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Talk:99 questions/1 to 10

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[[User:Chrycheng|Chrycheng]] 12:02, 12 September 2007 (UTC)
[[User:Chrycheng|Chrycheng]] 12:02, 12 September 2007 (UTC)
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I think the answer to problem 6 is kind of "cheating". I think something like this would be a nice alternate solutions:
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<haskell>
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isPalindrome :: Eq (a) => [a] -> Bool
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isPalindrome [] = True
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isPalindrome [x] = True
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isPalindrome (x:xs) = x == last xs && isPalindrome (init xs)
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</haskell>
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[[User:Michael|Michael]] 16:40, 17 January 2008 (UTC)

Revision as of 16:40, 17 January 2008

Does the problem 1 example need correction?

The problem refers us to
last
as a
Prelude
function providing the same functionality. However,
last
has type
[a] -> a
which differs from the Lisp example's type
[a] -> [a]
. Should we revise the example or should we rephrase the reference to
last
to highlight the difference?

Chrycheng 12:02, 12 September 2007 (UTC)

I think the answer to problem 6 is kind of "cheating". I think something like this would be a nice alternate solutions: 

isPalindrome        :: Eq (a) => [a] -> Bool
isPalindrome []     = True
isPalindrome [x]    = True
isPalindrome (x:xs) = x == last xs && isPalindrome (init xs)

Michael 16:40, 17 January 2008 (UTC)

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