list of Data to String
So I have a list of Data, I know newtype is currently better but I will add more things to it. I would like to convert a list of Pack to a String.
unpack [Pack ('a','b'), Pack ('c','d') , Pack (' ', 'e') ] = "abcd e"
I was thinking about using a foldl but am stuck trying to figure it out.
data Pack= Pack (Char, Char) deriving ( Show)
unPack:: [Pack] -> String
unpack list = foldr (Pack (a,b) -> show a + show b -> concat) "" list
Thx for the helping
haskell functional-programming
asked Nov 13 '18 at 16:37
swaffelayswaffelay
325
add a comment |
So I have a list of Data, I know newtype is currently better but I will add more things to it. I would like to convert a list of Pack to a String.
unpack [Pack ('a','b'), Pack ('c','d') , Pack (' ', 'e') ] = "abcd e"
I was thinking about using a foldl but am stuck trying to figure it out.
data Pack= Pack (Char, Char) deriving ( Show)
unPack:: [Pack] -> String
unpack list = foldr (Pack (a,b) -> show a + show b -> concat) "" list
Thx for the helping
haskell functional-programming
asked Nov 13 '18 at 16:37
swaffelayswaffelay
325
add a comment |
So I have a list of Data, I know newtype is currently better but I will add more things to it. I would like to convert a list of Pack to a String.
unpack [Pack ('a','b'), Pack ('c','d') , Pack (' ', 'e') ] = "abcd e"
I was thinking about using a foldl but am stuck trying to figure it out.
data Pack= Pack (Char, Char) deriving ( Show)
unPack:: [Pack] -> String
unpack list = foldr (Pack (a,b) -> show a + show b -> concat) "" list
Thx for the helping
haskell functional-programming
asked Nov 13 '18 at 16:37
swaffelayswaffelay
325
So I have a list of Data, I know newtype is currently better but I will add more things to it. I would like to convert a list of Pack to a String.
unpack [Pack ('a','b'), Pack ('c','d') , Pack (' ', 'e') ] = "abcd e"
I was thinking about using a foldl but am stuck trying to figure it out.
data Pack= Pack (Char, Char) deriving ( Show)
unPack:: [Pack] -> String
unpack list = foldr (Pack (a,b) -> show a + show b -> concat) "" list
Thx for the helping
haskell functional-programming
haskell functional-programming
asked Nov 13 '18 at 16:37
swaffelayswaffelay
325
asked Nov 13 '18 at 16:37
swaffelayswaffelay
325
edited Nov 13 '18 at 16:38
swaffelay
asked Nov 13 '18 at 16:37
swaffelayswaffelay
325
asked Nov 13 '18 at 16:37
swaffelayswaffelay
325
asked Nov 13 '18 at 16:37
swaffelayswaffelay
325
325
add a comment |
add a comment |
2 Answers
2
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oldest
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Your Pack type is isomorphic to two-character strings:
pack2String :: Pack -> String
pack2String (Pack (a,b)) = a:[b]
string2Pack :: String -> Pack -- partial, since String isn't limited in length
string2Pack (a:[b]) = Pack (a, b)
(Note that Pack (a,b) already adds an unnecessary level of wrapping; data Pack = Pack Char Char is also isomorphic to Pack (Char, Char).)
As such, you don't actually need foldr; you can use the list monad instead.
unpack :: [Pack] -> String
unpack xs = xs >>= pack2String
If you aren't yet comfortable with monads, you can just use the concatMap function directly:
unpack :: [Pack] -> String
unpack = concatMap pack2String
answered Nov 13 '18 at 16:48
chepnerchepner
247k32233326
1
unpack xs = xs >>= pack2Stringwould make more sense to me as the list monad example.
– 4castle
Nov 13 '18 at 19:53
@4castle I'll comment later if I can think of a good excuse for not saying that in the first place :) (Right now, I'll blame lack of coffee and/or good breakfast.)
– chepner
Nov 13 '18 at 20:21
add a comment |
To pattern-patch on Pack xyz in a lambda, you need to put it in parentheses:
foldr ((Pack (a,b)) -> ...)
What you wrote would actually parse as two separate arguments
foldr ((Pack) -> (a,b) -> ...)
Next, you can't concatenate strings with +, that's for numbers. ++ or <> are for lists / strings.
Then, the -> concat isn't valid syntax. What you want to do is concatenate the remainder of the foldr computation to the shown a and b. That remainder is the second argument of the folding function:
foldr ((Pack (a,b)) rest -> show a ++ show b ++ rest)
...or shorter,
foldr ((Pack (a,b)) -> shows a . shows b)
answered Nov 13 '18 at 16:47
leftaroundaboutleftaroundabout
79.3k3117233
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Your Pack type is isomorphic to two-character strings:
pack2String :: Pack -> String
pack2String (Pack (a,b)) = a:[b]
string2Pack :: String -> Pack -- partial, since String isn't limited in length
string2Pack (a:[b]) = Pack (a, b)
(Note that Pack (a,b) already adds an unnecessary level of wrapping; data Pack = Pack Char Char is also isomorphic to Pack (Char, Char).)
As such, you don't actually need foldr; you can use the list monad instead.
unpack :: [Pack] -> String
unpack xs = xs >>= pack2String
If you aren't yet comfortable with monads, you can just use the concatMap function directly:
unpack :: [Pack] -> String
unpack = concatMap pack2String
answered Nov 13 '18 at 16:48
chepnerchepner
247k32233326
1
unpack xs = xs >>= pack2Stringwould make more sense to me as the list monad example.
– 4castle
Nov 13 '18 at 19:53
@4castle I'll comment later if I can think of a good excuse for not saying that in the first place :) (Right now, I'll blame lack of coffee and/or good breakfast.)
– chepner
Nov 13 '18 at 20:21
add a comment |
Your Pack type is isomorphic to two-character strings:
pack2String :: Pack -> String
pack2String (Pack (a,b)) = a:[b]
string2Pack :: String -> Pack -- partial, since String isn't limited in length
string2Pack (a:[b]) = Pack (a, b)
(Note that Pack (a,b) already adds an unnecessary level of wrapping; data Pack = Pack Char Char is also isomorphic to Pack (Char, Char).)
As such, you don't actually need foldr; you can use the list monad instead.
unpack :: [Pack] -> String
unpack xs = xs >>= pack2String
If you aren't yet comfortable with monads, you can just use the concatMap function directly:
unpack :: [Pack] -> String
unpack = concatMap pack2String
answered Nov 13 '18 at 16:48
chepnerchepner
247k32233326
1
unpack xs = xs >>= pack2Stringwould make more sense to me as the list monad example.
– 4castle
Nov 13 '18 at 19:53
@4castle I'll comment later if I can think of a good excuse for not saying that in the first place :) (Right now, I'll blame lack of coffee and/or good breakfast.)
– chepner
Nov 13 '18 at 20:21
add a comment |
Your Pack type is isomorphic to two-character strings:
pack2String :: Pack -> String
pack2String (Pack (a,b)) = a:[b]
string2Pack :: String -> Pack -- partial, since String isn't limited in length
string2Pack (a:[b]) = Pack (a, b)
(Note that Pack (a,b) already adds an unnecessary level of wrapping; data Pack = Pack Char Char is also isomorphic to Pack (Char, Char).)
As such, you don't actually need foldr; you can use the list monad instead.
unpack :: [Pack] -> String
unpack xs = xs >>= pack2String
If you aren't yet comfortable with monads, you can just use the concatMap function directly:
unpack :: [Pack] -> String
unpack = concatMap pack2String
answered Nov 13 '18 at 16:48
chepnerchepner
247k32233326
Your Pack type is isomorphic to two-character strings:
pack2String :: Pack -> String
pack2String (Pack (a,b)) = a:[b]
string2Pack :: String -> Pack -- partial, since String isn't limited in length
string2Pack (a:[b]) = Pack (a, b)
(Note that Pack (a,b) already adds an unnecessary level of wrapping; data Pack = Pack Char Char is also isomorphic to Pack (Char, Char).)
As such, you don't actually need foldr; you can use the list monad instead.
unpack :: [Pack] -> String
unpack xs = xs >>= pack2String
If you aren't yet comfortable with monads, you can just use the concatMap function directly:
unpack :: [Pack] -> String
unpack = concatMap pack2String
answered Nov 13 '18 at 16:48
chepnerchepner
247k32233326
edited Nov 13 '18 at 20:22
answered Nov 13 '18 at 16:48
chepnerchepner
247k32233326
answered Nov 13 '18 at 16:48
chepnerchepner
247k32233326
answered Nov 13 '18 at 16:48
chepnerchepner
247k32233326
247k32233326
1
unpack xs = xs >>= pack2Stringwould make more sense to me as the list monad example.
– 4castle
Nov 13 '18 at 19:53
@4castle I'll comment later if I can think of a good excuse for not saying that in the first place :) (Right now, I'll blame lack of coffee and/or good breakfast.)
– chepner
Nov 13 '18 at 20:21
add a comment |
1
unpack xs = xs >>= pack2Stringwould make more sense to me as the list monad example.
– 4castle
Nov 13 '18 at 19:53
@4castle I'll comment later if I can think of a good excuse for not saying that in the first place :) (Right now, I'll blame lack of coffee and/or good breakfast.)
– chepner
Nov 13 '18 at 20:21
1
1
unpack xs = xs >>= pack2String would make more sense to me as the list monad example.– 4castle
Nov 13 '18 at 19:53
unpack xs = xs >>= pack2String would make more sense to me as the list monad example.– 4castle
Nov 13 '18 at 19:53
@4castle I'll comment later if I can think of a good excuse for not saying that in the first place :) (Right now, I'll blame lack of coffee and/or good breakfast.)
– chepner
Nov 13 '18 at 20:21
@4castle I'll comment later if I can think of a good excuse for not saying that in the first place :) (Right now, I'll blame lack of coffee and/or good breakfast.)
– chepner
Nov 13 '18 at 20:21
add a comment |
To pattern-patch on Pack xyz in a lambda, you need to put it in parentheses:
foldr ((Pack (a,b)) -> ...)
What you wrote would actually parse as two separate arguments
foldr ((Pack) -> (a,b) -> ...)
Next, you can't concatenate strings with +, that's for numbers. ++ or <> are for lists / strings.
Then, the -> concat isn't valid syntax. What you want to do is concatenate the remainder of the foldr computation to the shown a and b. That remainder is the second argument of the folding function:
foldr ((Pack (a,b)) rest -> show a ++ show b ++ rest)
...or shorter,
foldr ((Pack (a,b)) -> shows a . shows b)
answered Nov 13 '18 at 16:47
leftaroundaboutleftaroundabout
79.3k3117233
add a comment |
To pattern-patch on Pack xyz in a lambda, you need to put it in parentheses:
foldr ((Pack (a,b)) -> ...)
What you wrote would actually parse as two separate arguments
foldr ((Pack) -> (a,b) -> ...)
Next, you can't concatenate strings with +, that's for numbers. ++ or <> are for lists / strings.
Then, the -> concat isn't valid syntax. What you want to do is concatenate the remainder of the foldr computation to the shown a and b. That remainder is the second argument of the folding function:
foldr ((Pack (a,b)) rest -> show a ++ show b ++ rest)
...or shorter,
foldr ((Pack (a,b)) -> shows a . shows b)
answered Nov 13 '18 at 16:47
leftaroundaboutleftaroundabout
79.3k3117233
add a comment |
To pattern-patch on Pack xyz in a lambda, you need to put it in parentheses:
foldr ((Pack (a,b)) -> ...)
What you wrote would actually parse as two separate arguments
foldr ((Pack) -> (a,b) -> ...)
Next, you can't concatenate strings with +, that's for numbers. ++ or <> are for lists / strings.
Then, the -> concat isn't valid syntax. What you want to do is concatenate the remainder of the foldr computation to the shown a and b. That remainder is the second argument of the folding function:
foldr ((Pack (a,b)) rest -> show a ++ show b ++ rest)
...or shorter,
foldr ((Pack (a,b)) -> shows a . shows b)
answered Nov 13 '18 at 16:47
leftaroundaboutleftaroundabout
79.3k3117233
To pattern-patch on Pack xyz in a lambda, you need to put it in parentheses:
foldr ((Pack (a,b)) -> ...)
What you wrote would actually parse as two separate arguments
foldr ((Pack) -> (a,b) -> ...)
Next, you can't concatenate strings with +, that's for numbers. ++ or <> are for lists / strings.
Then, the -> concat isn't valid syntax. What you want to do is concatenate the remainder of the foldr computation to the shown a and b. That remainder is the second argument of the folding function:
foldr ((Pack (a,b)) rest -> show a ++ show b ++ rest)
...or shorter,
foldr ((Pack (a,b)) -> shows a . shows b)
answered Nov 13 '18 at 16:47
leftaroundaboutleftaroundabout
79.3k3117233
answered Nov 13 '18 at 16:47
leftaroundaboutleftaroundabout
79.3k3117233
answered Nov 13 '18 at 16:47
leftaroundaboutleftaroundabout
79.3k3117233
answered Nov 13 '18 at 16:47
leftaroundaboutleftaroundabout
79.3k3117233
79.3k3117233
add a comment |
add a comment |
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