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

Question

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 in process context. This is in contrast to the other bottom-half mechanisms, which all run in interrupt context. Code running in interrupt context is unable to sleep, or block, because interrupt context does not have a backing process with which to reschedule. Therefore, because interrupt handlers are not associated with a process, there is nothing for the scheduler to put to sleep and, more importantly, nothing for the scheduler to wake up..."

I don't get it. AFAIK, scheduler in the kernel is O(1), that is implemented through the bitmap. So what stops the scehduler from putting interrupt context to sleep and taking next schedulable process and passing it the control?

Answer 1

So what stops the scehduler from putting interrupt context to sleep and taking next schedulable process and passing it the control?

The problem is that the interrupt context is not a process, and therefore cannot be put to sleep.

When an interrupt occurs, the processor saves the registers onto the stack and jumps to the start of the interrupt service routine. This means that when the interrupt handler is running, it is running in the context of the process that was executing when the interrupt occurred. The interrupt is executing on that process's stack, and when the interrupt handler completes, that process will resume executing.

If you tried to sleep or block inside an interrupt handler, you would wind up not only stopping the interrupt handler, but also the process it interrupted. This could be dangerous, as the interrupt handler has no way of knowing what the interrupted process was doing, or even if it is safe for that process to be suspended.

A simple scenario where things could go wrong would be a deadlock between the interrupt handler and the process it interrupts.

1. Process1 enters kernel mode.
2. Process1 acquires LockA.
3. Interrupt occurs.
4. ISR starts executing using Process1's stack.
5. ISR tries to acquire LockA.
6. ISR calls sleep to wait for LockA to be released.

At this point, you have a deadlock. Process1 can't resume execution until the ISR is done with its stack. But the ISR is blocked waiting for Process1 to release LockA.