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由 Thomas Gleixner 提交于
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: NFredrik Markström <fredrik.markstrom@gmail.com> Signed-off-by: NThomas Gleixner <tglx@linutronix.de> Cc: stable@vger.kernel.org
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