- Beginners - Haskell.org

Forcing evalation in the IO Monad
by Philip Scott 16 Apr '10

16 Apr '10

Hi folks, This question is so trivial it must have been answered before, but I have been googling all morning to no avail so please bear with me. I've been wading through mires of $! and seq, deepSeq, $!! etc.. and I have leaned a lot of useful stuff about lazy evaluation - but I haven't been able to solve my problem.. What I would like to do is to force an expression to be fully evaluated to normal form, from inside the IO monad, right there and then. For example, in GHCI, calling putStrLn does just what I want: Debug.Trace Prelude> let a = trace "Hello" 42 Debug.Trace Prelude> putStrLn $ show a Hello 42 Except that it only takes showable things, and it puts an annoying message on the screen. What I want is something like putStrLn that doesn't print anything, and takes any type as its input, evaluates it to normal form and then does nothing else. It might be called something like 'evaluate' have a type like this evaluate :: a -> IO() It _must_ exist - and it _must_ be easier to find this sort of thing than I am making it, or else how did all you haskell gurus attain your status? Am I missing something :) All the best, Philip

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Re: calling polymorphic function in Selenium question
by MH 16 Apr '10

16 Apr '10

I tried that and I got the following error: A pattern type signature cannot bind scoped type variables `a' unless the pattern has a rigid type context In the pattern: x :: Selenium a -> IO (Either String a) In a stmt of a 'do' expression: (x :: Selenium a -> IO (Either String a)) <- start " https://www.google.com" In the expression: do { (x :: Selenium a -> IO (Either String a)) <- start " https://www.google.com"; Then I tried to add "forall a" but the Selenium library doesn't use forall so there was a conflict. Is there a way to work around the rigid type error you know of? Thanks On Apr 15, 3:14 pm, Brent Yorgey <byorgey(a)seas.upenn.edu> wrote: > On Sun, Apr 04, 2010 at 09:25:16PM -0400, MH wrote: > > I am running the following code that is using Selenium. If you look at > > the function "start", you will see that the return type of the > > function is polymorphic. Now in main function, I call start function > > to get selenium IO monad and sequentially call selenium commands > > (open, doCommand etc...). The problem that I have here is, while I can > > call all Selenium commands with signature (String -> Selenium > > String), I can't call commands with signature (String -> Selenium > > Bool). As I understand it, even though "start" function shall return > > IO (Selenium a -> IO (Either String a)), it actually return IO > > (Selenium String -> IO (Either String String)). > > Hmm, I must confess that I don't understand exactly what is going on > here. One thing to try is to give an explicit type annotation at > start's call site, something like > > > main = do > > (selenium :: Selenium a -> IO (Either String a)) <- start " http://www.google.com" > > selenium $ open "/" > > selenium $ doCommand SWindowMaximize [] > > selenium $ typeText (Name "q") "stuff" > > selenium $ clickAndWait (Name "btnG") > > return selenium > > Does that help? > > -Brent > _______________________________________________ > Beginners mailing list > Beginners@haskell.orghttp://www.haskell.org/mailman/listinfo/beginners

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Re: [Haskell-beginners] Compiling C into Haskell
by Hein Hundal 16 Apr '10

16 Apr '10

Thanks Allan! I was hoping the C-Haskell mix would work. I am glad to hear that you have good things to say about it. My main reason for thinking about the C-to-Haskell compiler was to address the question "Say you had a C program. Can you always convert it to Haskell in such a way that the compiled Haskell is not too slow and does not need too much memory?" Supposing that too slow means slower than 1/4 the speed of C and too much memory means twice the memory of C. Do you know the answer to this question? Cheers, Hein

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Trying to find an alternative StateMonad
by edgar klerks 14 Apr '10

14 Apr '10

Hi All, I wanted to write the statemonad in a different way, because I want to track mutations to the state. And also to be able to undo an done or redo an undone mutation. Therefore I wrote a small dsl with an evaluator, which finds out the current state. The problem is, I don't know how to create a get and put function for my monad. I have a function which evaluate the state and gives back the current State. Can someone help me a bit along? I have the feeling I do something wrong, but I am not sure what. With kind regards, Edgar {-# LANGUAGE GADTs,DeriveFunctor #-} import Control.Monad import Data.Maybe data StateCmd a where Put :: a -> StateCmd a Undo :: StateCmd a Redo :: StateCmd a deriving (Show, Functor) newtype State s a = State { unState :: ([StateCmd s],a) } deriving Show joinState :: State s a -> State s b -> State s b joinState (State (xs, a)) (State (ys, b)) = State (xs ++ ys, b) instance Monad (State s) where return a = State $ ([], a) (>>=) = bindState -- m a -> ( a -> m b) -> m b bindState st@(State (_,a)) f = st `joinState` st' where st' = f a unPut :: StateCmd a -> Maybe a unPut (Put a) = Just a unPut _ = Nothing test = State $ ([Put 4, Undo, Redo, Undo, Put 5, Undo, Redo, Put 6, Undo, Redo, Undo], ()) getCurrent = fromJust.unPut.head.snd.(foldr current' ([],[])).reverse.fst.unState where current' x (ul,cl) = case x of Put n -> (ul, Put n : cl) Undo -> (head cl : ul, tail cl) Redo -> (tail ul, head ul : cl)

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Re: [Haskell-beginners] Compiling C into Haskell
by Hein Hundal 13 Apr '10

13 Apr '10

Heinrich wrote: >What are the complex operations you want to do on matrices? There >is the hmatrix that binds to LAPACK and is a pleasure to use, >but it's mainly about matrix multiplication and other numerical >matrix operations. The main operations I need are Matrix Inversion, Transpose, Singular Value Decomposition, and QR Factoring. LAPACK takes care of all of these. I will have to look into hmatrix. Thanks! Cheers, Hein

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Compiling C into Haskell
by Hein Hundal 13 Apr '10

13 Apr '10

Hi, I have been playing around with Haskell for about a month now and reading the nice book "Real World Haskell." My main reason for learning Haskell is that I want to code up some machine learning projects (heavy use of matricies). Normally, I use Mathematica and mix it with C++, but Mathematica is proprietary, slow, and can't produce executable while C++ is verbose. Learning Haskell has been fun; however, I have been a little worried that I will sacrifice too much performance when coding in Haskell. So, I recently asked one of my friends the following question, "Say you had a C program. Can you always convert it to Haskell in such a way that the compiled Haskell is not too slow and does not need too much memory?" Supposing that too slow means slower than 1/4 the speed of C and too much memory means twice the memory of C. Hopefully, someone that knows Haskell well can comment on this question. I am not sure, but I think the answer is yes, such a conversion can always be done and creating a C to Haskell compiler with the above performance constraints is not extremely hard. I started thinking about how a compiler might convert a simple C program into Haskell. Below I will paste a C program and the compile-by-hand Haskell code. It seems to me that the ideas I used to create the Haskell code can be implemented in a compiler that converts a simple subset of C into Haskell. I was thinking about restricting the C to one data type 32bit-integers, arithmetic (+-*/%), assignment (=), comparison (<,>,==,<=,>=), the if-condition-codeblock construct, and the while-condition-codeblock construct. (I would also like to do integer arrays, but I have not read about mutable arrays and monads yet.) Any comments? Cheers, Hein H. --------------------C to Haskell------------------------ module Main where -- Convert a C-Program line by line into Haskell -- -- 1-- #include <stdio.h> -- 2-- void main() -- 3-- { -- 4-- long i,j; -- 5-- i=7; -- 6-- j=0; -- 7-- while(j<10000) -- 8-- { -- 9-- if ((i % 17)== 11) --10-- i = i*2; --11-- if ((i % 35)== 12) --12-- i = i+13; --13-- if (i> 1000) --14-- i = i - 1000; --15-- i++; --16-- j++; --17-- } --18-- printf("%ld", i); --19--} --assignment statements line5 (i,j) = (7,j) line6 (i,j) = (i,0) line10 (i,j) = (i*2,j) line12 (i,j) = (i+13,j) line14 (i,j) = (i-1000,j) line15 (i,j) = (i+1,j) line16 (i,j) = (i,j+1) --while statement line7test (i,j) = j >=100000000 line7 (i,j) = until line7test body9To16 (i,j) --if statements line9 (i,j) = if ((mod i 17) == 11) then line10 (i,j) else (i,j) line11 (i,j) = if ((mod i 35) == 12) then line12 (i,j) else (i,j) line13 (i,j) = if (i>1000) then line14 (i,j) else (i,j) -- code blocks mainprogram :: (Int, Int) -> (Int, Int) mainprogram = line7 . line6 . line5 body9To16 = line16 . line15 . line13 . line11 . line9 main :: IO () main = putStrLn(show(mainprogram (0,0))) -----------------end of program-------------------------

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Re: Space leak in a fold even after the usual fixes
by Travis Erdman 13 Apr '10

13 Apr '10

> I take it that size of the array does not depend on arg ? > Apparently, despite your attempts to force evaluation, the array still > contains many unevaluated expressions. Documentation for the vector > pacakge reveals that Data.Vector is a *boxed* array, i.e. the entries > are only evaluated on demand, so that's where the unevaluated > expressions linger around. > Unless you need the extra laziness - which you probably don't, given how > you've structured your algorithm - you might want to switch to > Data.Vector.Unboxed or Data.Vector.Storable. Correct. The array size doesn't depend on arg (arg only controls how much of the input list to process. I did not realize that the boxed arrays would have this extra-lazy behavior. The reason I chose boxed is because I needed to store these very large trees (custom data type) that I referenced in an earlier question. Since D.V.Unboxed and Storable don't appear to support non-standard data types, is there another package that might better fit my needs? Data.IntMap perhaps? I originally chose an array over a map because I need fast random access to individual elements (and cheap modification). Thanks again, Travis Erdman

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Space leak in a fold even after the usual fixes
by Travis Erdman 12 Apr '10

12 Apr '10

The driver of my algorithm looks like this: foldl' processfn nullarray (take arg infinitelist) where processfn takes an array and the next set of inputs, processes, and delivers a new updated array (using the Data.Vector library). Apparently, I have a space leak ... the "maximum residency" varies linearly with the size of "arg" supplied, garbage collection consumes ~75% of CPU time, and, if the arg is too big, the whole thing crashes with an out of memory error. The algorithm should operate in constant space. As you can see, I'm using a strict left-fold and also have made the accumulating array strict in the processfn definition with a bang pattern. So, I'm sort of at a loss as to how to resolve this. The help provided on this list has been outstanding, thanks to all of you; hope you have something left in the tank for this one!

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hmatrix, and Float
by MAN 12 Apr '10

12 Apr '10

I noticed that hmatrix defines the auxiliary type 'Element', and provides instances for Double and Complex, but not Float. Then Vector and Matrix are defined in terms of Element... is there any reason why Float is intentionally left out. I'd like to use Numeric.LinearAlgebra with both single and double precision so as to compare results and times... could I?

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When to use ByteString rather than [Char] ... ?
by James Fisher 11 Apr '10

11 Apr '10

Hi, After working through a few Haskell tutorials, I've come across numerous recommendations to use the Data.ByteString library rather than standard [Char], for reasons of "performance". I'm having trouble swallowing this -- presumably the standard String is default for good reasons. Nothing has answered this question: in what case is it better to use [Char]? Could anyone point me to a good resource showing the differences between how [Char] and ByteString are implemented, and giving good a heuristic for me to decide which is better in any one case? Best, James Fisher

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