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Javier Neira

The Haskell 98 Library Report: Arrays - 0 views

  • 16.2  Incremental Array Updates The operator (//) takes an array and a list of pairs and returns an array identical to the left argument except that it has been updated by the associations in the right argument. (As with the array function, the indices in the association list must be unique for the updated elements to be defined.) For example, if m is a 1-origin, n by n matrix, then m//[((i,i), 0) | i <- [1..n]] is the same matrix, except with the diagonal zeroed.
  • -- A rectangular subarray subArray :: (Ix a) => (a,a) -> Array a b -> Array a b subArray bnds = ixmap bnds (\i->i) -- A row of a matrix row :: (Ix a, Ix b) => a -> Array (a,b) c -> Array b c row i x = ixmap (l',u') (\j->(i,j)) x where ((_,l'),(_,u')) = bounds x -- Diagonal of a matrix (assumed to be square) diag :: (Ix a) => Array (a,a) b -> Array a b diag x = ixmap (l,u) (\i->(i,i)) x        where           ((l,_),(u,_)) = bounds x -- Projection of first components of an array of pairs firstArray :: (Ix a) => Array a (b,c) -> Array a b firstArray = fmap (\(x,y)->x)
mountain

Pipes: del.icio.us popular links for programming - 0 views

  •  
    del.icio.us popular links for programming, inculding java, c++, smalltalk, ruby, haskell, ml, schema, calm, ocaml, php, erlang, rails, perl, STM, metaprograming, and DSL
Javier Neira

Existential type - HaskellWiki - 0 views

  • First of all, it is now impossible for a function to demand a Worker having a specific type of buffer. Second, the type of foo can now be derived automatically without needing an explicit type signature. (No monomorphism restriction.) Thirdly, since code now has no idea what type the buffer function returns, you are more limited in what you can do to it.
  • This illustrates creating a heterogeneous list, all of whose members implement "Show", and progressing through that list to show these items: data Obj = forall a. (Show a) => Obj a   xs :: [Obj] xs = [Obj 1, Obj "foo", Obj 'c']   doShow :: [Obj] -> String doShow [] = "" doShow ((Obj x):xs) = show x ++ doShow xs With output: doShow xs ==> "1\"foo\"'c'"
  • Existential types in conjunction with type classes can be used to emulate the dynamic dispatch mechanism of object oriented programming languages. To illustrate this concept I show how a classic example from object oriented programming can be encoded in Haskell.
Javier Neira

Haskell: The Confusing Parts - 0 views

  • f $ x = f x(f . g) x = f (g x)-- Some quick examples of using these. The following line...putStrLn (take 12 (map foo (bar ++ "ack")))-- ...can be rewritten as...putStrLn $ take 12 $ map foo $ bar ++ "ack"(putStrLn . take 12 . map foo) (bar ++ "ack")putStrLn . take 12 . map foo $ bar ++ "ack" In terms of good Haskell style, the last example above is preferable to the others. By the way, note that ($) has the lowest precedence (zero) of any operator, so you can almost always use arbitrary syntax on either side of it.
  • -- Actual definitions:f $ x = f x(f . g) x = f (g x)-- Some quick examples of using these. The following line...putStrLn (take 12 (map foo (bar ++ "ack")))-- ...can be rewritten as...putStrLn $ take 12 $ map foo $ bar ++ "ack"(putStrLn . take 12 . map foo) (bar ++ "ack")putStrLn . take 12 . map foo $ bar ++ "ack"
  • putStrLn (take 12 (map foo (bar ++ "ack")))-- ...can be rewritten as...putStrLn $ take 12 $ map foo $ bar ++ "ack"(putStrLn . take 12 . map foo) (bar ++ "ack")putStrLn . take 12 . map foo $ bar ++ "ack"
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  • putStrLn (take 12 (map foo (bar ++ "ack")))
  • anywhere you might write \x -> foo x [1..10], you should instead write flip foo [1..10], where flip is a standard Prelude function that flips the first two arguments of whatever function you give it
  • two arguments of whatever
  • anywhere you might write \x -> foo x [1..10], you should instead write flip foo [1..10], where flip is a standard Prelude function that flips the first
  • The curry and uncurry functions stand in for \f x y -> f (x, y) and \f (x, y) -> f x y, respectively.
  • The curry and uncurry functions stand in for \f x y -> f (x, y)
  • I will point out that return is, in fact, not a return statement. It’s a function, and an inappropriately named function, at that. Writing return () in your do block will not cause the function to return.
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    If you're used to the C family of languages, or the closely related family of "scripting languages," Haskell's syntax (mainly) is a bit baffling at first. For some people, it can even seem like it's sneaking out from under you every time you think you understand it. This is sort of a FAQ for people who are new to Haskell, or scared away by its syntax.
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