[Haskell-cafe] Re: [Haskell] GHC inferred type different for pointed/point-free defs?

[Neil Mitchell]

Hi,

You probably want to read up on:

Monomorphism restriction (I don't think this applies here, but I'm
never too sure!):

http://www.haskell.org/hawiki/MonomorphismRestriction

Defaulting:

http://www.haskell.org/onlinereport/decls.html#sect4.3.4

I don't understand either to any great degree, but when points free
starts changing things, its usually one of those two kicking in.

And as a side note, haskell@ tends to be used more for announcements,
and haskell-cafe@ is used more for asking questions etc.

Thanks

Neil



On 11/3/06, Dan Weston <westondan at imageworks.com> wrote:
> Help! One of two things is going on:
>
> 1) I don't understand what I'm doing
> 2) GHC is inferring different types for pointed and
>     point-free function definition.
>
> I wanted to define Haskell equivalents to the C ternary operator.
> Basically, predicate ??? doIfTrue ||| doIfFalse
>
> For some reason, though, in GHC 6.4 and 6.7, the types
> inferred for abs1 and ab2 (defined below) are different:
>
> *Main> :t abs1
> abs1 :: (Num a, Ord a) => a -> a
>
> *Main> :t abs2
> abs2 :: Integer -> Integer
>
> How is abs2 less general just because I use point-free
> notation instead of pointed notation? They should have the same
> type, no?  I'm stumped...
>
>  > import Control.Arrow
>  > import Data.Either
>  >
>  > infix 1 ?, ??, ???, ????
>  >
>  > (?)  True  = Left
>  > (?)  False = Right
>  >
>  > (??) True  = Left  . fst
>  > (??) False = Right . snd
>  >
>  > p ???  q = (p &&& arr id) >>> uncurry (?)  >>> q
>  > p ???? q = (p &&& arr id) >>> uncurry (??) >>> q
>  >
>  > abs1 x = (< 0) ??? negate ||| id $ x
>  > abs2   = (< 0) ??? negate ||| id
>  >
>  > checks :: [Bool]
>  > checks =
>  >  [
>  >   (True   ?                              3 ) == (Left   3),
>  >   (False  ?                              4 ) == (Right  4),
>  >   (True  ??                           (3,4)) == (Left   3),
>  >   (False ??                           (3,4)) == (Right  4),
>  >   (even ??? (`div` 2) ||| (+1).(*3) $  3   ) == (      10),
>  >   (even ??? (`div` 2) ||| (+1).(*3) $    4 ) == (       2),
>  >   (uncurry (==) ????  (+1) ||| (*3) $ (3,3)) == (       4),
>  >   (uncurry (==) ????  (+1) ||| (*3) $ (3,4)) == (      12),
>  >   (uncurry (==) ????  (+1) +++ (*3) $ (3,3)) == (Left   4),
>  >   (uncurry (==) ????  (+1) +++ (*3) $ (3,4)) == (Right 12),
>  >   (abs1   5                                ) == (       5),
>  >   (abs1 (-5)                               ) == (       5),
>  >   (abs2   5                                ) == (       5),
>  >   (abs2 (-5)                               ) == (       5)
>  >  ]
>  >
>  > main = print (and checks)
>
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