如何使用线程在迅速?

dispatchOnMainThread:^{

    NSLog(@"Block Executed On %s", dispatch_queue_get_label(dispatch_get_current_queue()));

}];

当前回答

在Swift 4.2中,这是可行的。

import Foundation

class myThread: Thread
{
    override func main() {
        while(true) {
            print("Running in the Thread");
            Thread.sleep(forTimeInterval: 4);
        }
    }
}

let t = myThread();
t.start();

while(true) {
    print("Main Loop");
    sleep(5);
}

其他回答

斯威夫特5

为了简化,创建一个“DispatchQueue+Extensions.swift”文件,内容如下:

import Foundation

typealias Dispatch = DispatchQueue

extension Dispatch {

    static func background(_ task: @escaping () -> ()) {
        Dispatch.global(qos: .background).async {
            task()
        }
    }

    static func main(_ task: @escaping () -> ()) {
        Dispatch.main.async {
            task()
        }
    }
}

用法:

Dispatch.background {
    // do stuff

    Dispatch.main { 
        // update UI
    }
}

下面的代码是否有缺点(当需要启动前台屏幕之后)?

import Foundation
import UIKit

class TestTimeDelay {

    static var connected:Bool = false
    
    static var counter:Int = 0

    static func showAfterDelayControl(uiViewController:UIViewController) {
        NSLog("TestTimeDelay", "showAfterDelayControl")
    }
    
    static func tryReconnect() -> Bool {
        counter += 1
        NSLog("TestTimeDelay", "Counter:\(counter)")
        return counter > 4
    }

    static func waitOnConnectWithDelay(milliseconds:Int, uiViewController: UIViewController) {
        
        DispatchQueue.global(qos: .background).async {
            DispatchQueue.main.asyncAfter(deadline: DispatchTime.now() + DispatchTimeInterval.milliseconds(milliseconds), execute: {
                waitOnConnect(uiViewController: uiViewController)
            })
        }
    }

    static func waitOnConnect(uiViewController:UIViewController) {

        connected = tryReconnect()
        if connected {
            showAfterDelayControl(uiViewController: uiViewController)
        }
        else {
            waitOnConnectWithDelay(milliseconds: 200, uiViewController:uiViewController)
        }
     }
}   

来自Jameson Quave的教程

斯威夫特2

dispatch_async(dispatch_get_global_queue(DISPATCH_QUEUE_PRIORITY_DEFAULT, 0), {
    //All stuff here
})

我真的很喜欢Dan Beaulieu的答案,但是它在Swift 2.2中不起作用,我认为我们可以避免那些讨厌的强制打开!

func backgroundThread(delay: Double = 0.0, background: (() -> Void)? = nil, completion: (() -> Void)? = nil) {

    dispatch_async(dispatch_get_global_queue(QOS_CLASS_USER_INITIATED, 0)) {

        background?()

        if let completion = completion{
            let popTime = dispatch_time(DISPATCH_TIME_NOW, Int64(delay * Double(NSEC_PER_SEC)))
            dispatch_after(popTime, dispatch_get_main_queue()) {
                completion()
            }
        }
    }
}

快4.倍

把这些放到文件中:

func background(work: @escaping () -> ()) {
    DispatchQueue.global(qos: .userInitiated).async {
        work()
    }
}

func main(work: @escaping () -> ()) {
    DispatchQueue.main.async {
        work()
    }
}

然后在需要的地方调用它:

background {
     //background job
     main {
       //update UI (or what you need to do in main thread)
     }
}