So I have a data type
data SomeType a =
Type a |
Mix (SomeType a) (SomeType a)
This my show instance for SomeType
instance (Show a) => Show (SomeType a) where
show (Type a) = show a
show (Mix a b) = "(" ++ show a ++ " " ++ show b ++ ")"
So
Mix (Type 5) (Type 4)
would give me
(5 4)
Now I want to have
read "(3 4)" :: SomeType Int
produce
(3 4)
or
read "(a b)" :: SomeType Char
produce
(a b)
I am lost at how to use the Read class.
Here's an example based on the documentation which should be able to parse everything that show renders (assuming the type has a compatible Read instance defined), that is read . show should be more or less the identity:
instance (Read a) => Read (SomeType a) where
readsPrec d r = readMix r ++ readType r
where
readMix = readParen True $ \r -> do
(v1, r'') <- readsPrec d r
(v2, r') <- readsPrec d r''
return (Mix v1 v2, r')
readType r = do
(v, r') <- readsPrec d r
return (Type v, r')
Thus,
> read "(3 4)" :: SomeType Int
(3 4)
it :: SomeType Int
But note, that for SomeType Char the default Show instance of Char surrounds the character with single quotes:
> read "('a' ('b' 'c'))" :: SomeType Char
('a' ('b' 'c'))
it :: SomeType Char
hope this helps
d is the operator precedence (which I accidentally incremented -- I've edited this) of the enclosing parsing context and r is the current remaining string to be parsed.
SomeType Charis not whatshow (Mix (Type 'a') (Type 'b'))would have generated.deriving (Show, Read)after the type delcaration. While they won't produce your output and input, they are a standard way of showing (and sometimes reading) values, which is why you should use them instead of your own instances. If you want to show them in a different way, use a separate function and call itrenderor so.