JSON output with Haskell

Question

I am trying to use the show function and get back output that looks like JSON The type I have to work with is

data JSON = JNum Double
          | JStr String

I am looking for

JNum 12, JStr"bye", JNum 9, JStr"hi" to return

[12, "bye", 9, "hi"]

I have attempted:

instance Show JSON where
  show ans = "[" ++ ans ++ "]"

but fails with a compile error. I have also tried

instance Show JSON where
  show ans = "[" ++ ans ++ intercalate ", " ++ "]"

but failed with "Not in scope: data constructor 'JSON' Not sure how to use "ans" to represent whatever type JSON receives as input in the ouput, be it a string, double..etc... Not very good with Haskell so any hints would be great.

Thx for reading

chi · Accepted Answer · 2015-01-31 19:54:12Z

2

You can have GHC automatically derive a show function for you by adding deriving (Show) to your data declaration, e.g.:

data JSON = ... deriving (Show)

As for your code, in order for show ans = "[" ++ ans ++ "]" to type check ans needs to be a String, but ans has type JSON.

To write your own show function you have to write something like:

instance Show JSON where
   show (JNum d) = ... code for the JNum constructor ...
   show (JObj pairs) = ... code for the JObj constructor ...
   show (JArr arr) = ... code for the JArr constructor ...
   ...

Here d will have type Double, so for the first case you might write:

   show (JNum d) = "JNum " ++ show d

or however you want to represent a JSON number.

edited Jan 31, 2015 at 19:54

chi

117k4 gold badges143 silver badges230 bronze badges

answered Jan 31, 2015 at 18:08

ErikR

52.2k9 gold badges78 silver badges136 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

Sibi · Accepted Answer · 2015-01-31 18:09:33Z

1

If you want to write your own instance, you can do something like this:

instance Show JSON where
  show (JNum x) = show x
  show (JStr x) = x
  show (JObj xs) = show xs
  show (JArr xs) = show xs

Note that for JObj and JArr data constructor, the show will use the instance defined for JObj and JArr.

Demo:

λ> JArr[JNum 12, JStr"bye", JNum 9, JStr"hi"] 
[12.0,bye,9.0,hi]

answered Jan 31, 2015 at 18:09

Sibi

49k18 gold badges105 silver badges172 bronze badges

3 Comments

SumYungGai Over a year ago

How come you wrote 'show (JStr x) = x' rather than 'show (JStr x ) = show x '? Thank you

Sibi Over a year ago

@SumYungGai They both are equivalent. show x will produce String but for JStr s data constructor, s is already a String

chi Over a year ago

@Sibi Actually, they are not equivalent: if s is a string, show s will add quotes around s, as well as escape any quotes which might be inside s, and any other "special" characters.

Collectives™ on Stack Overflow

JSON output with Haskell

2 Answers 2

Comments

3 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

Comments

3 Comments

Your Answer

Sign up or log in

Post as a guest

Related