How do I modify a variable in Haskell?

Re:How do I modify a variable in Haskell? by Ajedi32 on Tuesday August 15, 2017 @11:29AM (#15019130)

> I've no idea why the blog post is written in such long winded style. Mutation in Haskell is not difficult!

For a second there, I honestly thought you were being sarcastic. I'd consider the example you just gave to be extremely "long winded" and "difficult" for what it does. If I ported that code line-for-line to Ruby, for example, the whole thing would look like this:

  size = 10

  # Initialize the array to 0
  arr = Array.new(size) { Array.new(size, 0) }

  # Set the diagonal to 1
  (0...size).each{|i| arr[i][i] = 1 }

  # Print the array
  puts arr.map{|row| row.join}.join("\n")

[ Parent | on HN ]

Re:How do I modify a variable in Haskell? by WarDaft on Tuesday August 15, 2017 @04:39PM (#15022244)

That's not really a line for line port. Or even block for block since yours is never uninitialized. The original code was designed to basically mimic writing bad C in Perl, with an uninitialised array that you then explicitly looped over to initialise to zero, and then modified the diagonal. That sort of thing is difficult on purpose in Haskell.

On the other hand...

  main = mapM_ (putStrLn . foldMap show) [bool 0 1 . (==x) <$> [1..size]| x<-[1..size]]

  x .* y = replicate y x
  main = mapM_ (putStrLn . foldMap show) [0 .* x ++ 1 : 0 .* (size-x)| x<-[0..size-1]]

Or if you really do want mutation shorthand (because you're a crazy person) just build it:

  a .// l = mapM_ (uncurry $ writeArray a) l

  pickFn f g (b,v) | b = g v | otherwise = f v

  main = do
    arr <- newArray ((1,1), (size,size)) 0
    arr .// [((x,x),1)|x<-[1..size]]
    mapM_ (pickFn putStr putStrLn . (((==size).snd) *** show)) $ getAssocs arr

It's not pretty, but it's not supposed to be. Do you know what is pretty? Doing things the right way.

  sudo apt-get install haskell-stack
  stack setup
  stack ghci --package matrix
  import Data.Matrix
  print $ identity 10

And of course there's similar for Ruby.

[ Parent | on HN ]

Re:How do I modify a variable in Haskell? by tome on Tuesday August 15, 2017 @11:39AM (#15019219)

Pop quiz:

Do you genuinely believe I wrote the code the way that I did because it's simplest way of initialising an identity matrix in Haskell?

[ Parent | on HN ]

Re:How do I modify a variable in Haskell? by tome on Tuesday August 15, 2017 @11:47AM (#15019294)

FWIW. (Haskell doesn't have a convenient built-in function for printing arrays. Guilty as charged!)

    size = 10
    
    main = do
      -- Initialise the array to 0
      arr <- newArray ((1,1), (size, size)) 0
      let _ = arr :: IOArray (Int, Int) Int

      -- Set the diagonal to 1    
      forM_ [1..size] $ \i -> writeArray arr (i,i) 1
    
      -- Print the array
      forM_ [1..size] $ \i -> do
        forM_ [1..size] $ \j -> do
          a <- readArray arr (i, j)
          putChar (if a == 0 then '0' else '1')
          when (j == size) (putChar '\n')

[ Parent | on HN ]

Re:How do I modify a variable in Haskell? by Ajedi32 on Tuesday August 15, 2017 @12:02PM (#15019419)

Thanks, that looks much nicer and more in-line with what I'd expect from an example intended to demonstrate how easy a simple task is in a specific language.

As an aside, Ruby doesn't have a built-in function for printing arrays either (unless you count .inspect or .to_s, which let you print _anything_). The code I gave formatted the array as a string using some simple data manipulation functions before printing it; it was intended to match the output of the program you provided.

[ Parent | on HN ]