The fold function in fold takes as parameters the accumulator x, and the element y. So there is no z that is passed.
But even if that was somehow possible, there are still other issues. x is the accumulator here, so a list, that means that x : y makes no sense, since (:) :: a -> [a] -> [a] takes an element and a list, and constructs a new list.
You can however easily make use of foldr to implement a takeWhile function. Indeed:
takeWhile' p = foldr (\x -> if p x then (x :) else const []) []We here thus check if the predicate holds, if that is the case, we preprend the accumulator with x. If not, we return [], regardless of the value of the accumulator.
Due to the laziness of foldr, it will not look for elements after an element has failed the accumulator, since const [] will ingore the value of the accumulator.
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