[Haskell] Proposal: Allow "\=" for field update in record update syntax

[Keean Schupke]

Benjamin Franksen wrote:

I haven't read Daan's paper yet, but I think his translation is similar 
to the TIR (type indexed row)
part of the HList library...

    Keean.

>Dear Keean,
>
>you should read more carefully what people write. Nowhere have I stated 
>that I want higher-ranked *labels*. In fact, in my translation labels 
>always have the value bottom.
>
>My concern is with higher-ranked record fields. Stupid example:
>
> data R = R {
>  f :: (forall a. a -> a)
> }
>
>My translation doesn't work in this case, because the compiler doesn't 
>accept
>
> instance RecordField R Label_f (forall a. a->a) where
>  ...
>
>  
>
>>Here's an example of a higher ranked type used as a non-label which
>>works fine:
>>
>>---------------------------------------------------------------------
>>----------------------- --{-# OPTIONS -fglasgow-exts #-}
>>
>>module Main where
>>
>>class Test a b | a -> b where
>>        test :: a -> b -> Bool
>>
>>newtype I = I (forall a . Integral a => a)
>>newtype S = S (forall a . Show a => a)
>>
>>instance Test Int I where
>>        test _ _ = True
>>
>>instance Test String S where
>>        test _ _ = False
>>
>>main = do
>>        putStrLn $ show $ test (1::Int) (I undefined)
>>        putStrLn $ show $ test ("a"::String) (S undefined)
>>
>>---------------------------------------------------------------------
>>-----------
>>
>>Which shows that even though you cannot use higher ranked types as
>>labels, you can use them in other fields... Effectively they cannot
>>be on the LHS of a functional dependancy (for obvious reasons if you
>>think about it).
>>    
>>
>
>Yes, you can wrap higher-ranked types into a newtype and then you can 
>define instances for them.
>
>Again, that is what I already wrote in my previous message. With the 
>above stupid example:
>
> newtype Wrap_f = Wrap_f (forall a. a->a)
>
> unWrap_f (Wrap_f x) = x
>
>However, the result of 
>
> getField Label_f
>
>now has type Wrap_f and not (forall a. a->a). To really get the field, I 
>have to unwrap the newtype constructor manually:
>
> get_f :: R -> (forall a. a->a)
> get_f = unWrap_f . getField Label_f
>
>This means that a translation as proposed by Daan (i.e. without 
>first-class labels) is feasible even with higher-ranked field types, 
>but not my version.
>
>Ben
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