Hi Andrew,
Thanks for the comments, it really helps to have someone else's
opinion on my code. I'll be applying what you've said as soon as I
get a chance and I'm sure I'll have some more questions then. I'll
certainly look more closely at the Set interface and try and duplicate
all the parts
Hi,
Just a small comment on one of the comments.
On 5/1/07, [EMAIL PROTECTED] [EMAIL PROTECTED] wrote:
Also, rather than this:
add :: Bloom a - a - Bloom a
a better argument order is this:
insert :: a - Bloom a - Bloom a
That way, you can use it with foldr.
Hmmm. If you want to
Reminds me of this code from Data.Binary:
unroll :: Integer - [Word8]
unroll = unfoldr step
where
step 0 = Nothing
step i = Just (fromIntegral i, i `shiftR` 8)
roll :: [Word8] - Integer
roll = foldr unstep 0
where
unstep b a = a
G'day all.
Quoting Dom [EMAIL PROTECTED]:
But better than what is in Codec.Utils:
[deletia]
It seems a shame that everyone has to roll their own.
That and integer log base 2.
Cheers,
Andrew Bromage
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Haskell-Cafe mailing list
G'day all.
I wrote:
insert :: a - Bloom a - Bloom a
That way, you can use it with foldr.
Quoting Josef Svenningsson [EMAIL PROTECTED]:
Hmmm. If you want to create a Bloom using a fold wouldn't it make more
sense to use foldl'? I think the argument order is fine.
You're right that
Hi all,
I'm pretty new to Haskell, I've been working on a Bloom filter[1]
implementation as a learning exercise.
I'd really appreciate it if someone more experienced would comment on
the code. I'm sure there's plenty of places where I'm doing things in
silly or overly complex ways.
I've
G'day.
Quoting tom [EMAIL PROTECTED]:
I'm pretty new to Haskell, I've been working on a Bloom filter[1]
implementation as a learning exercise.
Excellent! Sounds like a fun test.
I'd really appreciate it if someone more experienced would comment on
the code. I'm sure there's plenty of
ajb:
Quoting tom [EMAIL PROTECTED]:
This looks cool:
bytes2int = foldr ((. (256 *)) . (+)) 0 . (map toInteger)
but I'm not smart enough to parse it. This is both more readable and
shorter:
bytes2int = foldr (\x r - r*256 + fromInteger x) 0
Integer log2's are probably better
