haskell

问题 Python's enumerate on lists can be written as zip [0..] . I looked at Control.Lens.Traversal and Control.Lens.Indexed, but I couldn't figure out how to use lenses to generalize this to any reasonable container (I hesitate to say "Traversable"). I'm guessing itraverse or itraverseOf is key. 回答1: If you're using a container that is an instance of FunctorWithIndex then you can simply use imap (,) : > imap (,) "abc" [(0,'a'),(1,'b'),(2,'c')] But if the index isn't the position this won't work: >

问题 I was wondering which are the benefits of using truncation towards minus infinity (Haskell, Ruby) instead of truncation towards zero (C, PHP), from the perspective of programming languages/compilers implementation. It seems that truncating towards minus infinity is the right way to go, but I haven’t found a reliable source for such claiming, nor how such decision impact the implementation of compilers. I’m particularly interested in possible compilers optimizations, but not exclusively.

问题 I'm trying this (for learning purposes): {-# LANGUAGE FlexibleInstances #-} instance Monoid (a -> a) where mempty = id mappend f g = f . g expecting id <> id to be equal to id . id However, with (id <> id) 1 I receive this error: Non type-variable argument in the constraint: Monoid (a -> a) What should I change to run it? It's just to understand monoids and Haskell typeclasses better, not for any practical usage . 回答1: This will need {-# OVERLAPPING #-} pragma since GHC.Base has an instance

问题 Given the following pseudocode for the bubble-sort procedure bubbleSort( A : list of sortable items ) repeat swapped = false for i = 1 to length(A) - 1 inclusive do: /* if this pair is out of order */ if A[i-1] > A[i] then /* swap them and remember something changed */ swap( A[i-1], A[i] ) swapped = true end if end for until not swapped end procedure Here is the code for Bubble Sort as Scala def bubbleSort[T](arr: Array[T])(implicit o: Ordering[T]) { import o._ val consecutiveIndices = (arr

问题 Given the following pseudocode for the bubble-sort procedure bubbleSort( A : list of sortable items ) repeat swapped = false for i = 1 to length(A) - 1 inclusive do: /* if this pair is out of order */ if A[i-1] > A[i] then /* swap them and remember something changed */ swap( A[i-1], A[i] ) swapped = true end if end for until not swapped end procedure Here is the code for Bubble Sort as Scala def bubbleSort[T](arr: Array[T])(implicit o: Ordering[T]) { import o._ val consecutiveIndices = (arr

问题 I'd love to get the following example to type-check: {-# LANGUAGE AllowAmbiguousTypes #-} {-# LANGUAGE RankNTypes #-} {-# LANGUAGE TypeApplications #-} {-# LANGUAGE TypeFamilies #-} module Foo where f :: Int -> (forall f. Functor f => Secret f) -> Int f x _ = x g :: (forall f. Functor f => Secret f) -> Int g = f 4 type family Secret (f :: * -> *) :: * where Secret f = Int I get it that it's probably not possible to infer and check g s type (even though in this specific case it's obvious

问题 This is the textbook zip function: zip :: [a] -> [a] -> [(a,a)] zip [] _ = [] zip _ [] = [] zip (x:xs) (y:ys) = (x,y) : zip xs ys I asked on #haskell earlier wether "zip" could be implemented using "foldr" alone, no recursion, no pattern matching. After some thinking, we noticed the recursion could be eliminated using continuations: zip' :: [a] -> [a] -> [(a,a)] zip' = foldr cons nil where cons h t (y:ys) = (h,y) : (t ys) cons h t [] = [] nil = const [] We are still left with pattern matching

问题 I have this data on my Module Formula : data Formula = Formula { typeFormula :: String, nbClauses :: Int, nbVars :: Int, clauses :: Clauses } And I want to export it but I don't know the right syntax : module Formula ( Formula ( Formula ), solve ) where Someone can tell me the right syntax please ? 回答1: Some of your confusion is coming from having the same module name as the constructor you're trying to export. module Formula ( Formula ( Formula ), solve ) where Should be module Formula (

问题 How does one combine using $ and point-free style? A clear example is the following utility function: times :: Int -> [a] -> [a] times n xs = concat $ replicate n xs Just writing concat $ replicate produces an error, similarly you can't write concat . replicate either because concat expects a value and not a function. So how would you turn the above function into point-free style? 回答1: You can use this combinator: (The colon hints that two arguments follow) (.:) :: (c -> d) -> (a -> b -> c) -

问题 Initially I thought I would get install Haskell with couple of commands using apt-get but its seems somehow complex. As I look at the haskell org download page, I downloaded haskell-platform-2013.2.0.0.tar.gz . Then next step is somehow confusing. It ask to install GHC before installing platform but at the same time if one opens GHC download page , it shows some warning e.g Stop ! ..... we recommend installing the Haskell Platform instead of GHC . Please guide me how to install Haskell on