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fetch: make the code more understandable

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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>
This commit is contained in:
Felipe Contreras 2019-06-03 21:13:29 -05:00 committed by Junio C Hamano
parent a8363b5719
commit 9528b80b1a
1 changed files with 9 additions and 7 deletions
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16
builtin/fetch.c
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@ -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;
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... */
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);