5

我已经阅读了彼得森的互斥算法。然后有一个问题,如果我们重新排序 do...while 循环中的第一个和第二个命令会发生什么?如果我们这样做,我看不到会发生什么……有人能告诉我我错过了什么吗?

do {
     flag[me] = TRUE;
     turn = other;
     while (flag[other] && turn == other)
             ;
     critical section
     flag[me] = FALSE;
     remainder section
   } while (TRUE);
4

1 回答 1

4

如果您重新排序这两个命令,您将让这两个进程同时运行:

turn = 1;                       |      turn = 0;
flag[0] = true;                 |      flag[1] = true;
while(flag[1] && turn==1)       |      while(flag[0] && turn==0)
    ;                           |          ;
critical section 0              |      critical section 1
flag[0] = false;                |      flag[1] = false;

这可能会按以下顺序执行:

Process 1: turn = 0;
Process 0: turn = 1;
Process 0: flag[0] = true;
Process 0: while(flag[1] && turn==1) // terminates, since flag[1]==false
Process 0: enter critical section 0
Process 1: flag[1] = true;
Process 1: while(flag[0] && turn==0) // terminates, since turn==1
Process 1: enter critical section 1
Process 0: exit critical section 0
Process 1: exit critical section 1

这违反了互斥标准。

于 2014-03-16T11:33:33.713 回答