ios - 分派(dispatch)到主线程时返回值

Question

我有一个从后台线程调用的函数 func getValue() -> Bool。这是有意的，也是必须的。现在，getValue() 需要在主线程上执行一些操作，在这种情况下它需要访问 UIApplication.shared.canOpenURL，它必须在主队列上运行.

这是我当前的功能:

func getValue() -> Bool {
    guard let url = URL(string: "someurl") else { return false }
    return UIApplication.shared.canOpenURL(url)
}

我如何将该函数转换为线程安全函数，即确保它始终在主线程上运行，而无需

• 从主线程开始调用函数
• 重构函数以在闭包中返回值

我已经试过了:

// This causes a deadlock, see https://stackoverflow.com/a/42484670/1531270
func getValue() -> Bool {
    var flag = false

    let group = DispatchGroup()
    group.enter()
    DispatchQueue.main.async {
        if let url = URL(string: "someurl"), UIApplication.shared.canOpenURL(url) {
            flag = true
        }
        group.leave()
    }
    group.wait()

    return flag
}

还有这个:

// This crashes with EXC_BREAKPOINT (SIGTRAP) dispatch_sync called on queue already owned by current thread
func getValue() -> Bool {
    return DispatchQueue.main.sync {
        guard let url = URL(string: "someurl") else { return false }
        return UIApplication.shared.canOpenURL(url)
    }
}

但它们都不起作用。有什么想法吗？

Answer 1

你正在寻找一个信号量 - 试试这个:

DispatchQueue.global(qos: .background).async {
    var value: Bool? = nil
    let semaphore = DispatchSemaphore(value: 0)
    DispatchQueue.main.async {
        let alert = UIAlertController(title: "Choose one", message: "Take your time, I'll wait", preferredStyle: .alert)
        alert.addAction(UIAlertAction(title: "true", style: .default, handler: { _ in
            value = true
            semaphore.signal()
        }))
        alert.addAction(UIAlertAction(title: "false", style: .default, handler: { _ in
            value = false
            semaphore.signal()
        }))
        self.present(alert, animated: true, completion: nil)
    }
    semaphore.wait()
    print("Choice: \(value!)")
}

或者使用上面的示例:

func getValue() -> Bool {
    var flag = false
    let semaphore = DispatchSemaphore(value: 0)

    DispatchQueue.main.async {
        if let url = URL(string: "someurl"), UIApplication.shared.canOpenURL(url) {
            flag = true
            semaphore.signal()
        }
    }
    semaphore.wait()
    return flag
}