[PATCH] x86: vm86() defaults to returning VM86_SIGNAL -- make this explicit

[Samuel Bronson <naesten@gmail.com>]

This code has apparently used "0" and not VM86_SIGNAL since Linux
1.1.9, when Linus added VM86_SIGNAL to vm86.h. This patch changes the
code to use the symbolic name.

The magic 0 tripped me up in trying to extend the vm86(2) manpage to
actually explain vm86()'s interface -- my greps for VM86_SIGNAL came up
fruitless.

Signed-off-by: Samuel Bronson <naesten@gmail.com>
---
 arch/x86/kernel/vm86_32.c |    4 ++--
 1 files changed, 2 insertions(+), 2 deletions(-)

diff --git a/arch/x86/kernel/vm86_32.c b/arch/x86/kernel/vm86_32.c
index d7ac84e..b8035a0 100644
--- a/arch/x86/kernel/vm86_32.c
+++ b/arch/x86/kernel/vm86_32.c
@@ -318,9 +318,9 @@ static void do_sys_vm86(struct kernel_vm86_struct *info, struct task_struct *tsk
 	}
 
 /*
- * Save old state, set default return value (%ax) to 0
+ * Save old state, set default return value (%ax) to 0 (VM86_SIGNAL)
  */
-	info->regs32->ax = 0;
+	info->regs32->ax = VM86_SIGNAL;
 	tsk->thread.saved_sp0 = tsk->thread.sp0;
 	tsk->thread.saved_fs = info->regs32->fs;
 	tsk->thread.saved_gs = get_user_gs(info->regs32);
-- 
1.6.3.rc3.24.g6ffd5