2

AIX(如果有人关心的话,还有HPUX)有一个很好的小特性,叫做msemaphores,它可以很容易地同步多个进程共享的内存映射文件的粒度片段(例如记录)。有人知道linux中有类似的东西吗?

为了清楚起见,msemaphore 函数通过此处的相关链接进行描述。

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3 回答 3

1

这可以使用 POSIX 共享内存互斥锁来完成:

pthread_mutexattr_t attr;
int pshared = PTHREAD_PROCESS_SHARED;
pthread_mutexattr_init(&attr);
pthread_mutexattr_setpshared(&attr, &pshared);

pthread_mutex_init(&some_shared_mmap_structure.mutex, &attr);
pthread_mutexattr_destroy(&attr);

现在,您可以使用普通的 pthread_mutex_lock() 等调用从映射了它的多个进程中解锁和锁定 &some_shared_mmap_structure.mutex。

实际上,您甚至可以按照以下方式实现 msem API:(未经测试)

struct msemaphore {
    pthread_mutex_t mut;
};

#define MSEM_LOCKED 1
#define MSEM_UNLOCKED 0
#define MSEM_IF_NOWAIT 1

msemaphore *msem_init(msemaphore *msem_p, int initialvalue) {
    pthread_mutex_attr_t attr;
    int pshared = PTHREAD_PROCESS_SHARED;

    assert((unsigned long)msem_p & 7 == 0); // check alignment

    pthread_mutexattr_init(&attr);
    pthread_mutexattr_setpshared(&attr, &pshared); // might fail, you should probably check
    pthread_mutex_init(&msem_p->mut, &attr); // never fails
    pthread_mutexattr_destroy(&attr);

    if (initialvalue)
        pthread_mutex_lock(&attr);

    return msem_p;
}

int msem_remove(msemaphore *msem) {
    return pthread_mutex_destroy(&msem->mut) ? -1 : 0;
}

int msem_lock(msemaphore *msem, int cond) {
    int ret;
    if (cond == MSEM_IF_NOWAIT)
        ret = pthread_mutex_trylock(&msem->mut);
    else
        ret = pthread_mutex_lock(&msem->mut);

    return ret ? -1 : 0;
}

int msem_unlock(msemaphore *msem, int cond) {
    // pthreads does not allow us to directly ascertain whether there are
    // waiters. However, a unlock/trylock with no contention is -very- fast
    // using linux's pthreads implementation, so just do that instead if
    // you care.
    //
    // nb, only fails if the mutex is not initialized
    return pthread_mutex_unlock(&msem->mut) ? -1 : 0;
}
于 2009-05-08T02:39:51.887 回答
1

如果POSIX 信号量sem_init(3)的第二个参数" pshared" 为真,则可以将 POSIX 信号量放置在进程之间共享的内存中。这似乎与msem所做的相同。

#include <semaphore.h>
#include <stdio.h>
#include <stdlib.h>
#include <sys/mman.h>
#include <time.h>
#include <unistd.h>
int main() {
    void *shared;
    sem_t *sem;
    int counter, *data;
    pid_t pid;
    srand(time(NULL));
    shared = mmap(NULL, sysconf(_SC_PAGE_SIZE), PROT_READ | PROT_WRITE,
            MAP_ANONYMOUS | MAP_SHARED, -1, 0);
    sem_init(sem = shared, 1, 1);
    data = shared + sizeof(sem_t);
    counter = *data = 0;
    pid = fork();
    while (1) {
        sem_wait(sem);
        if (pid)
            printf("ping>%d %d\n", data[0] = rand(), data[1] = rand());
        else if (counter != data[0]) {
            printf("pong<%d", counter = data[0]);
            sleep(2);
            printf(" %d\n", data[1]);
        }
        sem_post(sem);
        if (pid) sleep(1);
    }
}

这是一个非常愚蠢的测试,但它有效:

$ cc -o test -lrt test.c
$ ./test
ping>2098529942 315244699
pong<2098529942 315244699
pong<1195826161 424832009
ping>1195826161 424832009
pong<1858302907 1740879454
ping>1858302907 1740879454
ping>568318608 566229809
pong<568318608 566229809
ping>1469118213 999421338
pong<1469118213 999421338
ping>1247594672 1837310825
pong<1247594672 1837310825
ping>478016018 1861977274
pong<478016018 1861977274
ping>1022490459 935101133
pong<1022490459 935101133
...

因为信号量是在两个进程之间共享的,所以尽管有s, pongs 也不会从 s 获得交错的数据。pingsleep

于 2009-05-08T02:25:52.417 回答
0

在 Linux 下,你或许可以用 SysV 共享内存来实现你想要的;快速谷歌搜索出现了这个(相当旧的)指南,可能会有所帮助。

于 2009-05-07T20:48:28.453 回答