Why kernel code/thread executing in interrupt context cannot sleep?

前端 未结 11 1758
余生分开走
余生分开走 2020-11-30 17:20

I am reading following article by Robert Love

http://www.linuxjournal.com/article/6916

that says

\"...Let\'s discuss the fact that work queues run i

11条回答
  •  感情败类
    2020-11-30 18:11

    Disallowing an interrupt handler to block is a design choice. When some data is on the device, the interrupt handler intercepts the current process, prepares the transfer of the data and enables the interrupt; before the handler enables the current interrupt, the device has to hang. We want keep our I/O busy and our system responsive, then we had better not block the interrupt handler.

    I don't think the "unstable states" are an essential reason. Processes, no matter they are in user-mode or kernel-mode, should be aware that they may be interrupted by interrupts. If some kernel-mode data structure will be accessed by both interrupt handler and the current process, and race condition exists, then the current process should disable local interrupts, and moreover for multi-processor architectures, spinlocks should be used to during the critical sections.

    I also don't think if the interrupt handler were blocked, it cannot be waken up. When we say "block", basically it means that the blocked process is waiting for some event/resource, so it links itself into some wait-queue for that event/resource. Whenever the resource is released, the releasing process is responsible for waking up the waiting process(es).

    However, the really annoying thing is that the blocked process can do nothing during the blocking time; it did nothing wrong for this punishment, which is unfair. And nobody could surely predict the blocking time, so the innocent process has to wait for unclear reason and for unlimited time.

提交回复
热议问题