I see. But I want CPB to reflect both the J expression and the architecture.
+./y does not depend on the value of every byte in the argument. Others of that ilk are (> i. 1:) (> i. 0:) etc. A CPB value of less than 1 is exceptional and requires explanation. ----- Original Message ----- From: Oleg Kobchenko <[EMAIL PROTECTED]> Date: Friday, December 22, 2006 10:38 am Subject: Re: [Jprogramming] Cycles per Byte > Oops, for size of y. > > But the calibration argument still stands. > The same expression needs to get the same > CPB, but clock rate does not reflect that. > Instead it relects CPU architectures, not J expressions. > > I hope this is right > > (#y) %~ 1.45e9 * 6!:2 '+./y' > 0.00210641 > > > --- Roger Hui <[EMAIL PROTECTED]> wrote: > > > Huh? > > > > a. In this sequence of msgs y has been a Boolean > > vector, in which case you need to divide by > > (#y) rather than (4*#y). If you are using a > > different y please tell me what it is. > > > > b. I don't understand how your results show that > > calibration is required. If anything it shows > > that calibration gives the wrong answers. > > > > > > > > ----- Original Message ----- > > From: Oleg Kobchenko <[EMAIL PROTECTED]> > > Date: Friday, December 22, 2006 8:54 am > > Subject: Re: [Jprogramming] Cycles per Byte > > > > > OK. Next is N]\y, N<0 > > > > > > (4*#y) %~ 1.45e9 * 6!:2 '_2]\y' NB. 2.8 GHz calibrated > > > 0.737346 > > > > > > I specifically reproduced the results for > > > $ and + to show that the clock rate does not > > > reflect time without calibration. ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
