GHC 6.10.1 type puzzler
Here is an example illustrating a type problem I've having with GHC 6.10.1: module Main where newtype Box a = B a make :: a - Box a make x = B x val :: Box a - a val (B x) = x test1 :: Box a - a - [a] test1 box x = go box x where go :: Box a - a - [a] go b y = [(val b), y] test2 :: Box a - a - [a] test2 box x = go x where -- go :: a - [a] go y = [(val box), y] main :: IO () main = do print $ test1 (make 1) 2 If I uncomment the commented type declaration, I get the familiar error Couldn't match expected type `a1' against inferred type `a' On the other hand, the earlier code is identical except that it passes an extra argument, and GHC matches the types without complaint. This is a toy example to isolate the issue; in the actual code one wants a machine-checkable type declaration to help understand the function, which is local to save passing an argument. To the best of my understanding, I've given the correct type, but GHC won't make the inference to unify the type variables. I wonder if I found a GHC blind spot. However, it is far more likely that my understanding is faulty here. Any thoughts? Thanks, Dave ___ Glasgow-haskell-users mailing list Glasgow-haskell-users@haskell.org http://www.haskell.org/mailman/listinfo/glasgow-haskell-users
Re: GHC 6.10.1 type puzzler
test2 :: Box a - a - [a]
test2 box x = go x
where
-- go :: a - [a]
go y = [(val box), y]
Couldn't match expected type `a1' against inferred type `a'
You need to turn on {-# ScopedTypedVariables #-}.
See
http://www.haskell.org/ghc/docs/latest/html/users_guide/other-type-extensions.html#scoped-type-variables
Your first example works without this extension, because the local
definition 'go' is fully polymorphic. However, in the second example,
the type variable 'a' in the signature of 'go' is not fully polymorphic
- it must be exactly the _same_ 'a' as in the top-level signature.
Regards,
Malcolm
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Re: GHC 6.10.1 type puzzler
Dave Bayer wrote:
Here is an example illustrating a type problem I've having with GHC 6.10.1:
module Main where
newtype Box a = B a
make :: a - Box a
make x = B x
val :: Box a - a
val (B x) = x
test1 :: Box a - a - [a]
test1 box x = go box x
where
go :: Box a - a - [a]
go b y = [(val b), y]
test2 :: Box a - a - [a]
test2 box x = go x
where
-- go :: a - [a]
go y = [(val box), y]
If you really want to specify the type for a local declaration, you'll have to
turn on ScopedTypeVariables. Otherwise 'a' is test2 type signature and 'a' in
go type signature mean different unrelated things.
In test2 box remains with the first 'a', while go is with the second. There is
need for unification. In the case of test1 go has all the variables.
In other words, you get
test1 :: forall p. Box p - p - [p]
go :: forall q. Box q - q - [q]
and you perfectly can call go with test1 arguments.
On the other hand,
test2 :: forall p. Box p - p - [p]
go :: forall q. q - [q]
and obviously there's a problem.
Do this
{-# LANGUAGE ScopedTypeVariables, PatternSignatures #-}
test2 :: Box a - a - [a]
test2 box (x::a) = go x-- here you say, that x :: a
where
go :: a - [a] -- here you mention the same a
go y = [(val box), y]
PatternSignatures appears to be needed for pattern x::a
If I uncomment the commented type declaration, I get the familiar error
Couldn't match expected type `a1' against inferred type `a'
On the other hand, the earlier code is identical except that it passes
an extra argument, and GHC matches the types without complaint.
This is a toy example to isolate the issue; in the actual code one wants
a machine-checkable type declaration to help understand the function,
which is local to save passing an argument. To the best of my
understanding, I've given the correct type, but GHC won't make the
inference to unify the type variables.
I wonder if I found a GHC blind spot. However, it is far more likely
that my understanding is faulty here. Any thoughts?
Thanks,
Dave
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Re: GHC 6.10.1 type puzzler
Dave Bayer [EMAIL PROTECTED] wrote in article [EMAIL PROTECTED] in gmane.comp.lang.haskell.glasgow.user: test1 :: Box a - a - [a] test1 box x = go box x where go :: Box a - a - [a] go b y = [(val b), y] The type signature go :: Box a - a - [a] is equivalent to go :: Box b - b - [b] or go :: forall a. Box a - a - [a] or go :: forall b. Box b - b - [b] because the type variable a in the type signature for test1 does not scope over the type signature for go. Fortunately, go here does have that type. test2 :: Box a - a - [a] test2 box x = go x where -- go :: a - [a] go y = [(val box), y] Here go does not have the type forall a. a - [a]; it is not polymorphic enough. To write the signature for go inside test2, you need lexically scoped type variables: http://www.haskell.org/ghc/docs/latest/html/users_guide/other-type-extensions.html#scoped-type-variables -- Edit this signature at http://www.digitas.harvard.edu/cgi-bin/ken/sig 2008-11-20 Universal Children's Day http://unicef.org/ 1948-12-10 Universal Declaration of Human Rights http://everyhumanhasrights.org ___ Glasgow-haskell-users mailing list Glasgow-haskell-users@haskell.org http://www.haskell.org/mailman/listinfo/glasgow-haskell-users
Re: GHC 6.10.1 type puzzler
Thanks all. Sorting this out for my actual code was a bit harder, but
it worked. The manual sure wasn't kidding when it said
Read this section carefully!
Dave
On Nov 19, 2008, at 12:12 PM, Malcolm Wallace wrote:
You need to turn on {-# ScopedTypedVariables #-}.
On Nov 19, 2008, at 12:15 PM, Daniil Elovkov wrote:
If you really want to specify the type for a local declaration,
you'll have to turn on ScopedTypeVariables.
On Nov 19, 2008, at 12:29 PM, Chung-chieh Shan wrote:
To write the signature for go inside test2, you need lexically
scoped type variables:
http://www.haskell.org/ghc/docs/latest/html/users_guide/other-type-extensions.html#scoped-type-variables
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