GCD 和延时调用

// GCD 和延时调用

DispatchQueue.global().async {

print("耗时的。。。努力工作")

//

DispatchQueue.main.async {

print("结束工作，更新 UI")

}

}

/*

要实现以下功能

delay(2){

print("2秒输出")

}

想取消的话，保留一个 task 的引用，然后调用 cancel

let task = delay(5) {

print("拨打110")

}

仔细想了想，还是取消为妙

cancel(task)

*/

typealias Task = (_ cancel: Bool) -> Void

// 延时执行一个block， 返回这个一个可以取消的block

func delay(_ time : TimeInterval, task:@escaping()->()) -> Task? {

func dispatch_later(block: @escaping()->()){

let t = DispatchTime.now() + time

DispatchQueue.main.asyncAfter(deadline: t, execute: block)

}

var closure:(()->Void)? = task

var result : Task?

let delayedClosure: Task = {

cancel in

if let internalClosure = closure {

if cancel == false {

DispatchQueue.main.async(execute: internalClosure)

}

}

closure = nil

result = nil

}

result = delayedClosure

dispatch_later {

if let delayedClosure = result {

delayedClosure(false)

}

}

return result

}

func cancel(_ task: Task?){

task?(true)

}

let task = delay(5) {

print("拨打110")

}

cancel(task)