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Talk:99 questions/1 to 10

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Revision as of 16:40, 17 January 2008 by Michael (Talk | contribs)

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Does the problem 1 example need correction?

The problem refers us to
as a
function providing the same functionality. However,
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
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)