"mapping" an Enumerator

Kannan Goundan

21 Dec 2011 21 Dec '11

5:35 a.m.

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.)

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Michael Snoyman

21 Dec 21 Dec

5:37 a.m.

On Wed, Dec 21, 2011 at 12:35 PM, Kannan Goundan 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

Kannan Goundan

5:59 a.m.

Michael Snoyman wrote:

...

On Wed, Dec 21, 2011 at 12:35 PM, Kannan Goundan 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 :-)

Michael Snoyman

6:26 a.m.

On Wed, Dec 21, 2011 at 12:59 PM, Kannan Goundan wrote:

...

Michael Snoyman wrote:

...
On Wed, Dec 21, 2011 at 12:35 PM, Kannan Goundan 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

Kannan Goundan

7:14 a.m.

Michael Snoyman wrote:

...

On Wed, Dec 21, 2011 at 12:59 PM, Kannan Goundan wrote:

...
Michael Snoyman wrote:

...
On Wed, Dec 21, 2011 at 12:35 PM, Kannan Goundan 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!

Erik de Castro Lopo

7:55 a.m.

Kannan Goundan wrote:

...

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 :-)

I seem to be hitting exactly the same problem. I constrain my types too early and the compiler can't find a solutuon. If I then relax the cionstraints, GCH says "ok, here's something that fits",

...

Wonder if I can use a "forall" in the definition of MyRequestBody to fix things without introducing an externally visible type parameter...

That may indeed help. Erik -- ---------------------------------------------------------------------- Erik de Castro Lopo http://www.mega-nerd.com/

John Millikin

1:35 p.m.

The presence of (Step b m r) is an artifact of Haskell's type system. It can be removed through use of language extensions and 'forall' to give a more aesthetically pleasing signature, but there should be no behavioral difference. On Wed, Dec 21, 2011 at 03:26, Michael Snoyman wrote:

...

On Wed, Dec 21, 2011 at 12:59 PM, Kannan Goundan wrote:

...
Michael Snoyman wrote:

...
On Wed, Dec 21, 2011 at 12:35 PM, Kannan Goundan 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

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Erik de Castro Lopo
John Millikin
Kannan Goundan
Michael Snoyman