x86: get rid of calling do_notify_resume() when returning to kernel mode

If we end up calling do_notify_resume() with !user_mode(refs), it
does nothing (do_signal() explicitly bails out and we can't get there
with TIF_NOTIFY_RESUME in such situations).  Then we jump to
resume_userspace_sig, which rechecks the same thing and bails out
to resume_kernel, thus breaking the loop.

It's easier and cheaper to check *before* calling do_notify_resume()
and bail out to resume_kernel immediately.  And kill the check in
do_signal()...

Note that on amd64 we can't get there with !user_mode() at all - asm
glue takes care of that.

Acked-and-reviewed-by: Thomas Gleixner <tglx@linutronix.de>
Signed-off-by: Al Viro <viro@zeniv.linux.org.uk>

1 files changed, 0 insertions, 10 deletions

diff --git a/arch/x86/kernel/signal.c b/arch/x86/kernel/signal.c
index e8a89374d356..21af737053aa 100644
--- a/arch/x86/kernel/signal.c
+++ b/arch/x86/kernel/signal.c
@@ -737,16 +737,6 @@ static void do_signal(struct pt_regs *regs)
         siginfo_t info;
         int signr;
 
-        /*
-         * We want the common case to go fast, which is why we may in certain
-         * cases get here from kernel mode. Just return without doing anything
-         * if so.
-         * X86_32: vm86 regs switched out by assembly code before reaching
-         * here, so testing against kernel CS suffices.
-         */
-        if (!user_mode(regs))
-                return;
-
         signr = get_signal_to_deliver(&info, &ka, regs, NULL);
         if (signr > 0) {
                 /* Whee! Actually deliver the signal.  */