[Haskell-cafe] Performance help

[Ryan Ingram]

Sure, if the ring size is a power of two, and a is greater than or equal
to 0, then

a `mod` ringSize == a .&. (ringSize - 1)

that is:

a `mod` 8 == a .&. 7
a `mod` 256 == a .&. 255

etc.

On 11/13/07, Justin Bailey <jgbailey at gmail.com> wrote:
>
> On Nov 13, 2007 2:21 PM, Ryan Ingram <ryani.spam at gmail.com> wrote:
> > Never mind, I realized this is a ring buffer with `mod` s.   That's
> another
> > slow operation when you're doing code as tight as this.  If you can
> > guarantee the ring is a power of 2 in size you can use a mask instead,
> or
> > use my original suggestion of deriving rules from the previous rule and
> the
> > new bit; the initial state is determined by the last bits in the buffer
> and
> > you never wrap.
>
> I can't guarantee the ring is a power of 2 but do you feel like
> explaining the mask suggestion anyways?
>
> Thanks for the bits suggestion - I'll see if that helps performance at
> all. It looks like you have to be very careful in which concrete type
> you choose or you'll get a lot of  conversion going on.
>
> Justin
>
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