- Haskell-Cafe - Haskell.org

Re: [Haskell-cafe] Folding function composition: L or R?
by Graham Klyne 07 Jul '04

07 Jul '04

At 14:24 07/07/04 +0200, Henning Thielemann wrote: >On Wed, 7 Jul 2004, Graham Klyne wrote: > > > [[ > > sss :: [ShowS] > > sss = map (++) ["abc","def","ghi","jkl"] > > > > ssr = foldr (.) id sss > > ssl = foldl (.) id sss > > > > sr = ssr "" > > sl = ssl "" > > > > -- sl == sr == "abcdefghijkl" > > -- is there any efficiency difference? > > ]] > > > > Conventional wisdom suggests that foldr is better, but I'm not sure if it > > really matters in this case. > >foldl performs worse, e.g. for infinite lists it will need infinite >time :-) Of course! And your answer makes me notice an important touchstone when dealing with such questions: "what happens with an infinite list?". Thanks! #g ------------ Graham Klyne For email: http://www.ninebynine.org/#Contact

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Folding function composition: L or R?
by Graham Klyne 07 Jul '04

07 Jul '04

I have a sequence of functions that I wish to fold together using function composition (.). Does it matter if I use foldl or foldr? e.g. [[ sss :: [ShowS] sss = map (++) ["abc","def","ghi","jkl"] ssr = foldr (.) id sss ssl = foldl (.) id sss sr = ssr "" sl = ssl "" -- sl == sr == "abcdefghijkl" -- is there any efficiency difference? ]] Conventional wisdom suggests that foldr is better, but I'm not sure if it really matters in this case. #g ------------ Graham Klyne For email: http://www.ninebynine.org/#Contact

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RE: [Haskell-cafe] readMVar and the devils
by Simon Marlow 07 Jul '04

07 Jul '04

On 06 July 2004 11:29, Conor T McBride wrote: > OK, here's what I want, what I do with it, and my attempt to deliver > it. But I'm not an expert, so please let me know if there's some > disastrous flaw... > > The signature/spec: > > type Hole x > > hole :: IO (Hole x) > -- returns a fresh empty hole for an x > askHole :: Hole x -> IO (Maybe x) > -- inspects the current contents (or none) of the hole; mustn't > -- block > tellHole :: Hole x -> x -> IO (Maybe x) > -- tries to write the hole, returns what was previously there; > -- if the hole was empty, the value supplied is installed > -- if the hole was full, it's unchanged (too late, pal!); > -- mustn't block > readHole :: Hole x -> IO x > -- blocks until the hole has been filled, then returns its > -- value > instance Eq (Hole x) > -- a kind of higher-level `pointer equality' The following should do the trick (but I wouldn't rule out any lurking race conditions, and perhaps there's an even simpler way to achieve this): data Hole x = Hole (MVar ()) -- wait on this for readHole (MVar (Maybe x)) -- always full, Just x <=> hole contains x instance Eq (Hole x) where (Hole m1 _) == (Hole m2 _) = m1 == m2 hole = do wait <- newEmptyMVar val <- newMVar Nothing return (Hole wait val) askHole (Hole wait val) = readMVar val tellHole (Hole wait val) x = do modifyMVar val $ \mb -> case mb of Nothing -> do putMVar wait (); return (Just x, Nothing) Just y -> return (Just y, Just y) readHole (Hole wait val) = do readMVar wait mb <- readMVar val return (fromJust mb) Cheers, Simon

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problem with hsql on windows
by Shanon Fernald 07 Jul '04

07 Jul '04

I'm using: GHC 6.2.1 HSQL 1.4 on a Windows XP machine. I'm trying to the compile the HSQL example using the eclipse Haskell plug-in, but I get the following error message: Chasing modules from: Main.hs Compiling Queries ( ./Queries.hs, C:\workspace\dbtest\out/Queries.o ) ./Queries.hs:3: Failed to load interface for `Database.HSQL': Bad interface file: C:/ghc/ghc-6.2.1/imports/Database/HSQL/GHC/imports/Database/HSQL.hi mismatched interface file versions: expected 6021, found 6020 What it appears is that my HSQL has an interface matching GHC 6.2 maybe? I just downloaded the HSQL 1.4 from the sourceforge site. The windows exe version is HSQL-1.4.exe. Is this the correct version of HSQL? Am I matching the GHC and HSQL versions correctly? If so, do you know what else I might be doing wrong? I am able to compile other programs with no problem. Thanks for help. It is fun to learn a new language and my Haskell books are on order :)

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Re: Roman Numerals and Haskell Syntax abuse
by George Russell 07 Jul '04

07 Jul '04

Christian Maeder wrote (snipped) > btw, now 127: > roman=(!5);n!a|n<1=""|n>=t=s!!(a+1):(n-t)!a|c>=t=s!!(2*d):c!a|1>0=n!(a-1)where d=a`div`2;s="ivxlcdm";c=10^d+n;t=5^(d+1)*2^(a-d) This is amazing! When I posted my first attempt of 181 characters I thought it might be minimal.

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Random Numbers for the beginner ?
by Crypt Master 06 Jul '04

06 Jul '04

Hi I have tried I swear, even googled for 45 minutes, but I cant seem to get random numbers working. In the documentation is has: rollDice :: IO Int rollDice = getStdRandom (randomR (1,6)) But if I type "getStdRandom (randomR (1,6))" into hugs in the context of module which imports Random, I get get errors. ERROR - Unresolved overloading *** Type : (Random a, Num a) => IO a *** Expression : getStdRandom (randomR (1,6)) Roll dice takes no parameters and returns an IO Int. So in thoery (mine at least ;-) ) running this as an expresion should work. I should get an IO Int back from the interpreter ? So I added RollDice to my module. This doesnt error, but it doesnt return anything except blank spaces: HasGal> rollDice HasGal> Integers or nums should automatically have show correct? So this should show me something ? Ultimatly I want to get randomRs infinite list working so I can build randNums = (take (length popList) [1..]) where the length of pop list is how many random numbers I want. My code works as it, just need to replace [1..] with some random numbers. Thanks, C _________________________________________________________________ Add photos to your e-mail with MSN 8. Get 2 months FREE*. http://join.msn.com/?page=features/featuredemail

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Re: [Haskell-cafe] Random Numbers for the beginner ?
by Jon Cast 06 Jul '04

06 Jul '04

sholderm(a)students.cs.uu.nl wrote: <snip> > Use > > (getStdRandom (randomR (1,6))) :: IO Int This much is fine, but... > instead. > > \begin{code} > module Main where > import Random > > > main :: IO () > main = do n <- (getStdRandom (randomR (1,6))) :: IO Int > ; print n > \end{code} Layout error: the semicolon must be indented further than the `n' is. Furthermore, the semicolon is unnecessary; if it's deleted, yielding > main :: IO () > main = do n <- (getStdRandom (randomR (1,6))) :: IO Int > print n the code works just fine. Jon Cast

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Roman Numerals and Haskell Syntax abuse
by George Russell 06 Jul '04

06 Jul '04

The following declaration for a function for converting positive integers to Roman numerals is 181 characters long. Is there a shorter one? roman=f[(i-j,i)|i<-[2,4..8],j<-[2,1]]where f(d@((a,b):c))n|n<1=""|n>=t!!a=s!!a:f d(n-t!!a)|1>0=if n+t!!b>=t!!a then s!!b:f d(n+t!!b)else f c n;s="mdclxvi";t=[1000,500,100,50,10,5,1]

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RE: [Haskell-cafe] readMVar and the devils
by Simon Marlow 06 Jul '04

06 Jul '04

On 02 July 2004 10:16, Conor T McBride wrote: > Hi folks > > I'm having a play with Concurrent Haskell, and came across this in the > library. > > readMVar :: MVar a -> IO a > This is a combination of takeMVar and putMVar; ie. it takes the > value from the MVar, puts it back, and also returns it. > > I was just wondering if I understand things correctly. Suppose myMVar > is nonempty and I have one thread executing > > do x <- takeMVar myMVar -- (A) > putMVar myMVar x -- (B) > return x > > and another executing > > putMVar myMVar y -- (C) > > Questions > > (1) Is it possible that an evil scheduler will execute (A) then (C) > and block on (B)? Yes. In fact, you don't need an evil scheduler, an ordinary scheduler will do this :-) > (2) Is it the case that if myMVar is nonempty and readMVar is chosen > for execution, that readMVar myMVar takes the value > from myMVar, guaranteed that the _same_ value will be put > back immediately, and also returned? Only if there are no other threads executing putMVar on myMVar concurrently. > (3) If yes to both, what combination of takeMVar and putMVar manages > to maintain the lock in between the two? There isn't one. readMVar is only atomic with respect to other well-behaved threads; that is, other threads doing strictly take-followed-by-put operations. There are lots of similar cases in the world of MVars. Our general approach is that we provide combinators which help you build correct code (eg. withMVar, modifyMVar), but you can always shoot yourself in the foot by using the lower-level operations. > I'm only asking, because I'm trying to cook up some kind of partial > evaluator for programs which are being simultaneously edited by > fiercely argumentative devils. Kind of `demonic laziness', where > the computations decide when _they_ feel like running (eg sometime > after the program turns up), rather than the `angelic laziness' we're > used to. I'd like to use MVars in a write-once-read-many style, so > that once someone fills in a bit of program, it stays put. I'd hate > for another devil to sneak in a replacement program just in the > twinkling of an eye between taking something out to read and putting > it back to be read again. > > I guess I could use some kind of extra semaphore MVar to ensure that > the reader has the lock on the program MVar, but that's more like > hard work. Is readMVar what I want? It sounds like you want a different abstraction, but one that can almost certainly be built using MVars. I'm not sure exactly what it is you need, but if you provide the signatures of the operations then I could probably sketch an implementation. Cheers, Simon

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Re: [Haskell-cafe] readMVar and the devils
by MR K P SCHUPKE 06 Jul '04

06 Jul '04

at a guess the magic take put is: block ( do a <- takeMVar x putMVar x a ) return a Keean.

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