
On 7/19/07, Dan Weston
Oops, you got me. I didn't even look at the third line, I just took it from the previous post. My first instinct actually was to write:
allEqual x@(h:t) = and (zipWith (==) x t)
I prefer, allEqual [] = True allEqual xs = foldl1 (==) xs But, unfortunately, it's not a one liner like yours (unless you allow allEqual [] = undefined). Jason
but I don't think that zipWith is allowed in the question.
Dan
Antoine Latter wrote:
On 7/19/07, Dan Weston
wrote: I would define:
allEqual [] = True allEqual [_] = True allEqual (x1:x2:xs) = (x1 == x2) && allEqual xs
What does this function do for "allEqual [1, 1, 2]" ?
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