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Suggestions for creating a "bind" function that can be applied to different things?
by Costello, Roger L. 01 Aug '11

01 Aug '11

Hi Folks, When I say: "Let x be of datatype Int." I think the proper terminology is "bind". That is, I bind the name "x" to the Int datatype. Is that the correct terminology? Assuming it is, I would like to bind a name to many different things. I seek your suggestions on how to do this. First let dt1, dt2, dt3, dt4 be "DataTypes": dt1, dt2, dt3, dt4 :: DataType And let "Name" be synonymous with String: type Name = String And let title, author, date, cost be "Elements": title, author, date, cost, book :: Element I would like a "bind" function that binds a Name to a DataType to create an Element: bind :: Name -> DataType -> Element Here are a few examples of using bind to create Elements: title = bind "Title" dt4 author = bind "Author" dt4 date = bind "Date" dt4 I would also like to bind a Name to a DataType to create an "Attribute": bind :: Name -> DataType -> Attribute Here's an example of using bind to create an Attribute: currency = bind "Currency" dt3 And I'd like to bind an Attribute to an Element: bind :: Attribute -> Element -> Element Here's an example of binding the currency Attribute to a cost Element: cost = bind currency (bind "Cost" dt1) Finally, I'd like to create an Element by binding a Name to a list of Elements: bind :: Name -> [Element] -> Element Here's an example of this: book = bind "Book" [title, author, date, cost] As you see, I'd like to "bind" to lots of different things. But I can't create multiple functions with the same name, "bind". And I'd rather not create a function for each thing that I want to bind to, e.g., elementBind, attributeBind, attributeToElementBind, multipleElementsBind. There must be an elegant solution. What is your suggestion? /Roger

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A question on types
by C K Kashyap 01 Aug '11

01 Aug '11

Hi, To clarify my understanding, I created this table | | weak | strong | |---------+------+---------| | dynamic | perl | Ruby | | static | C | Haskell | |---------+------+---------| Could someone please ratify this? Regards, Kashyap

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Problem with catching error
by Alexey G 01 Aug '11

01 Aug '11

Hello. I have some problem with catching error. My function try to read value from string and if it's fails - return default (d). With this: >import Control.Exception >readIOWith :: Read a => a -> String -> IO a >readIOWith d x = catch tryRead errHandler > where errHandler :: SomeException -> IO a > errHandler _ = return d > tryRead = readIO x I have error: Could not deduce (a ~ a1) from the context (Read a) bound by the type signature for readIOWith :: Read a => a -> String -> IO a at BWClub/Common/Helpers/Packets.hs:(19,1)-(22,33) `a' is a rigid type variable bound by the type signature for readIOWith :: Read a => a -> String -> IO a at BWClub/Common/Helpers/Packets.hs:19:1 `a1' is a rigid type variable bound by the type signature for errHandler :: SomeException -> IO a1 at BWClub/Common/Helpers/Packets.hs:21:11 In the first argument of `return', namely `d' In the expression: return d In an equation for `errHandler': errHandler _ = return d But this code works: >readIOWith d x = catch tryRead (\e -> print (e :: SomeException) >> return d) > where tryRead = readIO x Sorry for my english.

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Haskell-style types in C or C++
by Christopher Howard 31 Jul '11

31 Jul '11

One of the things that I love about Haskell is the syntax for creating user defined types. E.g.: Data QueryResult = NoResult | DatabaseError String | Results [String] I can prototype an entire program around these self-made types and it is a lot of fun. Out of intellect curiosity, though: what would be the equivalent construction in C or C++? (Say, for the type defined above). -- frigidcode.com theologia.indicium.us

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No instance for (Show a)
by Ovidiu Deac 31 Jul '11

31 Jul '11

For some reason ghc complains about not being able to call show on an Integer (?!?!?) Please enlighten me! ovidiu This is the hspec: it "shows one element" ( show (push 1 EmptyStack) ≡ "Stack(1)") ...this is the code: data Stack a = EmptyStack | StackEntry a (Stack a) deriving(Eq) instance Show (Stack a) where show s = "Stack(" ⊕ (showImpl s) ⊕ ")" where showImpl EmptyStack = "" showImpl (StackEntry x _) = show x ...and this is the error: src/Stack.hs:12:22: No instance for (Show a) arising from a use of `showImpl' In the first argument of `(++)', namely `(showImpl s)' In the second argument of `(++)', namely `(showImpl s) ++ ")"' In the expression: "Stack(" ++ (showImpl s) ++ ")"

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Ambiguous type variable
by Ovidiu Deac 30 Jul '11

30 Jul '11

I'm playing with Haskell so I wrote a stack module (see the code below). I have a problem with the pop function which returns a tuple (Nothing, EmptyStack) if called with an EmptyStack. I kind of understand that the compiler cannot cannot figure out what type to use for a. But how could I tell the compiler that if the list is empty I don't care about that type? Thanks, Ovidiu ///////////////// This is the hspec ... it "pop empty stack gives Nothing" ( (pop EmptyStack) ≡ (Nothing, EmptyStack)) ... This is the code: module Stack where import Prelude data Stack a = EmptyStack | StackEntry a (Stack a) deriving(Show, Eq) ... pop :: Stack a → (Maybe a, Stack a) pop EmptyStack = (Nothing, EmptyStack) pop (StackEntry a s) = ((Just a), s) ...and this is the error I get: test/TestStack.hs:20:28: Ambiguous type variable `a0' in the constraint: (Eq a0) arising from a use of `==' Probable fix: add a type signature that fixes these type variable(s) In the second argument of `it', namely `((pop EmptyStack) == (Nothing, EmptyStack))' In the expression: it "pop empty stack gives Nothing" ((pop EmptyStack) == (Nothing, EmptyStack)) In the second argument of `describe', namely `[it "empty stack is empty" (isEmpty EmptyStack), it "non-empty stack is not empty" (not (isEmpty (push 10 EmptyStack))), it "push then pop retrieves the same value" ((pop $ push 10 EmptyStack) == (Just 10, EmptyStack)), it "push push then pop retrieves the last value" ((pop $ push 2 (push 1 EmptyStack)) == (Just 2, (push 1 EmptyStack))), ....]' make: *** [test] Error 1

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Re: [Haskell-beginners] minimal Haskell concepts subset
by Davi Santos 30 Jul '11

30 Jul '11

Hi Matt, the first and best I found was http://learnyouahaskell.com/chapters. The point is that I spent too much time (almost a year) learning, and I am not -productive- yet. Productive in the sense of time writing and debugging, not in the sense of number of characters written. Nothing in particular, it just was too much time. My background is, in order of experience: Java, C++, Delphi, Python, C#, Bash, Scala, Basic, Pascal and maybe others I forgot. The first two I really know and could learn the best practices. The rest I know only the minimal concepts but could do something important already. Haskell has been harder than Basic, my first programming language. When I was 13-year old, somebody told me: "each line is an order the computer follows in sequence". That was enough intuitive to me. After that I was taught: for, goto, if, print and input. The last concepts and harder were the while and the subroutines. Maybe I am get messy with the lots of alternatives to do a same thing in Haskell. After months in Java I started the habit of always search for the best practices - and I found them. Ok, it is long enough email. Maybe I have some kind of non-haskell-related addiction. Davi

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Building Haskell Platform with shared libraries?
by Mister Globules 30 Jul '11

30 Jul '11

Hi, I'm installing the Haskell Platform (2011.2.0.1) from source on Kubuntu. If I follow the recommended steps (configure, make, make install) it doesn't build any shared libraries. I've tried a few hacks (e.g. export EXTRA_CONFIGURE_OPTS=--enable-shared), but they result in build errors. Currently, I do a "cabal --reinstall install" for each of the packages in the platform, but I was wondering if there's a more "official" way to do it when building the platform itself. - Globules

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question on types
by Jake Penton 29 Jul '11

29 Jul '11

Yikes. I have been doing a fair bit of productive programming in Haskell, thinking that I am making a bit of progress. Then I hit something that is apparently *really simple* that I do not understand at all. How discouraging. Here is the code that makes me realize I don't understand types or type inference very well at all yet: f :: a f = 1 When I try to load the above, ghci gives me: No instance for (Num a) arising from the literal `2' In the expression: 2 In an equation for `f': f = 2 Failed, modules loaded: none. Ok, so then I try: g:: (Num a) => a g = 2 This compiles. Why? I mean, why is the first example (defining f) wrong, but the second example (defining g) ok? A slight variation on this is: h:: a h = 'a' to which ghci replies: Couldn't match type `a' with `Char' `a' is a rigid type variable bound by the type signature for c :: a at /Users/David/Project/EoP/ch04/weak.hs:114:1 In the expression: 'a' In an equation for `c': c = 'a' This last example is probably the most basic one which I need to understand. But, why is the problem apparently a different one than in the definition of "f" above? Of course, I cannot think of a reason to actually define things as shown above under ordinary circumstances. The code above is just boiled down to the simplest case I could find to illustrate my confusion. I guess I interpret "f::a" to mean "f is some (any) type a". So why can't it be whatever "1" is, which I suppose is Integer. What is the type system looking for? And why does the constraint (Num a) make things ok? Please point me in the direction of any reading I should do to clear up my confusion. TIA. - Jake -

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Re: [Haskell-beginners] lazyness and tail recursion
by Ovidiu Deac 29 Jul '11

29 Jul '11

Now I think I understand. Thanks for the explanation! ovidiu On Fri, Jul 29, 2011 at 12:04 AM, Will Ness <will_n48(a)yahoo.com> wrote: > ... the start of the function, so to speak. > > If we have (f x = a : f b), the distance between this entry into f, and the > next one, has constant length: (a :). If we have (f x = "123" ++ f y), it's > constant too, just bigger than 1 - no problem, it will all get consumed in > constant number of steps, there's no explosion in number of operations. > > But if we have (f x = f a ++ f b) e.g., then (f a) will add some, and its > calls will get some, and it all grows out of control - and all this needs to > be maintained by the system, to get to the final (f b). Not nice. > > But anyway I'm no expert; it's just how I explain it to myself, from > Prolog's "tail recursion modulo cons" perspective. We don't need to wait for > a function to return to prepend something onto its result; we prepent this > to a yet-to-be-filled list, and pass that stump into the tail-recursive > function. Presto, all it fine. Same in Haskell, no second function call will > get actually called until what's before it gets consumed - so there's no > proble if that's constant in size. But I repeat myself. :) > > Cheers, > Will > > > --- On *Thu, 7/28/11, Ovidiu Deac <ovidiudeac(a)gmail.com>* wrote: > > > From: Ovidiu Deac <ovidiudeac(a)gmail.com> > Subject: Re: [Haskell-beginners] lazyness and tail recursion > To: "Will Ness" <will_n48(a)yahoo.com> > Cc: beginners(a)haskell.org > Date: Thursday, July 28, 2011, 10:27 AM > > > If I got it right, the idea is to keep the number of operations > between the recursive call and the end of the function constant. > > On Wed, Jul 27, 2011 at 1:38 AM, Will Ness <will_n48(a)yahoo.com<http://us.mc1145.mail.yahoo.com/mc/compose?to=will_n48@yahoo.com>> > wrote: > > Ovidiu Deac <ovidiudeac <at> gmail.com> writes: > > > >> > >> I've read the paragraph below from Real World Haskell > >> > > ( > http://book.realworldhaskell.org/read/efficient-file-processing-regular-exp… > >> section "An important aside: writing lazy functions") > >> > >> I can get a feeling why lazyness compensates for the lack of tail > >> recursion but I don't really understand why it would work in general. > >> > >> My understanding is that that a function which returns a list with > >> non-tail recursion would only work this way with certain usage > >> patterns. In this case the caller of the funtion will have to use the > >> string in a stream-like fashion. > > > > The concept of tail recursion modulo cons is similar. Wikipedia has an > article. > > Basically it's about keeping one execution frame above the tail-recursive > one. > > Lazily accessing a list from the top is a lot like adding elements to it > one by > > one at the bottom. Prefixing a recursive call with an operation of fixed > number > > of operations is OK, because they will get consumed and the recursive > case > > called, in a fixed number of steps - the instance between current point > and next > > recursive invocation point won't grow, only get increased at times by a > set > > amount and then get consumed. > > > > The problem with non-tail recursion in lazy setting is that the distance > grows > > between the current point and the next invocation, and so the amount of > "what to > > do next?" information grows all the time. In imperative setting it gets > flipped > > into "what to do after the invocation" but it still grows. In > tail-recursion > > (even modulo cons, or (++) with fixed prefix) it's bounded, finite, > constant. > > Looking at Scheme example code in WP "continuation-passing style" article > might > > help. There we see as in translations of tail-recursive functions the > > constructed continuation does not grow in unlimited fashion, instead only > maybe > > getting prefixed by a fixed amount of operations at times. > > > > Does it make any sense? :) > > > > > > _______________________________________________ > > Beginners mailing list > > Beginners(a)haskell.org<http://us.mc1145.mail.yahoo.com/mc/compose?to=Beginners@haskell.org> > > http://www.haskell.org/mailman/listinfo/beginners > > > >

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