Reputation: 202
I have a function that is meant to combine strings in a list, adding a separator in between each and outputting a single string using foldl. Here is what I have and some expected behavior of the function -- It isn't working and I'm unsure why.
-- | `sepConcat sep [s1,...,sn]` returns `s1 ++ sep ++ s2 ++ ... ++ sep ++ sn`
--
-- >>> sepConcat "---" []
-- ""
--
-- >>> sepConcat ", " ["foo", "bar", "baz"]
-- "foo, bar, baz"
--
-- >>> sepConcat "#" ["a","b","c","d","e"]
-- "a#b#c#d#e"
sepConcat :: String -> [String] -> String
sepConcat sep [] = ""
sepConcat sep (x:xs) = foldLeft f base l
where
f a x = a ++ sep ++ x
base = ""
l = xs
Upvotes: 2
Views: 1297
Reputation: 16224
The huge problem is your pattern matching:
sepConcat sep [] = ""
sepConcat sep (x:xs) = foldLeft f base l
You don't need to divide the patterns again in []
and (x:xs)
because foldl and foldr take care of both cases. This is how foldl
can be defined to list with recursion:
foldLeft :: (b -> a -> b) -> b -> [a] -> b
foldLeft f base [] = base
foldLeft f base (x:xs) = f (foldLeft f base xs) x
you just need to apply correctly both cases:
sepConcat :: String -> [String] -> String
sepConcat sep xs = foldLeft (\rs s ->
if null rs
then s ++ rs
else s ++ sep ++ rs) "" xs
Here the case of the empty list is ""
and the function is for the recursive case of the list
with your example:
sepConcat ", " ["foo", "bar", "baz"]
=> "foo, bar, baz"
Upvotes: 0
Reputation: 3573
I think you can solve this simply by checking if the first argument is the empty string and handling it accordingly
sepConcat sep = foldl (\x y -> if x == "" then y else x ++ sep ++ y) ""
-- or
sepConcat sep = foldl combine ""
where combine "" x = x
combine x y = x ++ sep ++ y
Upvotes: 0