If the irq happens in user mode, our kernel stack is empty
(apart from the pt_regs themselves, of course), so there's no
need or advantage to switch.
And it really doesn't save any stack space, quite the reverse:
it means that a nested interrupt cannot switch irq stacks. So
instead of saving kernel stack space, it actually causes the
potential for *more* stack usage.
Also simplify the preemption count copy when we do switch
stacks: just copy the whole preemption count, rather than just
the softirq parts of it. There is no advantage to the partial
copy: it is more effort to get a less correct result.
Signed-off-by: Linus Torvalds <torvalds@linux-foundation.org>
Link: http://lkml.kernel.org/r/alpine.LFD.2.02.1202191139260.10000@i5.linux-foundation.org
Signed-off-by: Ingo Molnar <mingo@elte.hu>
irqctx->tinfo.task = curctx->tinfo.task;
irqctx->tinfo.previous_esp = current_stack_pointer;
- /*
- * Copy the softirq bits in preempt_count so that the
- * softirq checks work in the hardirq context.
- */
- irqctx->tinfo.preempt_count =
- (irqctx->tinfo.preempt_count & ~SOFTIRQ_MASK) |
- (curctx->tinfo.preempt_count & SOFTIRQ_MASK);
+ /* Copy the preempt_count so that the [soft]irq checks work. */
+ irqctx->tinfo.preempt_count = curctx->tinfo.preempt_count;
if (unlikely(overflow))
call_on_stack(print_stack_overflow, isp);
if (unlikely(!desc))
return false;
- if (!execute_on_irq_stack(overflow, desc, irq)) {
+ if (user_mode_vm(regs) || !execute_on_irq_stack(overflow, desc, irq)) {
if (unlikely(overflow))
print_stack_overflow();
desc->handle_irq(irq, desc);