线程1：当使用带有两个参数的选择器时EXC_BAD_ACCESS

Question

由于两个参数，它在runTimer()内部崩溃。如果我只放一个或什么都没有 - 它正常。崩溃前往AppDelegate，错误是

线程1：EXC_BAD_ACCESS（代码= 1，地址= 0x4）

var timer = Timer()
var seconds = Int()

@IBOutlet weak var twoMinView: UIView!
@IBOutlet weak var twoMinLabel: UILabel!

@objc func twoMinTimer() {
        seconds = 120
        runTimer(view: twoMinView, label: twoMinLabel)
}

@objc func runTimer(view: UIView, label: UILabel) {
    timer = Timer.scheduledTimer(timeInterval: 1, target: self, selector: #selector(MyViewController.updateTimer(view:withLabel:)), userInfo: nil, repeats: true)
}

@objc func updateTimer(view: UIView, withLabel label: UILabel) {
    seconds -= 1

    if seconds < 1 {
        view.isHidden = true
        timer.invalidate()

        if view == twoMinView {
            streamView.isHidden = false
        }
    }

    label.text = timeString(time: TimeInterval(seconds))
}

@objc func timeString(time: TimeInterval) -> String {
    let minutes = Int(time) / 60 % 60
    let seconds = Int(time) % 60

    return String(format: "%02i : %02i", minutes, seconds)
}

Answer 1

只需删除updateTimer并将runTimer替换为下一个

@objc func runTimer(view: UIView, label: UILabel) {
        timer = Timer.scheduledTimer(withTimeInterval: 1, repeats: true, block: { (timer) in

            if self.seconds < 1 {
                view.isHidden = true
                timer.invalidate()

                if view == self.twoMinView {
                    self.streamView.isHidden = false
                }
            }

            label.text = self.timeString(time: TimeInterval(self.seconds))

            self.seconds -= 1
        })
    }