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 <[email protected]>
Signed-off-by: Junio C Hamano <[email protected]>

Author: felipec

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;
+		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);