August 2012 - Beginners

Difference between Monad composition and transformation
by Song Zhang 25 Aug '12

25 Aug '12

When I use a State Monad transformer to combine with a Writer Monad StateT s (Writer w) a. is it different from composition of State Monad and Writer Monad. It is State s (Writer w a) ? StateT is defined as (s -> m (a, s)), so StateT s (Writer w) a can be regarded as (s -> Writer w a) , which is (s -> ((a,w),s) and on the other hand State s (Writer w a) is (s -> ((a,w),s). I suppose the are similar and if so, what is the point we still get Monad transformers? Thanks

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shared files / environment / portability
by Christopher Howard 25 Aug '12

25 Aug '12

I'm coding a resource tracker for a game that loads images, sounds, etc. from files, and wondering how the program will know where the files are located on the installation system. (With C/C++ programs I usually had the build system hardcode the appropriate share/ directory into the config.h file.) Assuming I package for cabal distribution, what approaches to this problem are available to me? (P.S. Thanks for the answer to the previous question about associative arrays.) -- frigidcode.com indicium.us

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cabal: updating a package
by Obscaenvs 24 Aug '12

24 Aug '12

Hi there! Excuse me, but I am a bit bewildered; should I really have to give cabal --force-reinstalls to upgrade yesod to the new version 1.1.0.2? It says a lot of packages are likely to be broken, as per the paste http://pastebin.com/cYdZkShu. I was under the impression I had RTFM, but maybe I have missed something? /Fredrik

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Good style on "." and "$" pipeline?
by Carlos J. G. Duarte 22 Aug '12

22 Aug '12

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The Holy Trinity of Functional Programming
by Costello, Roger L. 22 Aug '12

22 Aug '12

Hi Folks, In Richard Bird's book (Introduction to Functional Programming using Haskell) he has a fascinating discussion on three key aspects of functional programming. Below is my summary of his discussion. /Roger --------------------------------------------------------- The Holy Trinity of Functional Programming --------------------------------------------------------- These three ideas constitute the holy trinity of functional programming: 1. User-defined recursive data types. 2. Recursively defined functions over recursive data types. 3. Proof by induction: show that some property P(n) holds for each element of a recursive data type. Here is an example of a user-defined recursive data type. It is a declaration for the natural numbers 0, 1, 2, ...: data Nat = Zero | Succ Nat The elements of this data type include: Zero, Succ Zero, Succ (Succ Zero), Succ (Succ (Succ Zero)), ... To understand this, when creating a Nat value we have a choice of either Zero or Succ Nat. Suppose we choose Succ Nat. Well, now we must choose a value for the Nat in Succ Nat. Again, we have a choice of either Zero or Succ Nat. Suppose this time we choose Zero, to obtain Succ Zero. The ordering of the elements in the Nat data type can be specified by defining Nat to be a member of the Ord class: instance Ord Nat where Zero < Zero = False Zero < Succ n = True Succ m < Zero = False Succ m < Succ n = (m < n) Here is how the Nat version of the expression 2 < 3 is evaluated: Succ (Succ Zero) < Succ (Succ (Succ Zero)) -- Nat version of 2 < 3 = Succ Zero < Succ (Succ Zero) -- by the 4th equation defining order = Zero < Succ Zero -- by the 4th equation defining order = True -- by the 2nd equation defining order Here is a recursively defined function over the data type; it adds two Nat elements: (+) :: Nat -> Nat -> Nat m + Zero = m m + Succ n = Succ(m + n) Here is how the Nat version of 0 + 1 is evaluated: Zero + Succ Zero -- Nat version of 0 + 1 = Succ (Zero + Zero) -- by the 2nd equation defining + = Succ Zero -- by the 1st equation defining + Here is another recursively defined function over the data type; it subtracts two Nat elements: (-) :: Nat -> Nat -> Nat m - Zero = m Succ m - Succ n = m - n If the Nat version of 0 - 1 is executed, then the result is undefined: Zero - Succ Zero The "undefined value" is denoted by this symbol: _|_ (also known as "bottom") Important: _|_ is an element of *every* data type. So we must expand the list of elements in Nat: _|_, Zero, Succ Zero, Succ (Succ Zero), Succ (Succ (Succ Zero)), ... There are still more elements of Nat. Suppose we define a function that returns a Nat. Let's call the function undefined: undefined :: Nat undefined = undefined It is an infinitely recursive function: when invoked it never stops until the program is interrupted. This function undefined is denoted by the symbol _|_ Recall how we defined the ordering of values in Nat: instance Ord Nat where Zero < Zero = False Zero < Succ n = True Succ m < Zero = False Succ m < Succ n = (m < n) >From that ordering we can see that Succ _|_ is an element of Nat: Zero < Succ undefined -- Note: same as Zero < Succ _|_ = True -- by the 2nd equation defining order And Succ (Succ _|_ ) is an element of Nat: Succ Zero < Succ (Succ undefined) -- Note: same as Zero < Succ (Succ _|_ ) = Zero < Succ undefined -- by the 4th equation defining order = True -- by the 2nd equation defining order And Succ (Succ (Succ _|_ ) is an element of Nat, and so forth. So the list of elements in Nat expands: ..., Succ (Succ (Succ _|_ )), Succ (Succ _|_ ), Succ _|_, _|_, Zero, Succ Zero, Succ (Succ Zero), Succ (Succ (Succ Zero)), ... One can interpret the extra values in the following way: _|_ corresponds to the natural number about which there is absolutely no information Succ _|_ to the natural number about which the only information is that it is greater than Zero Succ (Succ _|_ ) to the natural number about which the only information is that it is greater than Succ Zero And so on There is one further value of Nat, namely the "infinite" number: Succ (Succ (Succ (Succ ...))) It can be defined by this function: infinity :: Nat infinity = Succ infinity It is different from all the other Nat values because it is the only number for which Succ m < infinity for all finite numbers m: Zero < infinity = True Succ Zero < infinity = True Succ (Succ Zero) < infinity = True Thus, the values of Nat can be divided into three classes: - The finite values, Zero, Succ Zero, Succ (Succ Zero), and so on. - The partial values, _|_, Succ _|_, Succ (Succ _|_ ), and so on. - The infinite value. Important: the values of *every* recursive data type can be divided into three classes: - The finite values of the data type. - The partial values, _|_, and so on. - The infinite values. Although the infinite Nat value is not of much use, the same is not true of the infinite values of other data types. Recap: We have discussed two aspects of the holy trinity of functional programming: recursive data types and recursively defined functions over those data types. The third aspect is induction, which is discussed now. Induction allows us to reason about the properties of recursively defined functions over a recursive data type. In the case of Nat the principle of induction can be stated as follows: in order to show that some property P(n) holds for each finite number n of Nat, it is sufficient to treat two cases: Case (Zero). Show that P(Zero) holds. Case (Succ n). Show that if P(n) holds, then P(Succ n) also holds. As an example, let us prove this property (the identity for addition): Zero + n = n Before proving this, recall that + is defined by these two equations: m + Zero = m m + Succ n = Succ(m + n) Proof: The proof is by induction on n. Case (Zero). Substitute Zero for n in the equation Zero + n = n. So we have to show that Zero + Zero = Zero, which is immediate from the first equation defining +. Case (Succ n). Assume P(n) holds; that is, assume Zero + n = n holds. This equation is referred to as the induction hypothesis. We have to show that P(Succ n) holds; that is, show that Zero + Succ n = Succ n holds. We do so by simplifying the left-hand expression: Zero + Succ n = Succ (Zero + n) -- by the 2nd equation defining + = Succ (n) -- by the induction hypothesis We have proven the two cases and so we have proven that Zero + n = n holds for all finite numbers of Nat. Full Induction. To prove that a property P holds not only for finite members of Nat, but also for every partial (undefined) number, then we have to prove three things: Case ( _|_ ). Show that P( _|_ ) holds. Case (Zero). Show that P(Zero) holds. Case (Succ n). Show that if P(n) holds, then P(Succ n) holds also. To just prove that a property P holds for every partial (undefined) number, then we omit the second case. To illustrate proving a property that holds for every partial number (not for the finite numbers), let us prove the rather counterintuitive result that m + n = n for all numbers m and all partial numbers n. For easy reference, we repeat the definition of + m + Zero = m m + Succ n = Succ(m + n) Proof: The proof is by partial number induction on n. Case ( _|_ ). Substitute _|_ for n in the equation m + n = n. So we have to show that m + _|_ = _|_, which is true since _|_ does not match either of the patterns in the definition of +. Case (Succ n). We assume P(n) holds; that is, assume m + n = n holds. This equation is the induction hypothesis. We have to show that P(Succ n) holds; that is, show that m + Succ n = Succ n holds. For the left-hand side we reason m + Succ n = Succ(m + n) -- by the second equation for + = Succ(n) -- by the induction hypothesis Since the right-hand side is also Succ n, we are done. The omitted case (m + Zero = Zero) is false, which is why the property does not hold for finite numbers. As an added bonus, having proved that an equation holds for all partial (undefined) numbers, we can assert that it holds for the infinite number too; that is, P(infinity) holds. Thus, we can now assert that m + infinity = infinity for all numbers m.

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[Real World Haskell] Question
by Yang 22 Aug '12

22 Aug '12

Hi all, In chapter 4 in Real World Haskell comes there the first example of system I/O, namely `interactWith`. According to author, the right way is to compile and run program in system shell, though I have loaded source in GHCi and can't figure out how to invoke `main` interactively. Therefore the question is simply how can I call `main` that takes arguments in GHCi? Thanks. BTW: This is my first message to the group and if there is any restriction I violate in topics, I'd like to know. The example I mentioned can be found at the beginning of the web page followed by link http://book.realworldhaskell.org/read/functional-programming.html -- Best Regards Yang Zhao

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some cabal file values
by Christopher Howard 20 Aug '12

20 Aug '12

I was getting reading to upload a module to hackage. I did that once before a long time ago, so I reused and modified an old .cabal file. Current there are two copied-in values that I am not sure about: Cabal-version: >=1.2 Build-depends: base >= 4 && < 5 What should these values be set to? code: -------- $ cabal --version cabal-install version 0.10.2 using version 1.10.2.0 of the Cabal library $ ghc --version The Glorious Glasgow Haskell Compilation System, version 7.4.1 -------- -- frigidcode.com indicium.us

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Error compiling OpenGL tutorial part 2
by Darren Grant 17 Aug '12

17 Aug '12

I am currently following the OpenGL tutorial here: http://www.haskell.org/haskellwiki/OpenGLTutorial2 But in the current incarnation of Display.hs (http://pastebin.com/g4NekDVh) I am getting the following error: $ cabal install Resolving dependencies... Configuring mygl-0.1... Preprocessing executables for mygl-0.1... Building mygl-0.1... [3 of 5] Compiling Display ( Display.hs, dist\build\mygl\mygl-tmp\Display.o ) Display.hs:13:9: Couldn't match expected type `IO a0' with actual type `()' In the return type of a call of `mapM_' In a stmt of a 'do' expression: mapM_ (\ (x, y, z) -> preservingMatrix $ do { color $ Color3 ((x + 1.0) / 2.0) ((y + 1.0) / 2.0) ((z + 1.0) / 2.0); translate $ Vector3 x y z; cube (0.1 :: GLfloat) }) points 7 In the expression: do { clear [ColorBuffer]; scale 0.7 0.7 (0.7 :: GLfloat); mapM_ (\ (x, y, z) -> preservingMatrix $ do { color $ Color3 ((x + 1.0) / 2.0) ((y + 1.0) / 2.0) ((z + 1.0) / 2.0); .... }) points 7; flush } cabal.exe: Error: some packages failed to install: mygl-0.1 failed during the building phase. The exception was: ExitFailure 1 Not being too familiar with the Haskell syntax I'm not sure how to go about decomposing the code to find the source of the problem. Is there any general advice that might help me with this sort of situation? Thank you!

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Backtrace when a certain location in the code is executed
by Nathan Hüsken 17 Aug '12

17 Aug '12

Hey, I am trying to understand haskell program not written by me. During runtime one function is called with parameters which makes the function throw an error. I want to know from where the function is called. In c++ I would set a breakpoint on the error throwing code, and let gdb print the backtrace. To I have similar options in haskell? What other options do I have? Thanks! Nathan

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99problems: different response from ghci <--> runhaskell / ghc
by GiGi 17 Aug '12

17 Aug '12

Hi all, trying to solve the problem 6 "reverse a list" I'm getting a different behaviour from ghci / runhaskell. main = do putStrLn $ concat (map show (myReverse [1,2,3])) -- putStrLn $ concat (map show (myReverse [])) myReverse :: [a] -> [a] myReverse [] = [] myReverse (x:xs) = go [] (x:xs) where go :: [a] -> [a] -> [a] go curr (x:xs) | null xs = [x] ++ curr go curr (x:xs) = go ([x] ++ curr) xs the second putStrLn (once uncommented) gives this error in the runhaskell/ghc: Ambiguous type variable `a0' in the constraint: (Show a0) arising from a use of `show' Probable fix: add a type signature that fixes these type variable(s) In the first argument of `map', namely `show' _but_ if I load the same source into ghci I can successfully enter&run the command: Prelude> :reload [1 of 1] Compiling Main ( 05.hs, interpreted ) Ok, modules loaded: Main. *Main> putStrLn $ concat (map show (myReverse [])) *Main> Then if I "help" the type inference changing the myReverse type declaration to: myReverse :: [Int] -> [Int] everything works well So, my question is: _where_ is the difference between ghci / runhaskell? thx

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