如何撤消对信号量的最新操作SEM_UNDO? [英] How to revoke SEM_UNDO of latest operation on semaphore?

问题描述

我的问题: A，B-信号灯.

My problem: A,B - semaphores.

过程是循环的:

1. 等待A> = 1，递减A
2. 进行一些计算
3. 增加B.
4. 回到1.

目标:进程终止后，我需要将A的减量等于B的增量.

Goal: After process terminates i need decrements of A be equal to increments of B.

所以我知道我将在1中添加标志SEM_UNDO，但是它将撤消该进程对A所做的每个减量.当进程达到4.时，如何从1.撤消/撤消SEM_UNDO?

So i figure out i will add flag SEM_UNDO in 1. but it undos every DEcrement ever made to A by that process. How to revoke / undo SEM_UNDO from 1. when process reaches 4.?

推荐答案

好的，当我想到将问题撤消" SEM_UNDO而不是撤消"该问题时，我感到很高兴.

ok, i was enlightened when thinking about problem as "undoing" SEM_UNDO rather then "revoking" it.

在使用SEM_UNDO的+ x操作调用semop()之后，您可以稍后通过以下两个操作再次调用 semop()来撤消该SEM_UNDO:

after calling semop() with operation +x with SEM_UNDO, you can later undo that SEM_UNDO by calling semop() again with two operations:

1. -x与SEM_UNDO
2. + x，不带SEM_UNDO

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