📅  最后修改于: 2023-12-03 15:42:15.527000             🧑  作者: Mango
这是GATE-CS-2003考试中的第4个问题,涉及有关门的问题。该问题有一个假设的门模型,其中开关请求被存储在“request”列表中,动作被模拟为“move”函数的调用结果。
以下是描述门模型的伪代码:
[initially, all doors are closed]
when [request comes to open door]:
if [door is already open]:
do nothing
else if [door is currently closed]:
move the door to the open position
else if [door is currently moving]:
add request to the queue of requests
when [request comes to close door]:
if [door is already closed]:
do nothing
else if [door is currently open]:
move the door to the closed position
else if [door is currently moving]:
add request to the queue of requests
on every time unit:
if [there are no pending requests]:
do nothing
else:
retrieve the next request from the queue
move the door accordingly
[the door can only be in one of three states:
closed, open, or moving]
任务是分析此伪代码,并回答以下问题:
request to close
request to open
request to close
request to close
request to open
request to open
request to close
request to open
这是一个由开关请求构成的序列。根据模型,当门处于移动状态时,新的请求将保存到请求队列中,直到当前门完成移动并处理队列中存储的请求。因此,我们可以将给定请求序列的行为划分为以下步骤:
因此,我们可以看到Door在以上请求序列中的行为如下:
cmd | state
----------------------
close | Closed
open | Moving
close | Moving
close | Closed
open | Moving
open | Moving
close | Moving
open | Moving
模型中未考虑两个请求同时到达的情况,这可能会导致一些异常情况。例如,如果同时存在open和close请求,则门可能会保持在同一位置,导致不确定的操作结果。同样,如果一系列请求包含过多的open或close请求,则请求队列可能会导致不必要的移动操作。
一种可能的解决方案是,在处理新的请求之前,检查请求队列中是否存在相同类型的请求。如果是,则可以从队列中删除旧的请求。这样可以确保当突然发生多个请求时门的状态不会受到不必要的影响。