为什么tasklet_action()函数中会调用BUG？答案

【问题标题】：Why is BUG called in the tasklet_action() function?为什么tasklet_action()函数中会调用BUG？
【发布时间】：2021-04-24 18:51:05
【问题描述】：

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();

                // ...
    }
}

我的理解是，如果一个 tasklet 已经安排好了，那么这段代码会抛出一个BUG()。这是否意味着同一个tasklet不能同时运行，也不能调度？

【问题讨论】：

标签： c linux-kernel

【解决方案1】：

这只是对 tasklet 的保证属性的健全性检查。你可以看到a comment in include/linux/interrupt.h中列出的tasklet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，tasklet 保证在调度后运行至少一次。这段代码：

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

确保此属性成立，否则有错误，BUG() 用于停止执行并导致运行时恐慌。

这是上面代码的注释版本，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，你不能有一个带有t->count == 0 和t->state & (1<<TASKLET_STATE_SCHED) == 0 的tasklet。如果发生这种情况，则存在错误。

【讨论】：