如何使用线程在迅速?
dispatchOnMainThread:^{
NSLog(@"Block Executed On %s", dispatch_queue_get_label(dispatch_get_current_queue()));
}];
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
由于OP问题已经在上面得到了回答,我只想添加一些速度方面的考虑:
我不建议使用.background线程优先级运行任务,尤其是在iPhone X上,因为任务似乎是分配在低功耗内核上的。
下面是一个计算密集型函数的一些真实数据,该函数从XML文件中读取(带缓冲)并执行数据插值:
设备名/ .background / .utility / .default / .userInitiated / .userInteractive
iPhone X: 18.7s / 6.3s / 1.8s / 1.8s iPhone 7: 4.6s / 3.1s / 3.0s / 2.8s / 2.6s iPhone 5s: 7.3s / 6.1s / 4.0s / 4.0s / 3.8s
注意,并不是所有设备的数据集都相同。iPhone X最大,iPhone 5s最小。
其他回答
最佳实践是定义一个可多次访问的可重用函数。
可重用的功能:
例如,AppDelegate.swift是一个全局函数。
func backgroundThread(_ delay: Double = 0.0, background: (() -> Void)? = nil, completion: (() -> Void)? = nil) {
dispatch_async(dispatch_get_global_queue(Int(QOS_CLASS_USER_INITIATED.value), 0)) {
background?()
let popTime = dispatch_time(DISPATCH_TIME_NOW, Int64(delay * Double(NSEC_PER_SEC)))
dispatch_after(popTime, dispatch_get_main_queue()) {
completion?()
}
}
}
注意:在Swift 2.0中替换QOS_CLASS_USER_INITIATED。使用QOS_CLASS_USER_INITIATED。rawValue相反
用法:
a .在后台运行一个进程,延迟3秒:
backgroundThread(3.0, background: {
// Your background function here
})
B.在后台运行一个进程,然后在前台运行一个补全:
backgroundThread(background: {
// Your function here to run in the background
},
completion: {
// A function to run in the foreground when the background thread is complete
})
C.延迟3秒-注意使用completion参数而不使用background参数:
backgroundThread(3.0, completion: {
// Your delayed function here to be run in the foreground
})
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
})
你必须将你想要在后台运行的更改与你想要在UI上运行的更新分开:
dispatch_async(dispatch_get_global_queue(DISPATCH_QUEUE_PRIORITY_DEFAULT, 0)) {
// do your task
dispatch_async(dispatch_get_main_queue()) {
// update some 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)
}
}
}
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()
}
}
}
}