Linux and Latency
geiger at epy.co.at
Sat Aug 21 14:58:30 CEST 1999
reine at mbox314.swipnet.se writes:
> Here are my guesses:
> The 2 in (2<<16) refers to number of fragments.
> The 14 refters to size of fragments. 14 bits is 16384.
> (correct me if im wrong).
this is exactely right.
The parameter is
fragmentnumber<<16 + log2(fragmentsize)
> It seems like if I stay on a total of 32768 (=2*16384)
> I can use a number of variations like svar=(4<<16) | (13);
This is funny behaviour and will get you rather high latency
(assuming that you can't use the GETOSPACE calls either)
BTW, what driver are you using ?
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