使用信号量实现N个过程障碍 [英] Implementing an N process barrier using semaphores

问题描述

我目前正在使用以前的迭代训练OS考试，并且碰到了这一点:

I'm currently training for an OS exam with previous iterations and I came across this:

实施"N个过程障碍"，即 是，请确保每个流程都已完成 他们中的一组在等待 指出其各自的执行情况， 其他过程达到他们的 给定点.

Implement a "N Process Barrier", that is, making sure that each process out of a group of them waits, at some point in its respective execution, for the other processes to reach their given point.

您具有以下内容 可用的操作:

You have the following ops available:

init(sem,value), wait(sem) and signal(sem)

N是任意数字.我可以做到这一点，以使其适用于给定数量的进程，而不适用于任何数量的进程.

N is an arbitrary number. I can make it so that it works for a given number of processes, but not for any number.

有什么想法吗?可以使用伪代码进行回复，这不是作业，而是个人学习.

Any ideas? It's OK to reply with the pseudo-code, this is not an assignment, just personal study.

推荐答案

《信号量小书》 .

n = the number of threads
count = 0
mutex = Semaphore(1)
barrier = Semaphore(0)


mutex.wait()
count = count + 1
mutex.signal()

if count == n: barrier.signal() # unblock ONE thread

barrier.wait()
barrier.signal() # once we are unblocked, it's our duty to unblock the next thread

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