[Haskell-beginners] permuting a list

[Jan Snajder]

Hi!

I'm trying to write a list permutation function, and there is in fact a
nice explanation of how to do it here:
http://sneakymustard.com/2008/12/23/shuffling-in-haskell

But for the start I wanted to keep things simple and avoid monad
transformers (since I'm not into this yet). Instead, I'd like to write a
function of type:

> permute :: [a] -> IO [a]

and so this is what I did:

> permute xs = do
>   let n = length xs - 1
>   arr0 <- newListArray (0, n) xs
>   arr <- foldM swap arr0 [n..1]
>   getElems arr
>   where swap arr n = do
>           x <- readArray arr n
>           r <- randomRIO (0, n)
>           y <- readArray arr r
>           writeArray arr n y
>           writeArray arr r x
>           return arr

Unfortunately, what I get is:

> permute :: (MArray a1 a IO) => [a] -> IO [a]

and so when I try to apply this function:

> permute [1,2,3]

this is what I get:

<interactive>:1:0:
    No instance for (MArray a1 t IO)
      arising from a use of `permute' at <interactive>:1:0-14
    Possible fix: add an instance declaration for (MArray a1 t IO)
    In the expression: permute [1, 2, 3]
    In the definition of `it': it = permute [1, 2, 3]

How can I fix this?

Thanx,
jan