June 2011 - Beginners

Bool -> a -> Maybe a, or something similar
by Tacit Sawk 17 Jun '11

17 Jun '11

|Hello! I tried Hoogle, but it didn't seem to have what I was after. I'm looking for something like this: > maybeWhen :: Bool -> a -> Maybe a > maybeWhen cond = > if cond > then Just > else const Nothing Is there a builtin that has the above sort of behaviour? I suppose it can even be generalised: > maybeWhen :: (MonadPlus m) => Bool -> a -> m a > maybeWhen cond = > if cond > then return > else const mzero Tacit |

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share a record type
by Kees Bleijenberg 17 Jun '11

17 Jun '11

I want to learn more about Haskell and I started with my Sudoku solver class SudokuClass s where initAreas :: s -> s createEmptySudoku :: s getBoardLength :: s -> Int setBoard :: s -> Board -> s getCellToAreas :: s ->CellToAreas ... other functions with default implementations I created a Sudoku solver for 9x9 sudoku's. data Sudoku9x9 = Sudoku9x9 { board :: Board, areas:: Areas, cellToAreas :: CellToAreas } deriving (Show) Board, Areas and CellToAreas are of type Map. After that I create an instance: instance SudokuClass Sudoku9x9 where initAreas s = ..... After implementing the 9x9 solver I wanted to create a second instance SudokuTwins, that solves a different kind of sudoku, but uses the same record: data SudokuTwins = SudokuTwins { board :: Board, areas:: Areas, cellToAreas :: CellToAreas } deriving (Show) This is almost a copy of Soduku9x9 (right side of the =) How can I express the fact that all sudoku solvers use the same recordtype? Can I make a type of the record structure {board:: Board.... } and use it in the constructor function or can I incorperate the recordtype in the class or ......? Kees

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Data.Foldable's foldr'
by Federico Mastellone 17 Jun '11

17 Jun '11

Hi, Looking at the source of Data.Foldable I found something I don't understand The Foldable class declares foldl like this: foldl :: (a -> b -> a) -> a -> t b -> a Them, outside of the class, foldr' is defined like this: -- | Fold over the elements of a structure,-- associating to the right, but strictly.foldr' :: Foldable t => (a -> b -> b) -> b -> t a -> bfoldr' f z0 xs = foldl f' id xs z0 where f' k x z = k $! f x z I don't understand this definition, foldl receives 3 parameters and here it is used with 4, how is it possible? Even the function passed to foldl has 3 parameters when a function of 2 is needed. What is the precedence and associativity involved here that makes foldr' a valid expression? Thanks! -- Federico Mastellone Computer Science Engineer - ITBA ".. there are two ways of constructing a software design: One way is to make it so simple that there are obviously no deficiencies, and the other way is to make it so complicated that there are no obvious deficiencies. The first method is far more difficult." Tony Hoare, 1980 ACM Turing Award Lecture.

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Haskell: The Craft of Functional Programming 3rd ed is out.
by KC 16 Jun '11

16 Jun '11

http://www.haskellcraft.com/craft3e/Home.html -- -- Regards, KC

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Is it possible to implement groupBy using folds?
by Zhi-Qiang Lei 16 Jun '11

16 Jun '11

Hi, The source code of groupBy shows that groupBy does not go through every element. Is it really possible to implement it using folds? Thanks. -- | The 'groupBy' function is the non-overloaded version of 'group'. groupBy :: (a -> a -> Bool) -> [a] -> [[a]] groupBy _ [] = [] groupBy eq (x:xs) = (x:ys) : groupBy eq zs where (ys,zs) = span (eq x) xs Best regards, Zhi-Qiang Lei zhiqiang.lei(a)gmail.com

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A neat, simple example of making code more efficient
by Costello, Roger L. 16 Jun '11

16 Jun '11

Hi Folks, This book [1] has a neat, simple example of making code more efficient. Suppose you want to create a function, quad, that raises a number to the 4th power. Example: quad 2 = 2 * 2 * 2 * 2 = 16 The function could be implemented like this: quad a = a * a * a * a That implementation requires three multiplications, which can be pretty expensive, especially with arbitrary numbers. Let's see if we can make it more efficient, by finding an implementation that uses less multiplications. By the associative law we can group the expression: (a * a) * (a * a) Each subexpression performs a square operation: (square a) * (square a) Which is itself a square operation: square (square a) That implementation requires only two multiplications. Let's take an example to see why: quad 2 = square (square 2) = square (2 * 2) -- First multiplication = square (4) -- Reduction = 4 * 4 -- Second multiplication = 16 -- Reduction to the answer Thus, by a little rework the number of multiplications is reduced from three to two. That is a significant efficiency improvement. Cool, aye? /Roger [1] http://www.amazon.com/Introduction-Functional-Programming-using-Haskell/dp/…

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Haskell User Group in Stockholm
by Obscaenvs 16 Jun '11

16 Jun '11

OK, this *might* be slightly off-topic, but I couldn't find a more suiting list to post to: Any body here living in Stockholm interested in starting a user group? /Fredrik, old-as-hell graduate mathematics student

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Options for creating a multiple select form via Yesod
by Michael Litchard 15 Jun '11

15 Jun '11

I'm trying to create a multiple select form, as illustrated on the following: http://api.jquery.com/selected-selector/ Here's the options I see possible: (1) Write a jQuery widget. (2) Use plain javascript via Julius (3) Use the low-level functions in Yesod.Form to write a widget (4) Use a pre-existing function that does what I need, but am not aware of this functionality (1) has appeal as it looks like something small I can contribute to the project. It will take me some extra time to figure out the details. But, I had a look at the other jQuery widgets and they seem to provide an approachable model to follow. (2) This looks like the most straight-forward approach. I'm just learning javascript so would have to figure out how to capture values in Haskell from the form. (3) This looks like the most difficult way. I don't think I know enough about the low-level functions in Yesod.Form to be able to accomplish this in a timely manner. (4) This is the best scenario. There's already a way to do this right now, and I just haven't identified it. If this is the case, I would appreciate being pointed in the right direction. Until informed otherwise, I'm evaluating options 1 and 2. All feedback welcomed. Thanks to all who made Yesod possible.

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merge two handles
by Manfred Lotz 15 Jun '11

15 Jun '11

Hi all, I have two handles where I get stdout resp. stderr from a command output. Is it possible to merge those two handles so that I get a new input handle in a way that data can be read from the new handle whenever it is available from either of the two original handles? -- Manfred

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How does the Deriving being implemented in Haskell?
by Haisheng Wu 15 Jun '11

15 Jun '11

Hi guys, What I understanding `deriving` is Haskell automatically make a type to be a instance of a particular typeclass. If that's the case, the question is How does Haskell do that underline? Take code below as sample, the type Day is now an instance of Bounded. data Day = Monday | Tuesday | Wednesday | Thursday | Friday | Saturday | Sunday deriving (Bounded, Enum) Therefore, *minBound :: Day* will produce *Monday*. Just be curious how it is being implemented. Thanks. Simon

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