[Haskell-beginners] Return a Foldable instance

[James Cook]

On May 4, 2011, at 3:26 PM, David McBride wrote:

> I still have a question about this.  If you look at the
> Text.Regex.Posix library you can do this:
>
>> "asdf" =~ "asdf" :: String
> "asdf"
>
>> "asdf" =~ "asdf" :: Int
> 1
>> :t "asdf" =~ "asdf"
> "asdf" =~ "asdf" :: (RegexContext Regex [Char] target) => target
>
> So, there is something similar to what the op was asking for.  Why is
> this possible, but his is not?

Because unlike the Foldable class, the RegexContext class does provide  
functions that return values of its type parameters.    RegexContext  
is defined as follows:

 > class RegexLike regex source => RegexContext regex source target  
where
 >   match :: regex -> source -> target
 >   matchM :: Monad m => regex -> source -> m target

The 'match' function returns a value of type 'target'.  For Foldable,  
though:

 > class Foldable t where
 >   fold :: Monoid m => t m -> m
 >   foldMap :: Monoid m => (a -> m) -> t a -> m
 >   foldr :: (a -> b -> b) -> b -> t a -> b
 >   foldl :: (a -> b -> a) -> a -> t b -> a
 >   foldr1 :: (a -> a -> a) -> t a -> a
 >   foldl1 :: (a -> a -> a) -> t a -> a

None of those functions return anything that mentions "t", so the  
Foldable class is not able to construct values of an unknown Foldable  
type.

For a simpler example, the same ideas applies to more-familiar  
typeclasses defined in the Prelude.  For example, you can have  
expressions of type "Num a => a", because the Num class defines a  
function "fromInteger :: Integer -> a" - so if you have an Integer,  
you can make a value of any type that is an instance of Num.  But you  
cannot have a useful expression of type "Show a => a", because Show  
doesn't define any functions that produce values.

-- James