prepared is assigned first, so it does not need to initialize the 
assignment.

Signed-off-by: Li zeming <zeming@nfschina.com>
---
 kernel/seccomp.c | 2 +-
 1 file changed, 1 insertion(+), 1 deletion(-)

diff --git a/kernel/seccomp.c b/kernel/seccomp.c
index d3e584065c7f..af672e03449a 100644
--- a/kernel/seccomp.c
+++ b/kernel/seccomp.c
@@ -1838,7 +1838,7 @@ static long seccomp_set_mode_filter(unsigned int flags,
 				    const char __user *filter)
 {
 	const unsigned long seccomp_mode = SECCOMP_MODE_FILTER;
-	struct seccomp_filter *prepared = NULL;
+	struct seccomp_filter *prepared;
 	long ret = -EINVAL;
 	int listener = -1;
 	struct file *listener_f = NULL;
-- 
2.18.2

On Sat, Aug 26, 2023 at 02:43:48AM +0800, Li zeming wrote:
> prepared is assigned first, so it does not need to initialize the 
> assignment.

I prefer to explicitly initialize variables so that later changes in
the code cannot suddenly leave a variable in a surprise state. :) The
compiler will do a good job removing redundant and dead stores.

-Kees

-- 
Kees Cook