Concatenate to a list of string another string
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))])
string list haskell concat
asked Nov 24 '18 at 17:30
Akos FajsziAkos Fajszi
445
add a comment |
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))])
string list haskell concat
asked Nov 24 '18 at 17:30
Akos FajsziAkos Fajszi
445
add a comment |
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))])
string list haskell concat
asked Nov 24 '18 at 17:30
Akos FajsziAkos Fajszi
445
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))])
string list haskell concat
string list haskell concat
asked Nov 24 '18 at 17:30
Akos FajsziAkos Fajszi
445
asked Nov 24 '18 at 17:30
Akos FajsziAkos Fajszi
445
asked Nov 24 '18 at 17:30
Akos FajsziAkos Fajszi
445
asked Nov 24 '18 at 17:30
Akos FajsziAkos Fajszi
445
asked Nov 24 '18 at 17:30
Akos FajsziAkos Fajszi
445
445
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
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)
answered Nov 24 '18 at 17:43
Jorge AdrianoJorge Adriano
2,220918
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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)
answered Nov 24 '18 at 17:43
Jorge AdrianoJorge Adriano
2,220918
add a comment |
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)
answered Nov 24 '18 at 17:43
Jorge AdrianoJorge Adriano
2,220918
add a comment |
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)
answered Nov 24 '18 at 17:43
Jorge AdrianoJorge Adriano
2,220918
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)
answered Nov 24 '18 at 17:43
Jorge AdrianoJorge Adriano
2,220918
answered Nov 24 '18 at 17:43
Jorge AdrianoJorge Adriano
2,220918
answered Nov 24 '18 at 17:43
Jorge AdrianoJorge Adriano
2,220918
answered Nov 24 '18 at 17:43
Jorge AdrianoJorge Adriano
2,220918
2,220918
add a comment |
add a comment |
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