Re: [Haskell-beginners] Can i define a record without defining access method.

9 Jul 2011


      On Sat, Jul 9, 2011 at 6:45 PM, Tom Murphy  wrote:
...
On 7/9/11, yi huang  wrote:
...
I'm trying to create a haskell implementation of json rpc, I try to
define
protocol using record like this:
data Request = Request {
    version :: Integer
  , id      :: Integer
  , method  :: String
  , args    :: [Value]
} deriving (Typeable, Data, Show)
data Response = Response {
    version :: Integer
  , id      :: Integer
  , code    :: Integer
  , method   :: String
  , result  :: Value
} deriving (Typeable, Data, Show)
so i can use json library to encode/decode it.
But this code fails, because haskell will define access function
automaticlly, and function names conflicts.
My question is, is there a way i can define record without access
function,
so i can have same attribute name in multiple record.
If you don't want access functions defined, you can simply not name
...
your record fields:
...
data Request = Request Integer Integer String [Value]
...
deriving (Typeable, Data, Show)
...
If you want it to be more readable, you can define type synonyms:
...
type ID = Integer
...
type Version = Integer
...
[...]
...
data Request = Request Version ID Method Args
...
deriving (Typeable, Data, Show)
...
The two "instances" of ID won't conflict, then.
I have to define it as a record, so aeson can inspect the attribute names
and encode it to a json object automatically, for example:
"{\"args\":[],\"id\":1,\"method\":\"test\",\"version\":1}"
But normal data constructor would be encoded to an array.
Though i can define it as a normal data constructor, and implement ToJSON
class manually, but the code would be more verbose.
...
Tom
-- 
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