Michael Snoyman wrote: > On Wed, Dec 21, 2011 at 12:59 PM, Kannan Goundan <kan...@cakoose.com> > wrote: >> Michael Snoyman wrote: >> >>> On Wed, Dec 21, 2011 at 12:35 PM, Kannan Goundan <kan...@cakoose.com> >>> wrote: >>>> I'm using the Data.Enumerator library. I'm trying to write a "map" >>>> function that converts an Enumerator of one type to another. >>>> Something like: >>>> >>>> mapEnum :: Monad m => >>>> (a -> b) -> >>>> Enumerator a m r -> >>>> Enumerator b m r >>>> >>>> Any hints? >>>> >>>> (My exact use case is that I have a ByteString enumerator and I need >>>> to pass it to something that requires a Blaze.Builder enumerator.) >>>> >>>> >>>> _______________________________________________ Haskell-Cafe mailing >>>> list >>>> Haskell-Cafe@haskell.org >>>> http://www.haskell.org/mailman/listinfo/haskell-cafe >>> >>> You can use the Data.Enumerator.List.map function to create an >>> Enumeratee, and then the Data.Enumerator.$= operators to join them >>> together. Something like: >>> >>> mapEnum f enum = enum $= EL.map f >> >> I tried something like that but the resulting type isn't quite what I'm >> looking for. >> >> mapEnum :: Monad m => >> (a -> b) -> >> Enumerator a m (Step b m r) -> >> Enumerator a m r >> >> (BTW, Michael, my exact use case is that I have ByteString enumerators, >> but your HTTP-Enumerator library wants Blaze.Builder enumerators :-) >> >> >> _______________________________________________ Haskell-Cafe mailing >> list >> Haskell-Cafe@haskell.org >> http://www.haskell.org/mailman/listinfo/haskell-cafe > > Huh, I'm stumped ;). John: is this possible in enumerator? > > In general though: do you need precisely that type signature? Most of > the time, Enumerators have polymorphic return types. It might be a > problem from http-enumerator requiring (), but I *think* we can work > around that. > > Michael
You're right -- now that I think about it, I don't really care what the type of 'r' is. (Boy, I could have saved several hours today if I had realized that earlier :-) But yeah, I'm having trouble matching the () in http-enumerator's interface. (I wonder if it's because I'm trying to use a fully concrete helper type 'MyRequestBody'): data MyRequestBody = MyRequestBody Int64 (Enumerator ByteString IO ()) myHttp :: URL -> MyRequestBody -> IO () myHttp url (MyRequestBody len bsEnum) = ... where builderEnum = mapEnum BlazeBS.fromByteString bsEnum reqBody = RequestBodyEnum len builderEnum GHC tells me that mapEnum expects bsEnum to be: Enumerator ByteString IO (Step Blaze.Builder IO ()) But it is actually: Enumerator ByteString IO () Wonder if I can use a "forall" in the definition of MyRequestBody to fix things without introducing an externally visible type parameter... Also, thanks for your help so far! _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe