If you wanted to use enumerator, it isn't too tough. You would do something like this: import Data.Enumerator (run_, ($$), (=$)) import Data.Enumerator.Binary (enumHandle, iterHandle) import Data.Enumerator.List as EL (map) import Data.ByteString as B (map) import Data.Bits (complement) import System.IO (withFile, IOMode(..)) main = withFile "infile" ReadMode $ \inh -> withFile "outfile" WriteMode $ \outh -> do run_ (enumHandle 4096 inh $$ EL.map (B.map complement) =$ iterHandle outh) On Sat, Sep 24, 2011 at 2:00 PM, Kyle Murphy wrote:

It's not for the faint of heart, but the enumerator package is also supposed to provide very good performance for stream transformations. I've looked at it a bit myself but I'm still struggling to wrap my head around the types involved, which like most things in Haskell is the key to understanding the whole thing.

-R. Kyle Murphy -- Curiosity was framed, Ignorance killed the cat.

On Sat, Sep 24, 2011 at 05:01, Benjamin Edwards wrote:

...

You should look up bytestring and friends on hackage.

If it is something quite simple you can use the lazy variants and provided that you don't try to hold onto the input you should get nice constant space without trying too hard.

I recommend the early chapters on IO in real world haskell if you want more info on lazy IO.

On 24 Sep 2011 03:06, "anyzhen" wrote:

...

consider this : i want load a 4G file(or some bigger file) ,and the process data operation just like 010 to 101( XOR bits ) , is it some efficient function down it ? such hPutStr hPutChar is Char layer , is exist bit layer operations?

thanks for any help

jiangzhen3s@qq.com

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