On Thu, 2014-06-26 at 16:53 +0200, Bart Van Assche wrote:
> On 06/11/14 11:09, Sagi Grimberg wrote:
> > +   return xfer_len + (xfer_len >> ilog2(sector_size)) * 8;
> 
> Sorry that I just noticed this now, but why is a shift-right and ilog2()
> used in the above expression instead of just dividing the transfer
> length by the sector size ?

It's a performance thing.  Division is really slow on most CPUs.
However, we know the divisor is a power of two so we re-express the
division as a shift, which the processor can do really fast.

James


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