Question

I’m using Posix Message Queues on Linux. Basically, I have multiple threads that receive messages from the same queue by calls to mq_timedreceive.

If more than one thread is running at the same time and the queue is not empty, am I guaranteed that a message does not get received more than once (i.e. the message does not get delivered to multiple threads)?

To be sure, I could synchronize the receive with a mutex, but I’d like to avoid this lock if possible. I read all the man pages (man mq_overview(3)) but without finding anything explicit.

Thanks in advance.

Answer 1

The kernel does that locking for you.

Look at the implementation in ipc/mqueue.c:

SYSCALL_DEFINE5(mq_timedreceive, mqd_t, mqdes, char __user *, u_msg_ptr,
                size_t, msg_len, unsigned int __user *, u_msg_prio,
                const struct timespec __user *, u_abs_timeout)
{    
    ...   
    struct mqueue_inode_info *info;
    ...
    filp = fget(mqdes);
    if (unlikely(!filp)) {
        ret = -EBADF;
        goto out;
    }

    inode = filp->f_path.dentry->d_inode;
    ...
    spin_lock(&info->lock);
    if (info->attr.mq_curmsgs == 0) {
        if (filp->f_flags & O_NONBLOCK) {
            spin_unlock(&info->lock);
...
    } else {
        msg_ptr = msg_get(info);

        inode->i_atime = inode->i_mtime = inode->i_ctime =
                            CURRENT_TIME;

        /* There is now free space in queue. */
        pipelined_receive(info);
        spin_unlock(&info->lock);
        ret = 0;
    }

Each mqueue has a spinlock, which is acquired before checking for new messages.

The last else (pipelined_receive) is where the message is dequeued. This is protected by the info->lock so there’s no way two threads could get the same message.