On 7/19/07, Dan Weston <[EMAIL PROTECTED]> wrote:
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 <[EMAIL PROTECTED]> 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]" ?
>
> Antoine
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