Thank you guys. I cannot use a explicit type for there are quite a few of them. But from MigMit, I understand why my original cannot work. On Sat, Jun 22, 2013 at 4:46 AM, Vincent Ambo wrote:

Is there a reason why you can't use an explicit type variable?

{-# LANGUAGE OverloadedStrings, ExistentialQuantification #-}

import Data.Aeson import Control.Applicative import Control.Monad (mzero)

data ActionData j = (FromJSON j, ToJSON j) => AD j j

instance ToJSON (ActionData j) where toJSON (AD o n) = object [ "oldData" .= o , "newData" .= n ]

instance (ToJSON j, FromJSON j) => FromJSON (ActionData j) where parseJSON (Object v) = AD <$> v .: "oldData" <*> v .: "newData" parseJSON _ = mzero

2013/6/21 Miguel Mitrofanov

...

Forgot to reply all, as usual.

-------- Пересылаемое сообщение -------- 21.06.2013, 12:52, "Miguel Mitrofanov" :

Actually, this is not the real error you should care about. Try removing FromJSON instance completely, and you'll get a lot more. And these are fundamental: you have to decide what "j" to use when serializing. Haskell won't automagically substitute some suitable type for you.

So, that's a classic mismatch: for serializing (ToJSON) you need your "j" type to be known to the AD value (meaning: it should be quantified existentially), but for deserializing you need it to be any type (quantified universally).

All in all, AD seems to be the wrong type.

21.06.2013, 12:18, "Magicloud Magiclouds"

...

:

...

data ActionData = AD { oldData :: (FromJSON j, ToJSON j) => j , newData :: (FromJSON j, ToJSON j) => j} instance ToJSON ActionData where toJSON (AD o n) = object [ "oldData" .= o , "newData" .= n ] instance FromJSON ActionData where parseJSON (Object v) = AD <$> v .: "oldData" <*> v .: "newData" parseJSON _ = mzero

I got when compile: No instance for (FromJSON (forall j. (FromJSON j, ToJSON j) => j)) arising from a use of `.:' Possible fix: add an instance declaration for (FromJSON (forall j. (FromJSON j, ToJSON j) => j)) In the second argument of `(<$>)', namely `v .: "oldData"' In the first argument of `(<*>)', namely `AD <$> v .: "oldData"' In the expression: AD <$> v .: "oldData" <*> v .: "newData"

-- 竹密岂妨流水过 山高哪阻野云飞

And for G+, please use magiclouds#gmail.com. , _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe -------- Завершение пересылаемого сообщения --------

_______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe

_______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe

-- 竹密岂妨流水过 山高哪阻野云飞 And for G+, please use magiclouds#gmail.com.