为什么在tasklet_action()函数中调用BUG？

Question

static void tasklet_action(struct softirq_action *a)
{
    // ...

    while (list) {
        struct tasklet_struct *t = list;

        list = list->next;

        if (tasklet_trylock(t)) {
            if (!atomic_read(&t->count)) {
                if (!test_and_clear_bit(TASKLET_STATE_SCHED,
                            &t->state))
                    BUG();

                // ...
    }
}

我所理解的是，如果一个微线程已经被调度，那么这段代码会抛出一个BUG()。这是否意味着同一个微线程既不能同时运行，也不能被调度？

Answer 1

这只是对微线程的一个有保证的属性进行的健全性检查。您可以查看a comment in include/linux/interrupt.h中列出的微线程的属性

   Properties:
   * If tasklet_schedule() is called, then tasklet is guaranteed
     to be executed on some cpu at least once after this.
   * If the tasklet is already scheduled, but its execution is still not
     started, it will be executed only once.
   * If this tasklet is already running on another CPU (or schedule is called
     from tasklet itself), it is rescheduled for later.
   * Tasklet is strictly serialized wrt itself, but not
     wrt another tasklets. If client needs some intertask synchronization,
     he makes it with spinlocks.

根据定义，微线程保证在调度后至少运行一次。这段代码：

    if (!atomic_read(&t->count)) {
        if (!test_and_clear_bit(TASKLET_STATE_SCHED,
                    &t->state))
            BUG();

确保此属性有效，否则会出现错误，并使用BUG()停止执行并导致运行时死机。

以下是上述代码的注释版本，以使其更清晰：

    // If the tasklet never ran (t->count == 0)
    if (!atomic_read(&t->count)) {
        // And the tasklet is not scheduled for running (bit TASKLET_STATE_SCHED of t->state is 0)
        if (!test_and_clear_bit(TASKLET_STATE_SCHED,
                    &t->state))
            // There's something wrong, this should never happen!
            BUG();

换句话说，你不能在t->count == 0和t->state & (1<<TASKLET_STATE_SCHED) == 0中使用微线程。如果发生这种情况，则存在错误。