) not going to happen.")
            dg.leave()
            return
        }
        let workSecs = UInt32(Int.random(in: 1...3))
        print("\(who) went to work for \(workSecs) seconds on task #\(attemptNo)")
        sleep(workSecs)
        print("\(who) is sliding down the dinosaur tail. Task #\(attemptNo) all done!")
        s.signal()
        dg.leave()
    }
}

您的代码还奇怪地使用了信号量和 sleep 。我想这是模拟长时间运行的后台进程的尝试。

关于swift - DispatchGroup 通知 block 被提前调用，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/54367351/