Haskell使用分隔符“ |”将字符串列表转换为字符串

Question

How can I convert a list to string with divider "|", using foldl'?

ltos :: String -> [String] -> String
ltos  []    = ""
ltos  (m:n) = foldl' f acc xs
   where
     f a b          = 
     acc            = 
     xs             = 

for example,

ltos ["a", "b", "c"]

would output

"a|b|c"

Answer 1

Here's a way:

foldl' (\acc x -> if (null acc) then acc ++ x else acc ++ "|" ++ x) [] ["a", "b", "c"]

The first time accumulator is empty, so just append the first string from the list. Thereafter, append the separator before the variable. If you run scanl , you can see the string being built as follows:

["","a","a|b","a|b|c"]

Answer 2

This is a cute combination of pattern matching and HOF.

ltos :: [String] -> String
ltos []    = ""
ltos (m:n) = foldl' (\a b -> a ++ "|" ++ b) m n

You use the first element of the list as the starting string, then fold over the remaining elements (adding | between each of them).

(I'm assuming that you mistyped the signature to ltos ).