为什么我们需要在民意调查中调用poll_wait? [英] Why do we need to call poll_wait in poll?

问题描述

在LDD3中，我看到了这样的代码

In LDD3, i saw such codes

static unsigned int scull_p_poll(struct file *filp, poll_table *wait)
{
    struct scull_pipe *dev = filp->private_data;
    unsigned int mask = 0;

    /*
     * The buffer is circular; it is considered full
     * if "wp" is right behind "rp" and empty if the
     * two are equal.
     */
    down(&dev->sem);
    poll_wait(filp, &dev->inq,  wait);
    poll_wait(filp, &dev->outq, wait);
    if (dev->rp != dev->wp)
        mask |= POLLIN | POLLRDNORM;    /* readable */
    if (spacefree(dev))
        mask |= POLLOUT | POLLWRNORM;   /* writable */
    up(&dev->sem);
    return mask;
}

但是它说poll_wait不会等待，将立即返回.那我们为什么要调用它呢?为什么我们不能只返回面具?

But it says poll_wait won't wait and will return immediately. Then why do we need to call it? Why can't we just return mask?

推荐答案

poll_wait将您的设备(由结构文件"表示)添加到可以唤醒进程的设备列表中.

poll_wait adds your device (represented by the "struct file") to the list of those that can wake the process up.

这个想法是，该进程可以使用民意调查(或select或epoll等)将一堆文件描述符添加到希望等待的列表中.每个驱动程序的轮询条目都将被调用.每个人(通过poll_wait)将自己添加到服务员列表中.

The idea is that the process can use poll (or select or epoll etc) to add a bunch of file descriptors to the list on which it wishes to wait. The poll entry for each driver gets called. Each one adds itself (via poll_wait) to the waiter list.

然后，核心内核将进程阻塞在一处.这样，任何一台设备都可以唤醒该过程.如果返回非零掩码位，则意味着那些就绪"属性(可读/可写等)适用于 now .

Then the core kernel blocks the process in one place. That way, any one of the devices can wake up the process. If you return non-zero mask bits, that means those "ready" attributes (readable/writable/etc) apply now.

因此，在伪代码中，大致是这样的:

So, in pseudo-code, it's roughly like this:

foreach fd:
    find device corresponding to fd
    call device poll function to setup wait queues (with poll_wait) and to collect its "ready-now" mask

while time remaining in timeout and no devices are ready:
    sleep

return from system call (either due to timeout or to ready devices)

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