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fetch: make the code more understandable 

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>
maint
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);