How can I fix my replace haskell function?

Question

It eill work when : replace :: Eq a => a -> a -> [a] -> [a] will be. How can I convert az a to an [a] in my code ?

replace :: Eq a => a -> [a] -> [a] -> [a]
replace _ _ [] = []
replace a x (y:ys)
 | a == y = x : replace a x ys
 | otherwise = y : replace a x ys

Example:

replace '?' "a" "" == ""
replace 'a' "e" "alma" == "elme"
replace 'a' "e" "nincsbenne" == "nincsbenne"

Guru Stron · Accepted Answer · 2022-06-26 13:39:15Z

3

You are using wrong operator for the first guard (a == y) - : is used to prepend a head element to a list but x is a list not a single element, so you need to use ++ which concatenates two lists (x and one returned by recursive call):

replace :: Eq a => a -> [a] -> [a] -> [a]
replace _ _ [] = []
replace a x (y:ys)
 | a == y = x ++ replace a x ys -- ++ instead of :
 | otherwise = y : replace a x ys

Related - Haskell (:) and (++) differences

answered Jun 26, 2022 at 13:39

Guru Stron

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