Semop：当减少一组信号量时,所有信号量都会立即递减或者在第一次

发布时间：2020-12-15 21:46:00 所属栏目：安全来源：网络整理

导读：所以,如果我有一个信号量集semid与num_of_sems信号量和sembuf * deleter_searchers_down struct sembuf *deleter_searchers_down = malloc(sizeof (*deleter_searchers_down) * num_of_sems);for (i = 0; i num_of_sems; ++i) { (deleter_searchers_down + i

所以,如果我有一个信号量集semid与num_of_sems信号量和sembuf * deleter_searchers_down

struct sembuf *deleter_searchers_down 
                        = malloc(sizeof (*deleter_searchers_down) * num_of_sems);
for (i = 0; i < num_of_sems; ++i) {
            (deleter_searchers_down + i)->sem_op = -1;
            (deleter_searchers_down + i)->sem_num = i;
            (deleter_searchers_down + i)->sem_flg = SEM_UNDO;
        }
semop(semid,deleter_searchers_down,num_of_sems);

对semop的调用将尝试立即降低集合中的所有信号量,或者一旦它试图降低0的第一个信号量并且在某个特定信号量的某些其他进程之后继续,它是否会阻塞？

解决方法

在所有更新都可以作为一个单元进行之前不会发生更新

关于这一点,POSIX规范可能更清楚,尽管它确实说semop是原子的.

在Linux上,glibc中的semop(3)是semop(2)的简单包装器. semop(2)手册依次说

The set of operations contained in sops is performed in array order,and atomically,that is,the operations are performed either as a complete unit,or not at all.

HP-UX semop(2)联机帮助页更加清晰：

Semaphore array operations are atomic in that none of the semaphore operations are performed until blocking conditions on all of the semaphores in the array have been removed.

（编辑：李大同）

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