filter_refs(): simplify logic
Simplify flow within loop: first decide whether to keep the reference, then keep/free it. This makes it clearer that each ref has exactly two possible destinies, and removes duplication of the code for appending the reference to the linked list. Signed-off-by: Michael Haggerty <mhagger@alum.mit.edu> Signed-off-by: Junio C Hamano <gitster@pobox.com>
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@ -544,37 +544,36 @@ static void filter_refs(struct ref **refs, struct string_list *sought)
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sought_pos = 0;
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for (ref = *refs; ref; ref = next) {
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int keep = 0;
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next = ref->next;
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if (!memcmp(ref->name, "refs/", 5) &&
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check_refname_format(ref->name + 5, 0))
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; /* trash */
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else if (args.fetch_all &&
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(!args.depth || prefixcmp(ref->name, "refs/tags/") )) {
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(!args.depth || prefixcmp(ref->name, "refs/tags/")))
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keep = 1;
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else {
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while (sought_pos < sought->nr) {
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int cmp = strcmp(ref->name, sought->items[sought_pos].string);
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if (cmp < 0)
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break; /* definitely do not have it */
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else if (cmp == 0) {
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keep = 1; /* definitely have it */
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sought->items[sought_pos++].util = "matched";
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break;
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}
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else
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sought_pos++; /* might have it; keep looking */
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}
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}
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if (keep) {
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*newtail = ref;
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ref->next = NULL;
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newtail = &ref->next;
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continue;
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} else {
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free(ref);
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}
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else {
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int cmp = -1;
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while (sought_pos < sought->nr) {
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cmp = strcmp(ref->name, sought->items[sought_pos].string);
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if (cmp < 0) /* definitely do not have it */
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break;
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else if (cmp == 0) { /* definitely have it */
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sought->items[sought_pos++].util = "matched";
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*newtail = ref;
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ref->next = NULL;
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newtail = &ref->next;
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break;
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}
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else /* might have it; keep looking */
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sought_pos++;
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}
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if (!cmp)
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continue; /* we will link it later */
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}
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free(ref);
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}
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if (!args.fetch_all)
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