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ghc-mod: cannot satisfy yesod-static
by Denis Shevchenko 23 Jul '15

23 Jul '15

Hi all! I'm trying to use ghc-mod with my Yesod-project, but I get this error: $ ghc-mod ghc-mod: <command line>: cannot satisfy -package-id yesod-static-1.4.0.4-c893310d859222b1b20461f9fdb4d916 (use -v for more information) My project builds correctly. In my .cabal-file I see this dependency: build-depends: base >= 4 && < 5 , yesod >= 1.4.0 && < 1.5 , yesod-core >= 1.4.0 && < 1.5 , yesod-auth >= 1.4.0 && < 1.5 , yesod-static >= 1.4.0 && < 1.5 ... I use LTS-2.18, and I see this package in ~/.stack/snapshots/x86_64-osx/lts-2.18/7.8.4/lib/x86_64-osx-ghc-7.8.4/: ... yesod-static-1.4.0.4 ... So, what can I do? Please help me. Sincerely, Denis.

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ANNOUNCE: fgl-5.5.2.0 and fgl-arbitrary-0.2.0.0
by Ivan Lazar Miljenovic 22 Jul '15

22 Jul '15

I'm pleased to announce that I've (finally!) just released version 5.5.2.0 of [fgl]. Major changes in this release include a test suite, refactorings, code clean-ups and the following potentially breaking changes (in that they were previously unspecified or incorrect): - `nodeRange` and `nodeRangeM` for the various graph data structures erroneously returned `(0,0)` for empty graphs (making them indistinguishable from graphs containing the single node `0`). They now match the default implementation of throwing an error. - The behaviour of `delLEdge` when dealing with multiple edges was unspecified; it now deletes only a single edge and the new function `delAllLEdge` deletes all edges matching the one provided. All changes can be found in the changelog. [fgl]: https://hackage.haskell.org/package/fgl Along with this I'm releasing version 0.2.0.0 of [fgl-arbitrary] (i.e. the "I finally build against the version on Hackage" release). [fgl-arbitrary]: https://hackage.haskell.org/package/fgl-arbitrary -- Ivan Lazar Miljenovic Ivan.Miljenovic(a)gmail.com http://IvanMiljenovic.wordpress.com

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Fwd: Re: Strange "ambiguity" problems thanks to GADTs
by Alexander Eyers-Taylor 22 Jul '15

22 Jul '15

Sorry I forgot to reply to the list Alex -------- Forwarded Message -------- Subject: Re: [Haskell-cafe] Strange "ambiguity" problems thanks to GADTs Date: Wed, 22 Jul 2015 13:24:56 +0100 From: Alexander Eyers-Taylor <aeyerstaylor11(a)gmail.com> To: Andres Loeh <mail(a)andres-loeh.de> Hi In this case what happens in the type checker is that after checking the start of the let we give the variable e1 a type (say abt0a) Then we enter the case. When entering the case, as SFun introduces a constaint (SFun is actually of type (a ~ Maybe j) => Sing a). As we have a constraint we cannot then unify the variables. The reason we can't unify is that we cant tell that (a ~ Maybe j) does not imply (abt0 ~ abt'). In GHC 6.12 the old type inference algotrithm was used. This had some problems as it didn't always infer a priciple type for GADTs. The new inference algorithm OutsideIn does this in a more principled way. In the non-indexed versions when we do the pattern match no constaints are introduced so we can typecheck as usual Alex On 22/07/15 07:59, Andres Loeh wrote: > Hi. > > I've drastically reduced the example. This program triggers the same error: > >> {-# LANGUAGE GADTs, KindSignatures #-} >> >> data Sing :: * -> * where >> SFun :: Sing (Maybe j) >> >> class WellTypedABT (abt :: * -> *) where >> typeOf :: abt a -> Sing a >> >> inferType :: (WellTypedABT abt') => abt a -> abt' a >> inferType e1 = >> let e1' = inferType e1 >> in case typeOf e1' of >> SFun -> e1' > Btw, this compiles fine with ghc-6.12 (as does a version of the > original program if we replace the data kinds involved by equivalent > kind-*-datatypes). > > Cheers, > Andres > > On Wed, Jul 22, 2015 at 1:11 AM, wren romano <winterkoninkje(a)gmail.com> wrote: >> On Wed, Jul 8, 2015 at 9:14 PM, Richard Eisenberg <eir(a)cis.upenn.edu> wrote: >>> I think GADTs are a red herring here -- this is the classic show/read ambiguity dressed up. In the code below, how should GHC determine the type of e1' or e2'? There's no way to know. >> Except that when I erase the indices, everything works just fine: >> >> >> {-# LANGUAGE KindSignatures, GADTs #-} >> >> data Type = Fun Type Type >> >> data AST :: * -> * where >> App :: ast -> ast -> AST ast >> >> class ABT (abt :: *) where >> view :: abt -> AST abt >> unview :: AST abt -> abt >> >> class WellTypedABT (abt :: *) where >> unviewWT :: Type -> AST abt -> abt >> typeOf :: abt -> Type >> >> data TypeCheckMonad a = TCM { unTCM :: a } >> >> instance Functor TypeCheckMonad where >> fmap f = TCM . f . unTCM >> >> instance Applicative TypeCheckMonad where >> pure = TCM >> TCM f <*> TCM x = TCM (f x) >> >> instance Monad TypeCheckMonad where >> return = pure >> TCM x >>= k = k x >> >> inferType >> :: (ABT abt, WellTypedABT abt') >> => abt >> -> TypeCheckMonad abt' >> inferType e0 = >> case view e0 of >> App e1 e2 -> do >> e1' <- inferType e1 >> case typeOf e1' of >> Fun typ2 typ3 -> do >> e2' <- inferType e2 >> return . unviewWT typ3 $ App e1' e2' >> >> >> -- >> Live well, >> ~wren >> _______________________________________________ >> Haskell-Cafe mailing list >> Haskell-Cafe(a)haskell.org >> http://mail.haskell.org/cgi-bin/mailman/listinfo/haskell-cafe > _______________________________________________ > Haskell-Cafe mailing list > Haskell-Cafe(a)haskell.org > http://mail.haskell.org/cgi-bin/mailman/listinfo/haskell-cafe

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Why there is no instance Generic Word?
by Alexey Vagarenko 21 Jul '15

21 Jul '15

There is instance Generic Int, but not Generic Word. Also no Generic instances for Int8, Int16, Int32, Int64, Word8, Word16, Word32, Word64.

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Message passing library
by Corentin Dupont 21 Jul '15

21 Jul '15

Hello, what would you recommend as a message passing library? Essentially, I would to replace this library http://hackage.haskell.org/package/eprocess by something more mainstream. It is used in Hint-server: http://hackage.haskell.org/package/hint-server Thanks! Corentin

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Several Experienced Developers for Advanced Functional Programming - Ocaml, Haskell, Scala - compiler technology, LLVM, C, etcetera, and Martin-Löf type theory, static analysis, our main target is embedded software
by Johan Glimming 21 Jul '15

21 Jul '15

Hi folks, There are openings at Functor, Sweden, including also remote work, except for Scala consultants who must immediately relocate: Do you want to be part of the journey as the Functor startup research spin-off innovates the future of software engineering? Do you want to make a difference? Do you want to devote your talents where it can have incredible impact on the entire software industry? Functor, Sweden, is again looking for brilliant minds, essentially co-founders as a generous equity program will after a certain threshold time make sure developers, building the company indeed, also own what they are building. Please apply on LinkedIn, closing soon take action if interested: https://www.linkedin.com/jobs2/consumer/overview/60296037 We use Haskell, OCaml, LLVM and our tools are closely tied to the C programming language and indeed to functional programming. 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It’s Martin-Löf / dependent type theory inside our tools. Best, Johan — Dr Johan Glimming Chief Executive Officer at Functor AB, Chairman at Functor Group AB Request a meeting or schedule a call: www.doodle.com/glimming<http://www.doodle.com/glimming> Mobile: +46-76-7646000<tel:+46-76-7646000> Direct: +46-8-55005505<tel://+46-8-55005505> Main: +46-8-55005500<tel:+46-8-55005500> Assistant: +46-8-55005501<tel://+46-8-55005501> Skype: glimming<skype://glimming> Twitter: @glimming<https://www.twitter.com/glimming> LinkedIn: se.linkedin.com/in/glimming<http://se.linkedin.com/in/glimming> Functor AB, Box 7070, 164 07 Kista, SWEDEN<x-apple-data-detectors://3/1> Web: www.functorgroup.com<http://www.functorgroup.com/> Twitter: @functors<https://www.twitter.com/functors/> UK tel: +44-1223-911400<tel:+44-1223-911400> US tel: +1-(415)-513-0090 — ANY INFORMATION DISCLOSED DIRECTLY OR INDIRECTLY IN, OR IMPLIED BY, THIS EMAIL, INCLUDING ATTACHMENTS, EXTERNAL REFERENCES OR APPENDICES, IS STRICTLY CONFIDENTIAL AND/OR PROPRIETARY AND IS INTENDED ONLY FOR USE BY FUNCTOR GROUP AB, SWEDEN, REG. 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Where else is EclipseFP available since SourceForge seems to be partially down?
by KC 20 Jul '15

20 Jul '15

Where else is EclipseFP available since SourceForge seems to be partially down? -- -- Sent from an expensive device which will be obsolete in a few months! :D Casey

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Add IsNumeric (similar to IsString and IsList)
by Michael Baikov 20 Jul '15

20 Jul '15

Hi Cafe! Now that we have OverloadedList and OverloadedStrings extensions it might make sense to add something similar for numerals - right now if you want to be able to specify some items as numbers you have to implement Num instance, unfortunately not everything that can be represented as number can have sensible operations required by Num: negation makes no sense for set of natural numbers for example, multiplication for some applications of currency values which results in Num instance full of (*) = error "No multipication for Foos", (+) = error "No addition for Foos". I don't have any statistics from hackage or github, but in codebase I'm working with there are 3 such instances. Proposal: add OverloadedNumerals language pragma, IsNumeric typeclass and some methods to make that class useful with behavior similar to Lists and Strings.

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Re: [Haskell-cafe] The evil GADTs extension in ghci 7.8.4 (maybe in other versions too?)
by Edward Z. Yang 20 Jul '15

20 Jul '15

GHC used to always generalize let-bindings, but our experience with GADTs lead us to decide that let should not be generalized with GADTs. So, it's not like we /wanted/ MonoLocalBinds, but that having them makes the GADT machinery simpler. This blog post gives more details on the matter: https://ghc.haskell.org/trac/ghc/blog/LetGeneralisationInGhc7 Edward Excerpts from Ki Yung Ahn's message of 2015-06-04 20:37:27 -0700: > Such order dependent could be very confusing for the users. I thought I > turned off certain feature but some other extension turning it on is > strange. Wouldn't it be better to decouple GADT and MonoLocalBinds? > > 2015년 06월 04일 20:31에 Edward Z. Yang 이(가) 쓴 글: > > This is because -XGADTs implies -XMonoLocalBinds. > > > > Edward > > > > Excerpts from Ki Yung Ahn's message of 2015-06-04 20:29:50 -0700: > >> \y -> let x = (\z -> y) in x x > >> > >> is a perfectly fine there whose type is a -> a. > >> (1) With no options, ghci infers its type correctly. > >> (2) However, with -XGADTs, type check fails and raises occurs check. > >> (3) We can remedy this by supplying some additional options > >> (4) Howver, if you put -XGADTs option at the end, it fails again :( > >> > >> > >> kyagrd@kyahp:~$ ghci > >> GHCi, version 7.8.4: http://www.haskell.org/ghc/ :? for help > >> Loading package ghc-prim ... linking ... done. > >> Loading package integer-gmp ... linking ... done. > >> Loading package base ... linking ... done. > >> Prelude> :t \y -> let x = (\z -> y) in x x > >> \y -> let x = (\z -> y) in x x :: t -> t > >> Prelude> :q > >> Leaving GHCi. > >> > >> > >> kyagrd@kyahp:~$ ghci -XGADTs > >> GHCi, version 7.8.4: http://www.haskell.org/ghc/ :? for help > >> Loading package ghc-prim ... linking ... done. > >> Loading package integer-gmp ... linking ... done. > >> Loading package base ... linking ... done. > >> Prelude> :t \y -> let x = (\z -> y) in x x > >> > >> <interactive>:1:30: > >> Occurs check: cannot construct the infinite type: t0 ~ t0 -> t > >> Relevant bindings include > >> x :: t0 -> t (bound at <interactive>:1:11) > >> y :: t (bound at <interactive>:1:2) > >> In the first argument of ‘x’, namely ‘x’ > >> In the expression: x x > >> Prelude> :q > >> Leaving GHCi. > >> > >> > >> ~$ ghci -XGADTs -XNoMonoLocalBinds -XNoMonomorphismRestriction > >> GHCi, version 7.8.4: http://www.haskell.org/ghc/ :? for help > >> Loading package ghc-prim ... linking ... done. > >> Loading package integer-gmp ... linking ... done. > >> Loading package base ... linking ... done. > >> Prelude> :t \y -> let x = (\z -> y) in x x > >> \y -> let x = (\z -> y) in x x :: t -> t > >> Prelude> :q > >> Leaving GHCi. > >> > >> > >> ~$ ghci -XNoMonoLocalBinds -XNoMonomorphismRestriction -XGADTs > >> GHCi, version 7.8.4: http://www.haskell.org/ghc/ :? for help > >> Loading package ghc-prim ... linking ... done. > >> Loading package integer-gmp ... linking ... done. > >> Loading package base ... linking ... done. > >> Prelude> :t \y -> let x = (\z -> y) in x x > >> > >> <interactive>:1:30: > >> Occurs check: cannot construct the infinite type: t0 ~ t0 -> t > >> Relevant bindings include > >> x :: t0 -> t (bound at <interactive>:1:11) > >> y :: t (bound at <interactive>:1:2) > >> In the first argument of ‘x’, namely ‘x’ > >> > > _______________________________________________ > > Glasgow-haskell-users mailing list > > Glasgow-haskell-users(a)haskell.org > > http://mail.haskell.org/cgi-bin/mailman/listinfo/glasgow-haskell-users > > >

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Re: [Haskell-cafe] Type signature for function where function parameter is used twice?
by Ivan Lazar Miljenovic 19 Jul '15

19 Jul '15

(Re-CC'ing Cafe) I think what's happening here is that when you do "apply f1 f1", the two c's and d's don't actually match with each other. But when you do it with applyBoth, they *have* to match each other. As a comparison, compare: :t (read, read) (read, read) :: (Read a, Read a1) => (String -> a, String -> a1) and :t Control.Monad.join read Control.Monad.join (,) read :: Read a => (String -> a, String -> a) I'm not sure how you could define such a function to do what you're requesting; my attempt is this: applyBoth :: (forall a b a2 b2. arr a b -> arr a2 b2) -> D arr a b -> D arr a2 b2 applyBoth f = apply f f but whilst ghci accepts it, I couldn't actually seem to use it (lots of errors about being unable to match types). On 19 July 2015 at 17:15, Clinton Mead <clintonmead(a)gmail.com> wrote: > Thanks for your reply Ivan > > Here's a full code example ( http://ideone.com/1K3Yhw ): > > --- > > data D c a b = D (c a b) (c b a) > > apply :: (c a b -> c a2 b2) -> (c b a -> c b2 a2) -> D c a b -> D c a2 b2 > apply f1 f2 (D x y) = D (f1 x) (f2 y) > > applyBoth f = apply f f > > f :: Int -> String > f = show > > g :: String -> Int > g = read > > h :: Int -> Int > h = (*2) > > join :: (a -> b) -> (c -> d) -> ((a, c) -> (b, d)) > join f g (x, y) = (f x, g y) > > x1 = D f g > > f1 = join h > > y1 = apply f1 f1 x1 -- This compiles > y2 = apply g g x1 where g = f1 -- Also works > > z1 = applyBoth f1 x1 -- This fails > > main = return () > > --- > > You see, with "y1" and "y2", a different "f1" instantiation (is that the > right word) of "f1" is chosen. Even when I let "g = f1", each "g" passed to > apply is different, the first operates on "Int -> String", the second, > "String -> Int". > > On the face of it, if: > > f x = g x x > > Then well, you should be able to replace "g x x" with "f x" wherever you see > it. > > But it seems in this case this doesn't work. Surely Haskell doesn't require > me to "repeat myself", and there's a way to write "applyBoth", instead of > having to write "apply f f" (note the repetition of f) all the time? > > I assume there's some type signature I can give "applyBoth f" so it can be > used just like "apply f f" but I can't work out what it is. > > On Sun, Jul 19, 2015 at 10:23 AM, Ivan Lazar Miljenovic > <ivan.miljenovic(a)gmail.com> wrote: >> >> On 19 July 2015 at 10:13, Clinton Mead <clintonmead(a)gmail.com> wrote: >> > Lets say I've got the following data type: >> > >> > data D c a b = D (c a b) (c b a) >> > >> > And I define a function to manipulate it: >> > >> > apply :: (c a b -> c a2 b2) -> (c b a -> c b2 a2) -> D c a b -> D c a2 >> > b2 >> > apply f1 f2 (D x y) = D (f1 x) (f2 y) >> > >> > This is all fine. But I want a shorter function if (f1 = f2). So I >> > write: >> > >> > applyBoth f = apply f f >> > >> > I originally thought that if "apply f f" is valid, then logically >> > "applyBoth >> > f" should also be valid. But it seems that type inference results in >> > applyBoth only working for functions "c a a -> c a2 a2". >> >> applyBoth f = apply f f >> >> In apply, (f1 :: c a b -> c a2 b2) and (f2 :: c b a -> c b2 a2). >> >> So for "apply f f" to typecheck, we must have that a ~ b and a2 ~ b2 >> (as the above two type signatures must match since f1 = f2). >> >> So we must have that (f :: c a a -> c a2 a2). >> >> > >> > Is there a way to type "applyBoth" so it works for all functions that >> > would >> > work simply by repeating them twice in "apply"? >> >> I'm not sure what you mean; what else would make sense? Can you >> provide an example? >> >> -- >> Ivan Lazar Miljenovic >> Ivan.Miljenovic(a)gmail.com >> http://IvanMiljenovic.wordpress.com > > -- Ivan Lazar Miljenovic Ivan.Miljenovic(a)gmail.com http://IvanMiljenovic.wordpress.com

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