[理工] [OS] CS

[mqazz1 (無法顯示)]

suppose we have two processes with indices 0 and 1. Assume that count is a

shared variable between the two processes with initial value zero. (Suppose

count is implemented without limit in its range.) Also each process with index

p has a local variable ticket[p] with initial value zero We have the following

mutual-exclusion algorithm for the two processes in a distributed system.

        while(ture){
                ticket[p] = count = count+1;
                while(ticket[1-p] != 0  &&  ticket[1-p] < ticket[p]);
                        CS
                ticket[p] = 0;
                        RS
        }

請問這分別滿足mutual exclusion, progress, bounded waiting嗎?

97台大電機

--
※ 發信站: 批踢踢實業坊(ptt.cc)
◆ From: 218.166.117.109

補充一下洪捷的解答

滿足mutual exclusion, bounded waiting

不滿足progress

請問這要怎麼判斷?

有高手會解這題嗎?

謝謝
※ 編輯: mqazz1          來自: 140.118.110.186      (07/31 20:53)
推 wheels:count = count + 1 不能保證一步完成 若兩個processes都先
→ wheels:做+ 兩個再做assign則抓到的ticket值都一樣
推 wheels:等等...真的是因為不滿足progress嗎...?
推 wheels:我反而覺得是mutual exclusion出問題
其實我一開始也是覺得mutual exclusion不滿足 其他兩個滿足= =

我已經把回文都看完了 題目的確是少打了atomic..

感謝大家的回覆
※ 編輯: mqazz1          來自: 140.118.110.186      (08/01 10:03)
推 wheels:對不起，睡個覺起來後發現我解錯了，B大是正確的!