Minki/linux
1
0
Fork 1
mirror of https://github.com/torvalds/linux.git synced 2024-11-10 06:01:57 +00:00
Code Issues Packages Projects Releases Wiki Activity

net: x25: Use list_for_each_entry() to simplify code in x25_forward.c

Browse Source 
Convert list_for_each() to list_for_each_entry() where
applicable. This simplifies the code.

Reported-by: Hulk Robot <hulkci@huawei.com>
Signed-off-by: Wang Hai <wanghai38@huawei.com>
Signed-off-by: David S. Miller <davem@davemloft.net>
This commit is contained in:
Wang Hai 2021-06-08 13:30:07 +00:00 committed by David S. Miller
parent 78595dfcb2
commit 96bffe7023
1 changed files with 2 additions and 6 deletions
Show Stats Download Patch File Download Diff File Expand all files Collapse all files

8
net/x25/x25_forward.c
View File

@ -19,7 +19,6 @@ int x25_forward_call(struct x25_address *dest_addr, struct x25_neigh *from,
{
struct x25_route *rt;
struct x25_neigh *neigh_new = NULL;
struct list_head *entry;
struct x25_forward *x25_frwd, *new_frwd;
struct sk_buff *skbn;
short same_lci = 0;
@ -46,8 +45,7 @@ int x25_forward_call(struct x25_address *dest_addr, struct x25_neigh *from,
* established LCI? It shouldn't happen, just in case..
*/
read_lock_bh(&x25_forward_list_lock);
list_for_each(entry, &x25_forward_list) {
x25_frwd = list_entry(entry, struct x25_forward, node);
list_for_each_entry(x25_frwd, &x25_forward_list, node) {
if (x25_frwd->lci == lci) {
pr_warn("call request for lci which is already registered!, transmitting but not registering new pair\n");
same_lci = 1;
@ -92,15 +90,13 @@ out_no_route:
int x25_forward_data(int lci, struct x25_neigh *from, struct sk_buff *skb) {
struct x25_forward *frwd;
struct list_head *entry;
struct net_device *peer = NULL;
struct x25_neigh *nb;
struct sk_buff *skbn;
int rc = 0;
read_lock_bh(&x25_forward_list_lock);
list_for_each(entry, &x25_forward_list) {
frwd = list_entry(entry, struct x25_forward, node);
list_for_each_entry(frwd, &x25_forward_list, node) {
if (frwd->lci == lci) {
/* The call is established, either side can send */
if (from->dev == frwd->dev1) {