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我正在研究组合调度程序，我有来自 Raywenderlich 书中的示例代码

let queue = OperationQueue()

let subscription = (1...10).publisher
  .receive(on: queue)
  .sink { value in
    print("Received \(value) on thread \(Thread.current.number)")
  }

这本书解释了 OperationQueue 使用所有可用的线程，因此打印顺序和线程可以是随机的。我理解那部分，但是当我在操场上运行这段代码时，我只看到 10 个数字中的 5 个。

Received 1 on thread 7
Received 2 on thread 6
Received 3 on thread 5
Received 7 on thread 9
Received 6 on thread 4

为什么该代码不显示所有 10 个数字？

4

1 回答 1

1

您需要在文件顶部导入PlaygroundSupport并设置。PlaygroundPage.current.needsIndefiniteExecution = true

如果您使用另一种接收器方法，也就是您还获得完成处理程序块的方法，那么在该完成处理程序中您应该调用PlaygroundPage.current.finishExecution()

编辑：PlaygroundPage.current.finishExecution()在队列上调用障碍块，否则在您的打印语句全部执行之前调用。

这是我的测试代码：

PlaygroundPage.current.needsIndefiniteExecution = true

let queue = OperationQueue()

let subscription = (1...10).publisher
    .receive(on: queue)
    .sink(receiveCompletion: { _ in
        queue.addBarrierBlock {
            print("Finished")
            PlaygroundPage.current.finishExecution()
        }
    }, receiveValue: { value in
        print("Received \(value) on thread \(Thread.current)")
    })

和输出：

Received 1 on thread <NSThread: 0x7fd86f204080>{number = 2, name = (null)}
Received 2 on thread <NSThread: 0x7fd86f404340>{number = 3, name = (null)}
Received 3 on thread <NSThread: 0x7fd86f404430>{number = 4, name = (null)}
Received 7 on thread <NSThread: 0x7fd86f405240>{number = 5, name = (null)}
Received 6 on thread <NSThread: 0x7fd86f205c50>{number = 6, name = (null)}
Received 10 on thread <NSThread: 0x7fd86ce487f0>{number = 7, name = (null)}
Received 5 on thread <NSThread: 0x7fd86cf048a0>{number = 10, name = (null)}
Received 9 on thread <NSThread: 0x7fd86ce49b20>{number = 11, name = (null)}
Received 4 on thread <NSThread: 0x7fd86cf2a4c0>{number = 12, name = (null)}
Received 8 on thread <NSThread: 0x7fd86ce4ad60>{number = 13, name = (null)}
Finished
于 2020-11-04T05:28:45.377 回答