pid: remove redundant condition

From: Liu Ping Fan <[email protected]>

Signed-off-by Liu Ping Fan <[email protected]>

---
 kernel/pid.c |    3 +--

 1 files changed, 1 insertions(+), 2 deletions(-)

kernel/pid.c +0 -0

index 9f08dfa..e4ca244 100644

Quoted 12 lines · Liu Ping Fan

From: Liu Ping Fan <[email protected]>

Signed-off-by Liu Ping Fan <[email protected]>

---
 kernel/pid.c |    3 +--

 1 files changed, 1 insertions(+), 2 deletions(-)

kernel/pid.c +0 -0

index 9f08dfa..e4ca244 100644

kernel/pid.c void put_pid(struct pid *pid) · line 254

Quoted 9 lines

kernel/pid.c +1 -2

@@ -248,8 +248,7 @@ void put_pid(struct pid *pid)
                return;
 
        ns = pid->numbers[pid->level].ns;
-       if ((atomic_read(&pid->count) == 1) ||
-            atomic_dec_and_test(&pid->count)) {
+       if (atomic_dec_and_test(&pid->count) {
                kmem_cache_free(ns->pid_cachep, pid);
                put_pid_ns(ns);
        }

No, this isn't necessarily redundant.

The atomic_dec_and_test() instruction can be very expensive. So if

count was 1 before, we're the last user, and we can free it without the expensive atomic_dec_and_test().

However, if we are *not* the last user, then we can race with another user decrementing the count, so now we need to use the expensive version.

Now, whether that special case is really worth it or not, I don't know. But it's not redundant.

                   Linus