tcp: prepare fastopen code for upcoming listener changes

While auditing TCP stack for upcoming 'lockless' listener changes,
I found I had to change fastopen_init_queue() to properly init the object
before publishing it.

Otherwise an other cpu could try to lock the spinlock before it gets
properly initialized.

Instead of adding appropriate barriers, just remove dynamic memory
allocations :
- Structure is 28 bytes on 64bit arches. Using additional 8 bytes
  for holding a pointer seems overkill.
- Two listeners can share same cache line and performance would suffer.

If we really want to save few bytes, we would instead dynamically allocate
whole struct request_sock_queue in the future.

Signed-off-by: Eric Dumazet <edumazet@google.com>
Signed-off-by: David S. Miller <davem@davemloft.net>

1 files changed, 8 insertions, 1 deletions

diff --git a/net/core/request_sock.c b/net/core/request_sock.c
index b42f0e26f89e..e22cfa4ed25f 100644
--- a/net/core/request_sock.c
+++ b/net/core/request_sock.c
@@ -59,6 +59,13 @@ int reqsk_queue_alloc(struct request_sock_queue *queue,
 
 	get_random_bytes(&lopt->hash_rnd, sizeof(lopt->hash_rnd));
 	spin_lock_init(&queue->syn_wait_lock);
+
+	spin_lock_init(&queue->fastopenq.lock);
+	queue->fastopenq.rskq_rst_head = NULL;
+	queue->fastopenq.rskq_rst_tail = NULL;
+	queue->fastopenq.qlen = 0;
+	queue->fastopenq.max_qlen = 0;
+
 	queue->rskq_accept_head = NULL;
 	lopt->nr_table_entries = nr_table_entries;
 	lopt->max_qlen_log = ilog2(nr_table_entries);
@@ -174,7 +181,7 @@ void reqsk_fastopen_remove(struct sock *sk, struct request_sock *req,
 	struct sock *lsk = req->rsk_listener;
 	struct fastopen_queue *fastopenq;
 
-	fastopenq = inet_csk(lsk)->icsk_accept_queue.fastopenq;
+	fastopenq = &inet_csk(lsk)->icsk_accept_queue.fastopenq;
 
 	tcp_sk(sk)->fastopen_rsk = NULL;
 	spin_lock_bh(&fastopenq->lock);