I have a quick follow-up question, actually; pipes-group defines: Pipes.Group.folds :: Monad m => (x -> a -> x) -- ^ Step function -> x -- ^ Initial accumulator -> (x -> b) -- ^ Extraction function -> FreeT (Producer a m) m r -- ^ -> Producer b m r
If I'm reading this right, when my FreeT "list" consists of just one Producer, then Pipes.Groups.folds returns a Producer that yields one output, and preserves the original Producer's return type, r, in the returned Producer. This is in contrast to the similar function Pipes.Prelude.fold :: Monad m => (x -> a -> x) -> x -> (x -> b) -> Producer a m () -> m b which only works on Producers with return type (). You note in the documentation for Pipes.Prelude.fold that this type is required because it may stop drawing from the Producer early, so you don't necessarily get to compute the return type. I'm wondering if it's easy to define a function foldToProducer :: Monad m => (x -> a -> x) -> x -> (x -> b) -> Producer a m r -> Producer b m r that does what I think Pipes.Group.folds is doing, but without needing all the FreeT bits as well. As an exercise, I tried to write foldToProducer, but couldn't figure it out. Thanks again, Dylan On Thursday, September 24, 2015 at 11:20:06 PM UTC-4, Dylan Tisdall wrote: > > Right, I wasn't recognizing that `Producer` was an instance of `Functor` > since it's an instance of `Monad`, so I wasn't even looking there. Thanks > again for all your help! > > On Tuesday, September 22, 2015 at 6:56:49 PM UTC-4, Gabriel Gonzalez wrote: >> >> Use the `void` function from `Control.Monad` if you want to erase the >> return type of a `Producer`: >> >> void :: Functor f => f a -> f () >> void = fmap (\_ -> ()) >> >> I might even re-export this from `pipes` as a convenience since this >> question comes up a lot. >> >> Originally functions like `Pipes.Prelude.length` had a more general type >> like this: >> >> Pipes.Prelude.length :: Producer a m r -> m Int >> >> ... but then at the advice of others I restricted the type to this: >> >> Pipes.Prelude.length :: Producer a m () -> m Int >> >> ... so that the user would have to explicitly discard the return value to >> signal that they were okay with ignoring that data. This is similar in >> principle to the warning you get if you turn on the `-Wall` flag that >> (among other things) warns if you have an unused non-empty return value, >> like this: >> >> example = do >> getLine // Compiler warning because you didn't use the result >> ... >> >> ... and you usually have to explicitly ignore the value using something >> like this syntax to indicate that you are intentionally ignoring the value: >> >> example = do >> _ <- getLine >> ... >> >> So the requirement to explicitly discard the value using `void` is in the >> same spirit as that compiler warning. >> >> On 9/22/15 3:50 PM, Dylan Tisdall wrote: >> >> Hi Gabriel, >> >> Thanks again for your help. That really clarified that I should be using >> lift to keep everything inside the Producer transfomer. To make all the >> types work, I ended up with: >> >> type MDHAndScanLineProducer = P.Producer MDHAndScanLine IO (Either >> (P.DecodingError, P.Producer P.ByteString IO ()) ()) >> >> measDatMDHScanLinePairs :: Handle -> MDHAndScanLineProducer >> measDatMDHScanLinePairs h = do >> (hLen, leftovers) <- lift $ P.runStateT (P.decodeGet getWord32le) p >> case (hLen :: Either P.DecodingError Word32) of >> Left err -> return $ Left (err, leftovers) >> Right len -> do >> lift (hSeek h AbsoluteSeek (fromIntegral len)) >> view P.decoded p >> where >> p = PB.fromHandle h >> >> This seems to work exactly as I'd hoped. >> >> As a follow-up, I'm now wondering how to use this producer and ignore its >> return type; effectively how to turn it into a Producer MDHAndScanLine >> IO (). This seems to be necessary to access many library functions. For >> example, I can't use >> >> Pipes.Prelude.length :: Monad m => Producer a m () -> m Int >> >> directly on the output of measDatMDHScanLinePairs because the return >> type doesn't match. >> >> Thanks again for all your help as I get up to speed on this! >> >> >> Dylan >> >> >> On Monday, September 21, 2015 at 11:43:58 PM UTC-4, Gabriel Gonzalez >> wrote: >>> >>> You're definitely on the right track. The type I would aim for would be >>> something like this: >>> >>> example :: Handle -> Producer MDHAndScanLine IO (Either >>> DecodingError (Producer ByteString IO ())) >>> >>> Notice that this slightly differs from your type; I'm merging the outer >>> `IO (Either DecodingError ...)` into the first `Producer` to simplify the >>> type. >>> >>> The implementation for that type would be very similar to the one you >>> wrote in your second e-mail: >>> >>> example :: Handle -> Producer MDHAndScanLine IO (Either >>> DecodingError (Producer ByteString IO ())) >>> example handle = do >>> let p = Pipes.ByteString.fromHandle handle >>> x <- lift (evalStateT (decodeGet getWord32le) p) >>> case x of >>> Left err -> return (Left err) >>> Right len -> do >>> lift (hSeek handle AbsoluteSeek (fromIntegral l)) >>> view decoded p >>> >>> That will definitely run in constant memory, meaning that it won't ever >>> load more than one chunk of bytes at a time (where a chunk is something >>> like 32 kB, I think). You can profile the heap if you want to verify this >>> by following these instructions: >>> >>> >>> https://downloads.haskell.org/~ghc/latest/docs/html/users_guide/prof-heap.html >>> >>> Also, to answer your other question, `pipes-attoparsec` runs in constant >>> memory. The difference between `pipes-attoparsec` and `attoparsec` is that >>> `pipes-attoparsec` runs a separate parser for each element in the stream, >>> which is equivalent to "committing" after each parsed element. That means >>> that it can only backtrack while parsing a single element in the stream, >>> but no further back. This is why `pipes-attoparsec` runs in constant space >>> over a large file and why `attoparsec` does not, because `attoparsec` >>> backtracks indefinitely and `pipes-attoparsec` does not. >>> >>> On 9/21/15 12:10 PM, Dylan Tisdall wrote: >>> >>> Following up on my last question, my next issue is also probably a very >>> straight ahead example of pipes, but I've managed to get tangled up going >>> back and forth in the packages' documentation. >>> >>> I've got a file whose first 4 bytes give the offset into the file of a >>> series of binary data elements (called MDHs in my case). Given a Handle to >>> the start of such a file, I want to: >>> >>> 1. read the first Word32 in the file, to retrieve the offset; >>> 2. skip the Handle to that offset; and >>> 3. turn the rest of the file into a Producer MDH IO () >>> >>> Given that the file I'm reading may be large, I want to make sure this >>> process is going to run in constant memory. I thought I could use >>> pipes-attoparsec, but I couldn't get straight whether it would need to read >>> the whole file before it could produce anything (as I understand is >>> normally the case with attoparsec). >>> >>> So far I have the following, which isn't complete, but at least does the >>> skip and converts the remaining file to a ByteString producer. >>> >>> handleToMDHs :: Handle -> IO (Either P.DecodingError (P.Producer P. >>> ByteString IO ())) >>> handleToMDHs h = do >>> hLen <- P.evalStateT (P.decodeGet getWord32le) (PB.fromHandle h) >>> case (hLen :: Either P.DecodingError Word32) of >>> Left err -> return $ Left err >>> Right len -> fmap Right (skipAndProceed h len) >>> where >>> skipAndProceed :: Handle -> Word32 -> IO (P.Producer P.ByteString >>> IO ()) >>> skipAndProceed handle l = do >>> (hSeek handle AbsoluteSeek) (fromIntegral l) >>> return $ PB.fromHandle handle >>> >>> >>> My MDH type is an instance of Binary, so there is a get method >>> available. I'm wondering: >>> >>> a) What's the right way to turn this into a Producer of MDHs instead of >>> a Producer of ByteStrings while operating in constant memory? >>> b) Is there a more elegant way to deal with error handling here? I'm not >>> even dealing with possible failure in hSeek, and I already think this looks >>> pretty messy. I'm not wedded to my function type being >>> >>> handleToMDHs :: Handle -> IO (Either P.DecodingError (P.Producer MDH IO >>> ())) >>> >>> I just am not sure how else to express the possibility of failure in >>> this kind of operation. >>> >>> >>> Thanks, >>> Dylan >>> -- >>> You received this message because you are subscribed to the Google >>> Groups "Haskell Pipes" group. >>> To unsubscribe from this group and stop receiving emails from it, send >>> an email to haskell-pipe...@googlegroups.com. >>> To post to this group, send email to haskel...@googlegroups.com. >>> >>> >>> -- >> You received this message because you are subscribed to the Google Groups >> "Haskell Pipes" group. >> To unsubscribe from this group and stop receiving emails from it, send an >> email to haskell-pipe...@googlegroups.com. >> To post to this group, send email to haskel...@googlegroups.com. >> >> >> -- You received this message because you are subscribed to the Google Groups "Haskell Pipes" group. 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