[Haskell-cafe] Fwd: How to make this data type work?

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

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