Re: [Haskell-cafe] Performance help

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 wrote:

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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

On Nov 13, 2007 2:21 PM, Ryan Ingram wrote: the

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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