So I am doing exercises from real world haskell book and I wrote following code for takeWhile function using foldl.
myTakeWhile' :: (a->Bool) -> [a] -> [a]
myTakeWhile' f xs = foldl step [] xs
where step x xs | f x = x:(myTakeWhile' f xs)
| otherwise = []
This gives the following error, which I am unable to understand.
Couldn't match type ‘a’ with ‘[a]’
‘a’ is a rigid type variable bound by
the type signature for myTakeWhile' :: (a -> Bool) -> [a] -> [a]
at p1.hs:17:17
Expected type: [a] -> [a] -> [a]
Actual type: a -> [a] -> [a]
Relevant bindings include
step :: a -> [a] -> [a] (bound at p1.hs:19:11)
xs :: [a] (bound at p1.hs:18:16)
f :: a -> Bool (bound at p1.hs:18:14)
myTakeWhile' :: (a -> Bool) -> [a] -> [a] (bound at p1.hs:18:1)
In the first argument of ‘foldl’, namely ‘step’
In the expression: foldl step [] xs