Re: [libvirt] [PATCH] interface: list all interfaces with flags == 0

On 05/21/2013 06:49 AM, Guannan Ren wrote: Guannan's logic says when to drop: Jirka's logic says when to keep; so it would need a ! around the overall expression if we want to turn it into the condition that represents when to drop. Let's compare what happens for all four combinations of the two flags [active is flag 1, inactive is flag 2]: flags status Guannan's [0 means keep] Jirka's [1 means keep] 0 active 0&&!(0||0)=keep !0||0||0=keep 0 inactive 0&&!(0||0)=keep !0||0||0=keep 1 active 1&&!(1||0)=keep !1||1||0=keep 1 inactive 1&&!(0||0)=drop !1||0||0=drop 2 active 1&&!(0||0)=drop !1||0||0=drop 2 inactive 1&&!(0||1)=keep !1||0||1=keep 3 active 1&&!(1||0)=keep !1||1||0=keep 3 inactive 1&&!(0||1)=keep !1||0||1=keep Ultimately, the two expressions are equivalent (you can use deMorgan's law to prove the equivalence), but I find Jirka's positive logic a bit easier to reason with than Guannan's negative logic, even if Guannan's style copied from how we did it on other API. I do agree that for purposes of adding future filter groups, as well as minimizing nested conditionals, that the logic looks better in terms of drop checks: foreach item: if (filter group 1 drops) continue; if (filter group 2 drops) continue; item was kept by all filters, add it to return rather than logic in terms of keep checks: foreach item: if (filter group 1 keeps) if (filter group 2 keeps) add it to return -- Eric Blake eblake redhat com +1-919-301-3266 Libvirt virtualization library http://libvirt.org