x86-32/irq: Don't switch to irq stack for a user-mode irq

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.orgSigned-off-by: Ingo Molnar <mingo@elte.hu>

arch/x86/kernel/irq_32.c

@@ -100,13 +100,8 @@ execute_on_irq_stack(int overflow, struct irq_desc *desc, int irq)
 	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);
@@ -196,7 +191,7 @@ bool handle_irq(unsigned irq, struct pt_regs *regs)
 	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);