Re: [PATCH 1/2] genirq: introduce update_irq_devid()
From: Thomas Gleixner
Date: Thu Aug 08 2019 - 15:56:24 EST
On Thu, 8 Aug 2019, Ben Luo wrote:
> +int update_irq_devid(unsigned int irq, void *dev_id, void *new_dev_id)
> +{
> + struct irq_desc *desc = irq_to_desc(irq);
> + struct irqaction *action, **action_ptr;
> + unsigned long flags;
> +
> + WARN(in_interrupt(),
> + "Trying to update IRQ %d from IRQ context!\n", irq);
This is broken. The function needs to return on that condition. Actually it
cannot even be called from non-preemptible code.
What's worse is that if the interrupt in question is handled concurrently,
then it will either see the old or the new dev_id and because the interrupt
handler loop runs with desc->lock dropped even more crap can happen because
dev_id can be subject to load and store tearing.
Staring at that, I see that there is the same issue in setup_irq() and
free_irq(). It's actually worse there. I'll have a look.
> + /*
> + * There can be multiple actions per IRQ descriptor, find the right
> + * one based on the dev_id:
> + */
> + action_ptr = &desc->action;
> + for (;;) {
> + action = *action_ptr;
> +
> + if (!action) {
> + WARN(1, "Trying to update already-free IRQ %d\n", irq);
That's wrong in two aspects:
1) The warn should be outside of the locked region.
2) Just having the irq number is not useful for debugging either
when the interrupt is shared.
> + raw_spin_unlock_irqrestore(&desc->lock, flags);
> + chip_bus_sync_unlock(desc);
> + return -ENXIO;
> + }
> +
> + if (action->dev_id == dev_id) {
> + action->dev_id = new_dev_id;
> + break;
> + }
> + action_ptr = &action->next;
> + }
> +
> + raw_spin_unlock_irqrestore(&desc->lock, flags);
> + chip_bus_sync_unlock(desc);
> +
> + /*
> + * Make sure it's not being used on another CPU:
> + * There is a risk of UAF for old *dev_id, if it is
> + * freed in a short time after this func returns
> + */
> + synchronize_irq(irq);
> +
> + return 0;
> +}
> +EXPORT_SYMBOL(update_irq_devid);
EXPORT_SYMBOL_GPL() please.
Thanks,
tglx