连接到一个字符串列表，另一个字符串

Question

The below code gives back a list of String but I want it work on multiple cases. The problem is that I can't create the same exact result with recursion. The program gives back the following result:

replaceTabs 6 ["\thello world"]

=> ["      hello world"]

Now this should work with a longer list like:

replaceTabs 6 ["asd dsa","\thello world"]

    => ["asd dsa","      hello world"]

Simple concat doesn't work, because it will give back undefined pattern.

replaceTab' :: Int -> [[Char]] -> [Char]
replaceTab' n [[x]] =
 if x == '\t' then replicate n ' '
 else [x]

replaceTabs :: Int -> [String]-> [String]
replaceTabs n [""] = [""]
replaceTabs n (x:xs) = (return . concat $ [replaceTab' n [a] | a <- (map (:[]) (x))])

Answer 1

This

replaceTab' :: Int -> [[Char]] -> [Char]

is the same as,

replaceTab' :: Int -> [String] -> String

What you should focus on is implementing a function,

replaceTab :: Int -> String -> String

which "fixes" a single String . Then replaceTabs is simply,

replaceTabs :: Int -> [String] -> [String]
replaceTabs n = map (replaceTab n)