分布式学习之--6.824MITLab1记录

记录这个实验的人也不少了,我就简单记录下Part3和Part4

Part3

func schedule(jobName string, mapFiles []string, nReduce int, phase jobPhase, registerChan chan string) {
	var ntasks int
	var n_other int // number of inputs (for reduce) or outputs (for map)
	switch phase {
	case mapPhase:
		ntasks = len(mapFiles)
		n_other = nReduce
	case reducePhase:
		ntasks = nReduce
		n_other = len(mapFiles)
	}

	fmt.Printf("Schedule: %v %v tasks (%d I/Os)\n", ntasks, phase, n_other)

	// All ntasks tasks have to be scheduled on workers. Once all tasks
	// have completed successfully, schedule() should return.
	//
	// Your code here (Part III, Part IV).
	//
	
	var done sync.WaitGroup
	// workstate := make(map[int]bool)
	for i:=0; i < ntasks; i=i+1 {

		// workstate
		done.Add(1)	//有一个u说明有一个goroutine
		go func(i int, registerChan chan string){
			u := mapFiles[i]
			req := <- registerChan	//堵塞了怎么办
			
			state := call(req,"Worker.DoTask", DoTaskArgs{jobName,u,phase,i,n_other},nil)
			if state == false {
				req2 := <- registerChan
				call(req2,"Worker.DoTask", DoTaskArgs{jobName,u,phase,i,n_other},nil)
				go func(req string){
					registerChan <- req
				}(req2)
			}
		
			go func(req string){
				registerChan <- req
			}(req)
			
			done.Done()
			
		}(i,registerChan)
	
	}

	//所有完成后才返回，所以要记得用waitgroup
	done.Wait()

	fmt.Printf("Schedule: %v done\n", phase)
}

part3主要是实现多个任务分配给多个工人,但是工人的数量少于任务数量,一个时间段一个工人只能处理一个任务,所以要变通一点,不能像之前常见的结构一样,让多个工人重复做一件事情了.这里有点坑的地方是,这个函数里使用的channel是没有buffer的,是会发生阻塞的,将工人信息放回channel的时候(意味着此时该工人的工作已经完成,它的状态是闲置的),如果直接用registerChan <- req,则会发生阻塞,因此我参照了一位知友的办法,另起一个协程,完成信息的发送.

Part4

state := call(req,"Worker.DoTask", DoTaskArgs{jobName,u,phase,i,n_other},nil)
if state == false {
    req2 := <- registerChan
    call(req2,"Worker.DoTask", DoTaskArgs{jobName,u,phase,i,n_other},nil)
    go func(req string){
	registerChan <- req
    }(req2)
}		

part4是实现错误解决,总的来说按照文字描述就可以,将这个工作再交付给另一个worker,但是需要注意的是,应该先从channel中接收另一个worker再将原来出问题的worker发送到channel中.(但是我一直好奇如果第二个worker也出错了,不会有问题么?)