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honor the claim of sort_list(), that the result is in anti-intuitive order, and treat element zero as the furthest

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Coren[m] 2013-09-21 03:13:44 +02:00
parent 89005f1701
commit 4e76ca432f
1 changed files with 10 additions and 10 deletions
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20
toxcore/DHT.c
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@ -404,23 +404,23 @@ static int replace_good( Client_data *list,
/* either we got an ipv4 address, or we're "allowed" to push out an ipv4
* address in favor of an ipv6 one
*
* because the list is sorted, we can simply check the last client_id,
* either it is closer, then every other one is as well, or it is further,
* then it gets pushed out in favor of the new address, which will with
* the next sort() move to its "rightful" position
* because the list is sorted, we can simply check the client_id at the
* border, either it is closer, then every other one is as well, or it is
* further, then it gets pushed out in favor of the new address, which
* will with the next sort() move to its "rightful" position
*
* we should NOT replace the best worse client_id (original implementation),
* because that one was still better than any later in the field, and
* we don't want to lose it, but the worst
* CAVEAT: weirdly enough, the list is sorted DESCENDING in distance
* so the furthest element is the first, NOT the last (at least that's
* what the comment above sort_list() claims)
*/
if (id_closest(comp_client_id, list[length - 1].client_id, client_id) == 2)
replace = length - 1;
if (id_closest(comp_client_id, list[0].client_id, client_id) == 2)
replace = 0;
} else {
/* ipv6 case without a right to push out an ipv4: only look for ipv6
* addresses, the first one we find is either closer (then we can skip
* out like above) or further (then we can replace it, like above)
*/
for (i = length - 1; i < length; i--) {
for (i = 0; i < length; i++) {
Client_data *client = &list[i];
if (client->ip_port.ip.family == AF_INET6) {
if (id_closest(comp_client_id, list[i].client_id, client_id) == 2)