将DeriveGeneric用于参数化类型

Kam*_*yar 2 haskell aeson

我想为我的参数化类型使用自动DeriveGeneric.我收到错误.我想解码一个yaml文件ino FromJSON类型.

{-# LANGUAGE OverloadedStrings #-}
{-# LANGUAGE DeriveGeneric #-}
{-# LANGUAGE TypeFamilies  #-}

import Web.Scotty
import Data.ByteString.Char8 (pack, unpack)
import Data.ByteString.Lazy (toStrict, fromStrict)
import Data.List
import Data.Yaml
import GHC.Generics

data EPSG a = EPSG { epsg3857 :: a }

data Resolution = Resolution { max :: Int, items :: [Double]}

data Config = Config { minX :: EPSG Double, minY :: EPSG Double, maxX :: EPSG Double, maxY :: EPSG Double
                   , resolution :: EPSG Resolution
                   , metersPerUnit :: EPSG Double
                   , pixelSize :: EPSG Double
                   , scaleNames :: EPSG [String]
                   , tileWidth :: EPSG Double
                   , tileHeight :: EPSG Double
                   , subdirBit :: EPSG [Int]
                   , subdirShiftBit :: EPSG [Int]
                   , subdirNumSize :: EPSG [Int]
                   , fileNameNumSize :: EPSG [Int] } deriving Generic

instance FromJSON EPSG *
instance FromJSON Resolution
instance FromJSON Config
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线EPSG*引发错误.我该如何解决?

jke*_*len 5

您定义EPSG派生通用的需求的定义,然后您需要约束您的实例以获得FromJSON实例a.

data EPSG a = EPSG { epsg3857 :: a } deriving Generic
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...

instance FromJSON a => FromJSON (EPSG a)
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  • 如果您也打开了“DeriveAnyClass”,则可以直接跳转到“data EPSG a = EPSG { epsg3857 :: a } deriving (Generic, FromJSON)”。 (2认同)