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fetch: make the code more understandable
author
Felipe Contreras
<felipe.contreras@gmail.com>
Tue, 4 Jun 2019 02:13:29 +0000
(21:13 -0500)
committer
Junio C Hamano
<gitster@pobox.com>
Tue, 4 Jun 2019 18:28:58 +0000
(11:28 -0700)
The comment makes it seem as if the condition is the other way around.
The exception is when the oid is null, so check for that.
Signed-off-by: Felipe Contreras <felipe.contreras@gmail.com>
Signed-off-by: Junio C Hamano <gitster@pobox.com>
builtin/fetch.c
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a8363b5
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diff --git
a/builtin/fetch.c
b/builtin/fetch.c
index 547b25d206c3acbd65fd042b0287ffc5568fa7d0..0bf8fa7030fec46984503f80c9046b73f57fe648 100644
(file)
--- a/
builtin/fetch.c
+++ b/
builtin/fetch.c
@@
-366,19
+366,21
@@
static void find_non_local_tags(const struct ref *refs,
*/
for_each_string_list_item(remote_ref_item, &remote_refs_list) {
const char *refname = remote_ref_item->string;
*/
for_each_string_list_item(remote_ref_item, &remote_refs_list) {
const char *refname = remote_ref_item->string;
+ struct ref *rm;
item = hashmap_get_from_hash(&remote_refs, strhash(refname), refname);
if (!item)
BUG("unseen remote ref?");
/* Unless we have already decided to ignore this item... */
item = hashmap_get_from_hash(&remote_refs, strhash(refname), refname);
if (!item)
BUG("unseen remote ref?");
/* Unless we have already decided to ignore this item... */
- if (!is_null_oid(&item->oid)) {
- struct ref *rm = alloc_ref(item->refname);
- rm->peer_ref = alloc_ref(item->refname);
- oidcpy(&rm->old_oid, &item->oid);
- **tail = rm;
- *tail = &rm->next;
- }
+ if (is_null_oid(&item->oid))
+ continue;
+
+ rm = alloc_ref(item->refname);
+ rm->peer_ref = alloc_ref(item->refname);
+ oidcpy(&rm->old_oid, &item->oid);
+ **tail = rm;
+ *tail = &rm->next;
}
hashmap_free(&remote_refs, 1);
string_list_clear(&remote_refs_list, 0);
}
hashmap_free(&remote_refs, 1);
string_list_clear(&remote_refs_list, 0);