Re: [Haskell-beginners] function defenition. Do I understand it right?

Yes, you are totally correct. your guard for length zero should be ([],[]). The type of the function is halve :: [a] -> ([a], [a]) so all the code paths have to finish with that type. Incidentally taking the length of a linked list forces you to walk the entire list. You are better off checking as to whether it is empty or not. Regards, Ben On 12 July 2011 10:44, Roelof Wobben wrote:

Oke,

I have now this as function definition.

halve (xs) | length xs `mod` 2 == 0 = (take n xs, drop n xs) | otherwise = (take (n+1) xs, drop (n+1) xs) where n= length xs `div` 2

main = do putStrLn $ show $ halve [1,2,3,4] putStrLn $ show $ halve [1,2,3]

this one works except for empty lists.

So I thought this would work .

halve (xs) | length xs == 0 = []

| length xs `mod`2 == 0 = (take n xs, drop n xs)

| otherwise = (take (n+1) xs, drop (n+1) xs)

where n = length xs `div`2

but then I see this error :

Error occurred ERROR line 2 - Type error in guarded expression *** Term : (take n xs,drop n xs) *** Type : ([b],[b]) *** Does not match : [a]

So I assume that a function must always have the same output and can't have 1 or 2 lists as output.

Is this the right assumption.

Roelof

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Date: Tue, 12 Jul 2011 10:34:25 +0100 Subject: Re: [Haskell-beginners] function defenition. Do I understand it right? From: edwards.benj@gmail.com To: rwobben@hotmail.com CC: beginners@haskell.org

I don't even understand what you are trying to do :)

if you want to pattern match on the empty list

foo :: [a] -> [a] foo [] = 0 foo (x:xs) = undefined

if you want to use the guard syntax

foo xs | null xs = 0 | otherwise = undefined

Ben

On 12 July 2011 10:02, Roelof Wobben mailto:rwobben@hotmail.com> wrote:

hello

Everyone thanks for the help.

I'm now trying to make this work on a empty list.

But my question is.

When the definition is :

[a] -> [a] [a]

Is it correct that I don't can use.

length xs = 0 | []

Roelof

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Subject: Re: [Haskell-beginners] function defenition. Do I understand it right? From: d@vidplace.commailto:d@vidplace.com Date: Mon, 11 Jul 2011 18:56:42 -0400 CC: beginners@haskell.orgmailto:beginners@haskell.org To: rwobben@hotmail.commailto:rwobben@hotmail.com

On Jul 11, 2011, at 5:13 PM, Roelof Wobben wrote:

...

What I try to achieve is this:

[1,2,3,4] will be [1,2] [3,4]

[1,2,3,4,5] will be [1,2,3] [3,4,5]

So, I think what you want is this http://codepad.org/kjpbtLfR

Is that correct?

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