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genirq: Prevent chip buslock deadlock
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
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@ -1434,6 +1434,7 @@ static struct irqaction *__free_irq(unsigned int irq, void *dev_id)
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if (!desc)
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return NULL;
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chip_bus_lock(desc);
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raw_spin_lock_irqsave(&desc->lock, flags);
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/*
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@ -1447,7 +1448,7 @@ static struct irqaction *__free_irq(unsigned int irq, void *dev_id)
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if (!action) {
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WARN(1, "Trying to free already-free IRQ %d\n", irq);
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raw_spin_unlock_irqrestore(&desc->lock, flags);
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chip_bus_sync_unlock(desc);
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return NULL;
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}
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@ -1475,6 +1476,7 @@ static struct irqaction *__free_irq(unsigned int irq, void *dev_id)
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#endif
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raw_spin_unlock_irqrestore(&desc->lock, flags);
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chip_bus_sync_unlock(desc);
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unregister_handler_proc(irq, action);
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@ -1553,9 +1555,7 @@ void free_irq(unsigned int irq, void *dev_id)
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desc->affinity_notify = NULL;
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#endif
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chip_bus_lock(desc);
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kfree(__free_irq(irq, dev_id));
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chip_bus_sync_unlock(desc);
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}
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EXPORT_SYMBOL(free_irq);
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