CODEWITHSUNDEEP
haskell
beld
beld

Reputation: 81

Creating a simple array in Haskell

I'm a newbie regarding Haskell (and Java for that matter) and I tried to find a solution to my probably very simple problem. I need to create two arrays where it checks if the first array is a prefix of the second one, for example [5,6,7] [5,6,7,6,2] would give me the boolean true. This is what I tried, however I have no idea how to declare an array. n should be the highest slot number of array a and it counts down. As soon as one slot of a doesn't equal b it should return False. Is there another way to implement it without the n variable? 

isaprefix :: [Int] -> [Int] -> Int -> Bool
isaprefix a b n = if n==0
            then True
            else
                if a[n] == b [n]
                then isaprefix ((a[n-1]) (b[n-1]))
                else False

Upvotes: 2

Views: 807

Answers (1)

CR Drost
CR Drost

Reputation: 9807

Yes, there is. You use pattern matching:

-- type signature can be slightly more general: any list of equatables.
isaprefix :: (Eq a) => [a] -> [a] -> Bool

-- first are the two "boring" cases.
-- we define that empty lists are prefixes of everything.
isaprefix [] _ = True

-- then: we also define that you can't be the prefix of an empty list.
isaprefix _ [] = False

-- those 2 are technically in conflict, so you need to know that the empty list is defined,
-- by the order of those definitions, to be a prefix of the empty list.  This supports the
-- general property that `isprefixof x x == True` for all `x`.

-- ok, now here's the more interesting pattern: both lists are nonempty. We need them to
-- match on their first elements and also for the rest of the first list to prefix the 
-- rest of the second list.
isaprefix (a:as) (b:bs)
    | a == b     = isaprefix as bs
    | otherwise  = False

Upvotes: 8

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