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

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
})

其他回答

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()
            }
        }
    }
}

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
})

好的答案,无论如何,我想分享我的面向对象的解决方案,直到swift 5。

请查看它:AsyncTask

从概念上受到android的AsyncTask的启发,我用Swift写了自己的类

AsyncTask允许正确和简单地使用UI线程。该类允许执行后台操作并在UI线程上发布结果。

下面是一些用法示例

例1 -

AsyncTask(backgroundTask: {(p:String)->Void in//set BGParam to String and BGResult to Void
        print(p);//print the value in background thread
    }).execute("Hello async");//execute with value 'Hello async'

例2 -

let task2=AsyncTask(beforeTask: {
           print("pre execution");//print 'pre execution' before backgroundTask
        },backgroundTask:{(p:Int)->String in//set BGParam to Int & BGResult to String
            if p>0{//check if execution value is bigger than zero
               return "positive"//pass String "poitive" to afterTask
            }
            return "negative";//otherwise pass String "negative"
        }, afterTask: {(p:String) in
            print(p);//print background task result
    });
    task2.execute(1);//execute with value 1

它有两个泛型类型:

BGParam -执行时发送给任务的参数类型。 BGResult -后台计算结果的类型。 当你创建一个AsyncTask时,你可以将这些类型传递给任何你需要传入和传入后台任务的类型,但如果你不需要这些类型,你可以将其标记为未使用的,只需将其设置为:Void或使用更短的语法:()

当一个异步任务执行时,它经过3个步骤:

beforeTask:()->在任务执行之前在UI线程上调用的Void。 backgroundTask: (param:BGParam)->BGResult在后台线程立即调用 afterTask:(参数:BGResult)->在UI线程上调用后台任务结果的Void

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);
}