# [Cake] Choosing a tin to work on

xnor xnoreq at gmail.com
Wed Apr 12 08:48:52 EDT 2017

```Hey,

>>                 s64 tdiff = b->tin_time_next_packet - now;
>>                 if(tdiff <= 0 || tdiff <= best_time) {
>>                     best_time = tdiff;
>>                     best_tin = tin;
>>                 }
>
>I'll try to answer this based on a vague bit of understanding...and
>then Jonathan can shoot me down as well :-)
>
>So the 'best_time' is signed because we can have a packet in a tin that
>is not yet due to be sent (a positive result...ie. we have got here
>early) and/or we can have a tin that is due now/overdue (a
>zero/negative result, we've got here late)
>
>The complication is that we can have multiple tins overdue, so we want
>the highest priority *and* least overdue (least late) tin - this is the
>reason for searching in tin_order[oi] and as a result tdiff can be <=0
>and bigger than best_time.
>
>best_time is initialised to the 'most early' time possible.
that makes no sense to me.

best_time is initialized to some high value (though why is it not 0x7FFF
FFFF FFFF FFFFL, the highest possible positive s64?) such that no matter
how far tdiff is in the future, best_time will always be set to the
lower tdiff.
(Just like you would initialize a variable keeping track of the max to
the lowest possible value, you set a min variable to the highest
possible value.)

But if you wanted best_time to end up as the lowest value, then you
would have to only set it if tdiff < best_time (or <= if you actually
want to prefer the last tin if they happen to have the same tdiff), and
not also if tdiff <= 0.

For example, if tdiff values were 5000, -5000, 0 then best_time would be
set to 0. The last value less than or equal to zero will always win as
best_tin.
If all tdiff values are positive then best_time will end up as the
lowest value however.

```