How to use and initialize a variable haskell

Question

I have a function in Haskell that given a list, returns some of its permutations.

My problem is that I have to give the length of the list in order for the function to work. Is there any way to change that using let or where?

generate_permutations list 0 = my_permutation 1 4 list
generate_permutations list n = list (my_permutation (list!!n) 4 list) ++ generate_permutations list (n-1)

I want to initialize n with length of the list.

I don't know how to do it.
If I manage to do it how can I stop the recursion?

The word "initialisation" totally doesn't make sense for a Haskell variable, because if you give a variable a value "initially" that means it'll have this value forever. Why would you want to give something a particular value, if not to ensure that value can be used later on without having to worry somebody might have changed it? — leftaroundabout
– leftaroundabout, Commented Mar 17, 2014 at 15:08

Danny Navarro · Accepted Answer · 2014-03-17 12:10:20Z

2

There is no variable assignment in Haskell. However you could use a where clause to assign a name to an expression:

generate_permutations list =
    my_permutation (list!!n) 4 list ++ generate_permutations list (n-1)
  where
    n = length list

edited Mar 17, 2014 at 12:10

answered Mar 17, 2014 at 12:03

Danny Navarro

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1 Comment

Danny Navarro Over a year ago

I don't know what my_permutation does, but I took the liberty of removing the list name used as a function. I'm pretty sure you meant something else.

bheklilr · Accepted Answer · 2014-03-17 11:58:50Z

0

The usual way is to use a where:

generate_permutations list = go list (length list)
    where
        go xs 0 = my_permutation 1 4 list
        go xs n = list (my_permutation (list!!n) 4 list) ++ go list (n - 1)
        --        ^--- What is this doing here?  This will cause a type error

The name go is quite popular for functions like this, but you can use whatever you like. Assuming that my_permutation exits, this function will as well. Your recursion decrements from n = length list to n = 0 in a pretty common pattern.

answered Mar 17, 2014 at 11:58

bheklilr

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