#1749: forall type gives "not polymorphic enough" error incorrectly
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Reporter: guest | Owner:
Type: bug | Status: new
Priority: normal | Milestone:
Component: Compiler (Type checker) | Version: 6.6.1
Severity: normal | Resolution:
Keywords: | Difficulty: Unknown
Os: Windows | Testcase:
Architecture: x86 |
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Old description:
> > type SFST s a b = a -> ST s b
> > newtype SF a b = SF { runSF :: forall s. ST s (SFST s a b) }
>
> When making this into an instance of Arrow, compose works beautifully:
>
> > sfCompose :: SF b c -> SF c d -> SF b d
> > sfCompose bc cd = SF sf where
> > sf = do
> > runBc <- runSF bc
> > runCd <- runSF cd
> > return $ \b -> runBc b >>= runCd
>
> but pure gives an odd message:
>
> > sfPure :: (b -> c) -> SF b c
> > sfPure f = SF sf where
> > sf = return $ \b -> return (f b)
>
> frp_typebug.lhs:20:12:
> Inferred type is less polymorphic than expected
> Quantified type variable `s' is mentioned in the environment:
> sf :: ST s (b -> ST s c) (bound at frp_typebug.lhs:21:5)
> In the first argument of `SF', namely `sf'
> In the expression: SF sf
> In the definition of `sfPure':
> sfPure f = SF sf
> where
> sf = return $ (\ b -> return (f b))
>
> This can be worked around using scoped type variables:
> > sfPure2 :: (b -> c) -> SF b c
> > sfPure2 (f :: b -> c) = SF sf where
> > sf :: forall s. ST s (SFST s b c)
> > = return $ \b -> return (f b)
>
> but it's ugly.
New description:
{{{
> type SFST s a b = a -> ST s b
> newtype SF a b = SF { runSF :: forall s. ST s (SFST s a b) }
}}}
When making this into an instance of Arrow, compose works beautifully:
{{{
> sfCompose :: SF b c -> SF c d -> SF b d
> sfCompose bc cd = SF sf where
> sf = do
> runBc <- runSF bc
> runCd <- runSF cd
> return $ \b -> runBc b >>= runCd
}}}
but pure gives an odd message:
{{{
> sfPure :: (b -> c) -> SF b c
> sfPure f = SF sf where
> sf = return $ \b -> return (f b)
}}}
{{{
frp_typebug.lhs:20:12:
Inferred type is less polymorphic than expected
Quantified type variable `s' is mentioned in the environment:
sf :: ST s (b -> ST s c) (bound at frp_typebug.lhs:21:5)
In the first argument of `SF', namely `sf'
In the expression: SF sf
In the definition of `sfPure':
sfPure f = SF sf
where
sf = return $ (\ b -> return (f b))
}}}
This can be worked around using scoped type variables:
{{{
> sfPure2 :: (b -> c) -> SF b c
> sfPure2 (f :: b -> c) = SF sf where
> sf :: forall s. ST s (SFST s b c)
> = return $ \b -> return (f b)
}}}
but it's ugly.
Comment (by sorear):
Thanks for the report, but I think I can explain this.
Firstly, and quite unrelatedly, you can use triple braces to request
monospace, see my edit to the description.
'sf' has no arguments, so the monomorphism restriction applies. The body
of 'sf' is assigned a typing derivation of the form:
{{{
Monad m1, Monad m2, f :: b -> c |- return $ \b -> return (f b) : m1 (a ->
m2 b)
}}}
Because 'm1' and 'm2' are constrained by type classes, the monomorphism
restriction forces them to be assigned to fully non-polymorphic types;
unfortunately this means that it is insufficiently polymorphic for SF.
Possible workarounds include:
{{{
{-# LANGUAGE NoMonomorphismRestriction #-}
-- for older GHC, {-# OPTIONS_GHC -fno-monomorphism-restriction #-}
}}}
or:
{{{
sfPure :: (b -> c) -> SF b c
sfPure f = SF sf where
returnST :: a -> ST s a
returnST = return
sf = returnST $ \b -> returnST (f b) -- no type classes, no MR
}}}
or the aforementioned explicit annotation approach (explicit annotations
disable the MR).
Unfortunately, the monomorphism restriction is specified by the Haskell 98
Language Report, so you're more likely to get results from the Haskell-
prime committee than you are from specific implementors.
Stefan
--
Ticket URL: <http://hackage.haskell.org/trac/ghc/ticket/1749#comment:2>
GHC <http://www.haskell.org/ghc/>
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