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copied!<p>There are a lot of good answers already; I have very little to offer performance-wise but a few style points.</p> <p>First, I would write it this way:</p> <pre><code>import Prelude as P import Data.Iteratee import qualified Data.Iteratee as I import qualified Data.Iteratee.IO as I import qualified Data.ByteString as B import Data.Char import System.Environment -- numLines has a concrete stream type so it's not necessary to provide an -- annotation later. It could have a more general type. numLines :: Monad m => I.Iteratee B.ByteString m Int numLines = I.foldl' step 0 where --step :: Int -> Word8 -> Int step acc el = if el == (fromIntegral $ ord '\n') then acc + 1 else acc main = do f:_ <- getArgs words <- run =<< I.enumFile 65536 f numLines print words </code></pre> <p>The biggest difference is that this uses <code>Data.Iteratee.ListLike.foldl'</code>. Note that only the individual stream elements matter to the step function, not the stream type. It's exactly the same function as you would use with e.g. <code>Data.ByteString.Lazy.foldl'</code>.</p> <p>Using <code>foldl'</code> also means that you don't need to manually write iteratees with <code>liftI</code>. I would discourage users from doing so unless absolutely necessary. The result is usually longer and harder to maintain with little to no benefit.</p> <p>Finally, I've increased the buffer size significantly. On my system this is marginally faster than <code>enumerator</code>s default of 4096, which is again marginally faster (with iteratee) than your choice of 1024. YMMV with this setting of course.</p>

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