If a interrupt chip utilizes chip->buslock then free_irq() can
deadlock in the following way:
CPU0 CPU1
interrupt(X) (Shared or spurious)
free_irq(X) interrupt_thread(X)
chip_bus_lock(X)
irq_finalize_oneshot(X)
chip_bus_lock(X)
synchronize_irq(X)
synchronize_irq() waits for the interrupt thread to complete,
i.e. forever.
Solution is simple: Drop chip_bus_lock() before calling
synchronize_irq() as we do with the irq_desc lock. There is nothing to
be protected after the point where irq_desc lock has been released.
This adds chip_bus_lock/unlock() to the remove_irq() code path, but
that's actually correct in the case where remove_irq() is called on
such an interrupt. The current users of remove_irq() are not affected
as none of those interrupts is on a chip which requires buslock.
Reported-by: Fredrik Markström <fredrik.markstrom@gmail.com>
Signed-off-by: Thomas Gleixner <tglx@linutronix.de>
Cc: stable@vger.kernel.org
if (!desc)
return NULL;
+ chip_bus_lock(desc);
raw_spin_lock_irqsave(&desc->lock, flags);
/*
if (!action) {
WARN(1, "Trying to free already-free IRQ %d\n", irq);
raw_spin_unlock_irqrestore(&desc->lock, flags);
-
+ chip_bus_sync_unlock(desc);
return NULL;
}
#endif
raw_spin_unlock_irqrestore(&desc->lock, flags);
+ chip_bus_sync_unlock(desc);
unregister_handler_proc(irq, action);
desc->affinity_notify = NULL;
#endif
- chip_bus_lock(desc);
kfree(__free_irq(irq, dev_id));
- chip_bus_sync_unlock(desc);
}
EXPORT_SYMBOL(free_irq);