I come up with a case in our life , here is a time table about the buying bread case . we suppose to get one  bread but finally we get two bread. Because two person don't communicate with each other . they find there is no bread in the fridge and they go to buy one .So here is the problem . how to get one bread when two person is  involved in this case .

• time                                Person A                                               Person B


• 3:00                          查看冰箱，没有面包了


• 3:05                          离开家去商店


• 3:10                           到达商店                                                   查看冰箱，没有面包了


• 3:15                            购买面包                                                    离开家去商店


• 3:20                           到家，把面包放入冰箱                           到达商店


• 3:25                                                                                                购买面包


• 3:30                                                                                               到家，把面包放入冰箱

 

My SOLUTION: In order to get only one bread . we could define a critical-section.

                              if (nobread){

 buy bread

}

only one person could go to buy bread at a time.

and we should use lock fuction to implement.

Lock.acquire()

Lock.release()

 

So the bread problem can be solved easily!  (the code is as following )

breadlock.acquire()         <------enter the critical - section

if (nobread){

buy bread

}

breadlock.release()        <-------leave the critical - section