Yu Xiangning wrote:
>> If we want to keep the existing order, this would be
>>
>> switch (ipif->ipif_flags & (IPIF_POINTOPOINT|IPIF_UNNUMBERED)) {
>> case IPIF_POINTOPOINT:
>> case 0:
>> if (ipif->ipif_lcl_addr == addr)
>> return (ipif);
>>
>> /* Match both local and remote for pointopoint */
>> if ((ipif->ipif_flags & IPIF_POINTOPOINT) &&
>> ipif->ipif_pp_dst_addr == addr)
>> return (ipif);
>>
>> break;
>>
>> case (IPIF_POINTOPOINT|IPIF_UNNUMBERED):
>> /* Match just the remote address */
>> if (ipif->ipif_pp_dst_addr == addr)
>> return (ipif);
>> break;
>> default:
>> ASSERT(0);
>> }
>
> If there are two interfaces in the system and it happens to be the first
> one is with IPIF_POINTTOPOINT and the second one without
> IPIF_UNNUMBERED, also suppose the first one's ipif_pp_dst_addr is equal
> to the second one's ipif_lcl_addr (this can happen if one of the
> interface is DOWN), the above code will result in different behavior
> from what it does today. i.e. Nevada matches the second one while the
> proposed code matches the first one.
>
> I'm not sure the correct behavior under this situation. If either is OK
> then it's fine we can eliminate twice lookups.
Good point.
Erik
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