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haskellinfinite-sequence

Haskell - Negating even numbers in infinite stream

I am trying to generate a list of infinite numbers

0,1,-2,3,-4,5,-6...

So far I got

evenise x   | x == 0 = 0
            | even x = -x
            | otherwise = x

s = foldl (\x -> evenise x) 0 [1..]

However I am getting the error

Occurs check: cannot construct the infinite type: a0 = b0 -> a0
In the first argument of `evenise', namely `x'
In the expression: evenise x
In the first argument of `foldl', namely `(\ x -> evenise x)'

I do not understand the error since evenise takes in an element, and the anoymous function (\x -> evenise x) also takes in a single element.

Solution

  • You want to use map not foldl

    s = map evenise [0..]

    map goes through a list and applies the mapped function to each element. foldl is used for "reducing" a list into a value--for example, adding all the elements in a list can be done like

    foldl (+) 0

    also, foldl only works on finite lists, while foldr (also for reducing) will sometimes work on infinite lists.