swift - DispatchQueue会互相等待吗？

Question

我有一个函数要用 Swift (X-Code) 制作。我基本上是通过蓝牙发送一些命令，我​​对 Main Asyncafter 的工作原理有疑问。这是我的设置代码:

func tPodSetUp()
    {
        var delayForResponse = DispatchTime.now() + 0.4 //seconds to wait for response
        writeValue(data: "231") //Put t-Pod in command mode, burst mode is OFF returns OK
        DispatchQueue.main.asyncAfter(deadline: delayForResponse)
        {
            if receivedString1.lowercased() == "ok"
            {
                print("t-Pod burst mode is OFF")
            }
        }

        writeValue(data: "202") //load calibration constants of probe, returns ok or 0
        DispatchQueue.main.asyncAfter(deadline: delayForResponse)
        {
            if receivedString1.lowercased() == "ok"
            {
                print("calibration Loaded")
            }
            if receivedString1 == "0"
            {
                print("No probe connected")
            }
        }
    }

我希望它基本上执行以下操作(按此顺序):
写值
等待 .4 秒
读取响应/检查响应
writeValue(在它完成读取/检查响应之后)
读取响应/检查响应

我担心，如果代码是现在这样，它会在等待其他值时超过 writeValues，并将它们放在异步运行的单独线程上。
另外，让我感到困惑的是，我在一开始就声明了 delayForResponse，并说每次调用时都会改变吗？就像如果我在 12:00:00:00 执行它，它现在将完成 + .4 秒(所以它应该在 12:00:00:40 被调用，但是在 12:00:00 会发生什么: 41 当它运行第二部分时(有另一个对 delayForResponse 的标注)它会突然说“什么？这本应该在 .01 秒前完成的!？

编辑 这是根据一些反馈对代码进行的另一种处理，这会按照我的想法进行吗？

let setupQueue = DispatchQueue(label: "setupQueue")
    let delayForResponse = DispatchTime.now() + 0.4 //seconds to wait for response

setupQueue.sync {
    writeValue(data: String(UnicodeScalar(UInt8(231)))) //231: Put t-Pod in command mode, burst mode is OFF returns OK
    DispatchQueue.main.asyncAfter(deadline: delayForResponse)
    {
        if receivedString1.lowercased() == "ok"
        {
            print("t-Pod burst mode is OFF")
        }
    }

    writeValue(data: String(UnicodeScalar(UInt8(202)))) //202: load calibration constants of probe, returns ok or 0
    DispatchQueue.main.asyncAfter(deadline: delayForResponse)
    {
        if receivedString1.lowercased() == "ok"
        {
            print("t-Pod burst mode is OFF")
        }
        if receivedString1 == "0"
        {
            print("No probe connected")
        }
    }

    writeValue(data: String(UnicodeScalar(UInt8(204)))) //204: load t-Pod serial number
    DispatchQueue.main.asyncAfter(deadline: delayForResponse)
    {
        if (receivedString1.count > 5)
        {
            print("received t-Pod SN: \(receivedString1)")
            tPodSN = receivedString1
        }
    }

    writeValue(data: String(UnicodeScalar(UInt8(205)))) //205: load probe serial number
    DispatchQueue.main.asyncAfter(deadline: delayForResponse)
    {
        if (receivedString1.count > 3)
        {
            print("received Probe SN: \(receivedString1)")
            probeSN = receivedString1
        }
    }

    //AFTER FINISHING SETUP, RESET TPOD AND TURN BEACON OFF

    writeValue(data: String(UnicodeScalar(UInt8(202)))) //200: resets t-Pod
    writeValue(data: String(UnicodeScalar(UInt8(202)))) //211: turns beacon off (temperature output)
} 

Answer 1

你检查过DispatchGroup了吗？？

func myFunction() {
    var a: Int?

    let group = DispatchGroup()
    group.enter()

    DispatchQueue.main.async {
        a = 1
        group.leave()
    }

    // does not wait. But the code in notify() is run 
    // after enter() and leave() calls are balanced

    group.notify(queue: .main) {
        print(a)
    }
}