- Haskell-Cafe - Haskell.org

award-winning
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10 Aug '03

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Outlook
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09 Aug '03

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idiom for producing comma-seperated lists?
by Antony Courtney 08 Aug '03

08 Aug '03

I often need to format a list of strings using some character as a *seperator* rather than a terminator for the items. Is there some simple combinator or idiom from the Prelude or standard libraries that could be used for this purpose? I ended up defining my own variation on foldr1 to solve this (code below), but I'd be much happier using some more standard solution: -- A variation on foldr1 that takes an extra argument to be returned if -- the list is empty. foldrs :: (a -> a -> a) -> a -> [a] -> a foldrs f z [] = z foldrs f _ xs = foldr1 f xs -- Example: format a list of strings, using a comma as a seperator: mkSepStr :: [String] -> String mkSepStr xs = foldrs (\x s -> x ++ ", " ++ s) "" xs t0 = mkSepStr [] -- ==> "" t1 = mkSepStr ["hello"] -- ==> "hello" t2 = mkSepStr ["10","20","30"] -- ==> "10, 20, 30" What do the rest of you do to solve this particular problem? If there isn't a standard solution, perhaps something like foldrs would be a useful addition to the Prelude. Thanks, -antony -- Antony Courtney Grad. Student, Dept. of Computer Science, Yale University antony(a)apocalypse.org http://www.apocalypse.org/pub/u/antony

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Re: Another typing question
by oleg＠pobox.com 08 Aug '03

08 Aug '03

> Is there any way to parametrize a type by a value, rather than another > type? I believe the following web page answers your question: http://pobox.com/~oleg/ftp/Haskell/number-parameterized-types.html It uses parameterization by decimal numbers -- which seem more natural to read. > It would be interesting to know if/how you could, e.g., define a > concatenation function with a type like Yes, you can -- and the page referenced above does that.

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Re: HI && TypeCast
by Iavor Diatchki 07 Aug '03

07 Aug '03

hello, this is not supposed to be serious :-) Sebastian Sylvan wrote: > ... > List comprehensions are overrated =) > > How about this: > > concatMap ( (':':) . (:[]) . fst ) [('d',3),('f',3)] > or even better: [('d',3),('f',3)] >>= (':':) . return . fst bye iavor -- ================================================== | Iavor S. Diatchki, Ph.D. student | | Department of Computer Science and Engineering | | School of OGI at OHSU | | http://www.cse.ogi.edu/~diatchki | ==================================================

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why are you waiting
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07 Aug '03

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Re: Another typing question
by Jon Cast 06 Aug '03

06 Aug '03

I wrote: > Wolfgang Jeltsch <wolfgang(a)jeltsch.net> wrote: > > On Tuesday, 2003-08-05, 15:22, CEST, Nick Name wrote: > > <snip> > > > > This is called "dependent types" and is not a feature of haskell > > > (nor of any language that I know); there was "cayenne" (try a google > > > search) but I don't believe it is still mantained. > > <snip> > > > > BTW, why is there no general interest for such a thing as dependent > > > types? > > > What negative consequences does their implementation have? > > It depends on whether you just want to write functions to take in > values of such types, or whether you want to write functions to return > values of such types. Doing the former uses dependant products, which > would require a significant extension to GHC's guts, but not (AIUI) > inherently unpleasant. Doing the latter (AIUI) essentially requires > dependant sums. Basically, a family of types A_i parameterized by a > value i :: I is written, for long-honored traditional reasons too > complex to get into here, Sigma i::I.A_i. Now, if you have such > dependant sums, there's a bad interaction with polymorphism: either > you must have a type alpha which cannot appear in a family of types > A_i in a dependant sum, or, for any function f :: alpha -> alpha, > there exists a type beta such that f cannot be aplied at either beta > or Sigma i::Int.beta. Clear? Actually, on second thought, I think this could be simplified to: if we have dependant types, then either there is a type which cannot be used in a dependant type, or for every polymorphic function there is a type it cannot be applied at. Jon Cast

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Re: Another typing question
by Jon Cast 05 Aug '03

05 Aug '03

> On Tuesday, 2003-08-05, 15:22, CEST, Nick Name wrote: <snip> > > This is called "dependent types" and is not a feature of haskell > > (nor of any language that I know); there was "cayenne" (try a google > > search) but I don't believe it is still mantained. <snip> > > BTW, why is there no general interest for such a thing as dependent > > types? > What negative consequences does their implementation have? It depends on whether you just want to write functions to take in values of such types, or whether you want to write functions to return values of such types. Doing the former uses dependant products, which would require a significant extension to GHC's guts, but not (AIUI) inherently unpleasant. Doing the latter (AIUI) essentially requires dependant sums. Basically, a family of types A_i parameterized by a value i :: I is written, for long-honored traditional reasons too complex to get into here, Sigma i::I.A_i. Now, if you have such dependant sums, there's a bad interaction with polymorphism: either you must have a type alpha which cannot appear in a family of types A_i in a dependant sum, or, for any function f :: alpha -> alpha, there exists a type beta such that f cannot be aplied at either beta or Sigma i::Int.beta. Clear? The alternative, btw., is to have both dependant sums and an undecidable type-checking problem. > I think, sometimes they could be quite handy. Sure. <Sarcastic comment censored>. Jon Cast

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Another typing question
by Konrad Hinsen 05 Aug '03

05 Aug '03

Is there any way to parametrize a type by a value, rather than another type? What I would like to do is to define "list of length 3" and "list of length 4" as separate parametrization of the same type, such that I could write functions that accept lists (under my new typename) of any length as long as parameters have the same length. The goal is to remove the need for runtime checks all over the code. Konrad.

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Show with Strings
by Thomas L. Bevan 05 Aug '03

05 Aug '03

-----BEGIN PGP SIGNED MESSAGE----- Hash: SHA1 Is there any reason why show :: String -> String is not equivalent to 'id' ? At the moment, show "one" = "\"one\"" which leads to problems because it often requires String to be treated as a special case, rather than just a member of Show. Tom -----BEGIN PGP SIGNATURE----- Version: GnuPG v1.2.2 (GNU/Linux) iD8DBQE/Li8QYha8TWXIQwoRApCgAJ49iJcOuf88WTIKVJWBkHF/QyugOwCfS2Gu jcJFn48T4uc+ooeSAaCkAhs= =pIel -----END PGP SIGNATURE-----

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