dons:

crespi.albert:

...

I'm trying to write in Haskell a function that in Java would be something like this:

char find_match (char[] l1, char[] l2, char e){ //l1 and l2 are not empty int i = 0; while (l2){ char aux = l2[i]; char[n] laux = l2; while(laux){ int j = 0; if(laux[j] = aux) laux[j] = e; j++; } if compare (l1, laux) return aux; else i++; } return ''; }

Yikes!

...

compare function just compares the two lists and return true if they are equal, or false if they are not. it is really a simple function, but I've been thinking about it a lot of time and I can't get the goal. It works like this:

find_match "4*h&a" "4*5&a" 'h' ----> returns '5' (5 matches with the h) find_match "4*n&s" "4dhnn" "k" ----> returns '' (no match at all - lists are different anyway)

That's almost a spec there :)

Ah, I see I misread the spec :) Time for some tea. -- Don

Carajillu writes:

That works good, but I have a problem with the return type, I forgot to mention... can it be a [char]??

If that's what you want, how about this: import Maybe find_match l1 l2 c = fmap catMaybes . sequence $ zipWith match l1 l2 where match a b | a == c = Just (Just b) | a == b = Just Nothing | otherwise = Nothing although part of the reason for writing it like that is to make you work hard to understand it ;-) ... and it returns Maybe [Char] since I can't bring myself to use "" to indicate failure... but judicious use of catMaybes . concat. maybeToList might help with that. -- Jón Fairbairn Jon.Fairbairn@cl.cam.ac.uk

mailing_list:

On Wed, Sep 20, 2006 at 01:31:22AM -0700, Carajillu wrote:

...

compare function just compares the two lists and return true if they are equal, or false if they are not. it is really a simple function, but I've been thinking about it a lot of time and I can't get the goal.

I forgot, obviously, that lists are an instance of the Eq class... so, this is enough: comp l1 l2 = if l1 == l2 then True else False

You never stop learning! andrea

which you would just write as: comp = (==) and then you'd just use == anyway :) -- Don

Andrea Rossato writes:

I forgot, obviously, that lists are an instance of the Eq class... so, this is enough:

comp l1 l2 = if l1 == l2 then True else False

Or why not:

comp l1 l2 = l1 == l2

Or simply:

comp = (==)

:-) -k -- If I haven't seen further, it is by standing in the footprints of giants

Yes, they must be equal the whole way, I like this recursive solution :) Ketil Malde-3 wrote:

Carajillu writes:

...

compare function just compares the two lists and return true if they are equal, or false if they are not.

...

find_match "4*h&a" "4*5&a" 'h' ----> returns '5' (5 matches with the h) find_match "4*n&s" "4dhnn" "k" ----> returns '' (no match at all - lists are different anyway)

Must they be equal the whole way, or just up to the occurrence of the searched-for character?

find_match (x:xs) (y:ys) c | x==c = Just y | x/=y = Nothing | True = find_match xs ys c find_match [] [] _ = Nothing

Or, to check the whole list:

find_match (x:xs) (y:ys) c | x==c && xs == ys = Just y | x/=y = Nothing | True = find_match xs ys c find_match [] [] _ = Nothing

-k -- If I haven't seen further, it is by standing in the footprints of giants

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How about something like this? import Data.List findMatch xs ys k = lookup k . concat $ zipWith zip (substrings xs) (substrings ys) where substrings = nonempty . map (nonempty . inits) . tails where nonempty = filter (not . null) On 20/09/06, Matthias Fischmann wrote:

... and if you want to search strings not single characters:

findmatch s t e = take m . drop n $ t where m' = length e (n, m) = f 0 s f i s | take m' s == e = (i, m') | null s = (0, 0) | otherwise = f (i+1) (tail s)

findmatch "asdfasdf" "asdfxvdf" "fas" == "fxv"

(this one skips equality checks before *and* after the match. feel free post the necessary modifications. :)

matthias

On Wed, Sep 20, 2006 at 02:22:29AM -0700, Carajillu wrote:

...

To: haskell-cafe@haskell.org From: Carajillu Date: Wed, 20 Sep 2006 02:22:29 -0700 (PDT) Subject: Re: [Haskell-cafe] Java or C to Haskell

Yes, they must be equal the whole way, I like this recursive solution :)

Ketil Malde-3 wrote:

...

Carajillu writes:

...

compare function just compares the two lists and return true if they are equal, or false if they are not.

...

find_match "4*h&a" "4*5&a" 'h' ----> returns '5' (5 matches with the h) find_match "4*n&s" "4dhnn" "k" ----> returns '' (no match at all - lists are different anyway)

Must they be equal the whole way, or just up to the occurrence of the searched-for character?

find_match (x:xs) (y:ys) c | x==c = Just y | x/=y = Nothing | True = find_match xs ys c find_match [] [] _ = Nothing

Or, to check the whole list:

find_match (x:xs) (y:ys) c | x==c && xs == ys = Just y | x/=y = Nothing | True = find_match xs ys c find_match [] [] _ = Nothing

-k

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