DISPATCH_QUEUE_CONCURRENT 和 DISPATCH_QUEUE_SERIAL 有什么区别

What's the difference between DISPATCH_QUEUE_CONCURRENT and DISPATCH_QUEUE_SERIAL

我执行以下 class:

class GCDStudy {

    func asyncSerial(time:Double){
        let queue = dispatch_queue_create("DISPATCH_QUEUE_SERIAL",DISPATCH_QUEUE_SERIAL)

        dispatch_async(queue){

            var i:Double = 0
            while(i < 3){
                i++
                print("asyncSerial -wait:\(time)-\(i)")

            }
        }

    }

    func asyncConcurrent(time:Double){

        let queue = dispatch_queue_create("DISPATCH_QUEUE_CONCURRENT",DISPATCH_QUEUE_CONCURRENT)

        dispatch_async(queue){
            var i:Double = 0
            while(i < 3){
                i++
                print("asyncConcurrent -wait:\(time)-\(i)")
            }
        }
    }
}

和运行如下:

运行答:

gCDStudy = GCDStudy()
gCDStudy.asyncSerial(1)
gCDStudy.asyncSerial(2)

运行 B

vgCDStudy2 = GCDStudy()
gCDStudy2.asyncConcurrent(1)
gCDStudy2.asyncConcurrent(2)

结果运行A:

asyncSerial -wait:1.0-1.0
asyncSerial -wait:2.0-1.0
asyncSerial -wait:1.0-2.0
asyncSerial -wait:2.0-2.0
asyncSerial -wait:1.0-3.0
asyncSerial -wait:2.0-3.0

运行B 的结果:

asyncSerial -wait:1.0-1.0
asyncSerial -wait:2.0-1.0
asyncSerial -wait:1.0-2.0
asyncSerial -wait:2.0-2.0
asyncSerial -wait:1.0-3.0
asyncSerial -wait:2.0-3.0

这些结果是一样的,它们之间有什么不同?

有时这些结果是不同的。

谢谢

串行队列按照它们被添加到队列的顺序一次执行一个任务,而并发队列同时执行一个或多个任务。需要注意的重要一点是,即使对于 并发队列 ,任务也是按照它们被添加到队列中的顺序启动的。因此,在您的情况下,您可能看不到输出有任何变化,因为 print 将由并发队列中的每个线程执行得非常快,因此看起来好像它们是按顺序执行的。

如果您希望看到差异,也许您可​​以尝试按如下方式更新您的代码:

class GCDStudy {

  func asyncSerial(time: Double) {
    let queue = dispatch_queue_create("DISPATCH_QUEUE_SERIAL", DISPATCH_QUEUE_SERIAL)

    for i in 0..<3 {
      dispatch_async(queue) {
        print("asyncSerial -wait:\(time)-\(i)")
      }
    }
  }

  func asyncConcurrent(time: Double) {
    let queue = dispatch_queue_create("DISPATCH_QUEUE_CONCURRENT", DISPATCH_QUEUE_CONCURRENT)

    for i in 0..<3 {
      dispatch_async(queue) {
        print("asyncConcurrent -wait:\(time)-\(i)")
      }
    }
  }
}