老师，看了并发版的schedule的代码有一个疑问，代码如下：

func (s *QueuedScheduler) Run() {
	s.workerChan = make(chan chan engine.Request)
	s.requestChan = make(chan engine.Request)
	go func() {
		var requestQ []engine.Request
		var workerQ []chan engine.Request
		for {
			var activeRequest engine.Request
			var activeWorker chan engine.Request
			if len(requestQ) > 0 &&
				len(workerQ) > 0 {
				activeWorker = workerQ[0]
				activeRequest = requestQ[0]
			}

			select {
			case r := <-s.requestChan:
				requestQ = append(requestQ, r)
			case w := <-s.workerChan:
				workerQ = append(workerQ, w)
			case activeWorker <- activeRequest:
			//这个case中，如果schedule工作了一段时间后，某一时刻requestQ先变为0， 
			activeRequest保留的应该是最后一个request，
			这个request对应一个worker channel被保留在activeWorker中，
			当activeWorker 完成的时候，select会从这个case返回吧，
			activeWorker 得到的还是最后的activeRequest。但是我们看下面的代码，
			requestQ = requestQ[1:]这个不是要crash？
				
				workerQ = workerQ[1:]
				requestQ = requestQ[1:]
			}
		}
	}()
}