考虑以下针对临界截面问题的建议解决方案。有n个过程：P0…Pn-1。在代码中，函数pmax返回一个不小于其任何参数的整数。对于所有i，t [i]初始化为零。 %20%7C%20Question%2060_0.jpg)
关于上述解决方案，以下哪一项是正确的?
(A)随时都可以在关键部分进行一个流程
(B)满足有条件的等待条件
(C)满足进度条件
(D)它不会导致死锁答案： (A)
解释：
Mutual exclusion is satisfied:
All other processes j started before i must have value (i.e. t[j])
less than the value of process i (i.e. t[i]) as function pMax()
return a integer not smaller than any of its arguments. So if anyone
out of the processes j have positive value will be executing in its
critical section as long as the condition t[j] > 0 && t[j] <= t[i] within
while will persist. And when this j process comes out of its critical
section, it sets t[j] = 0; and next process will be selected in for loop.
So when i process reaches to its critical section none of the processes j
which started earlier before process i is in its critical section. This
ensure that only one process is executing its critical section at a time.
Deadlock and progress are not satisfied:
while (t[j] != 0 && t[j] <=t[i]); because of this condition deadlock is
possible when value of j process becomes equals to the value of process i
(i.e t[j] == t[i]). because of the deadlock progess is also not possible
(i.e. Progess == no deadlock) as no one process is able to make progress
by stoping other process.
Bounded waiting is also not satisfied:
In this case both deadlock and bounded waiting to be arising from the same
reason as if t[j] == t[i] is possible then starvation is possible means
infinite waiting.这种解释是由Dheerendra Singh贡献的。
这个问题的测验