On 10 March 2011 17:55, Bas van Dijk v.dijk@gmail.com wrote:
On 10 March 2011 18:24, Yves Parès limestr...@gmail.com wrote:
Why has the operator (.) troubles with a type like (forall s. ST s a)?
Why can't it match the type 'b' in (.) definition?
As explained by the email from SPJ that I
Excerpts from Max Bolingbroke's message of Fri Mar 11 05:15:34 -0500 2011:
AFAIK this decision was reversed because SPJ found a simple way to
support them. Indeed, they work fine in 7.0.2 and generate warnings.
Correct. About a week-ish ago I submitted a patch to update the docs.
Cheers,
On 11 March 2011 11:15, Max Bolingbroke batterseapo...@hotmail.com wrote:
On 10 March 2011 17:55, Bas van Dijk v.dijk@gmail.com wrote:
On 10 March 2011 18:24, Yves Parès limestr...@gmail.com wrote:
Why has the operator (.) troubles with a type like (forall s. ST s a)?
Why can't it match
On 11 March 2011 12:04, Bas van Dijk v.dijk@gmail.com wrote:
Unfortunately foo still doesn't type check in 7.0.2:
foo :: (forall s. ST s a) - a
foo st = ($) runST st
Note that the following does type check with ImpredicativeTypes:
bar = id runST
Bas
Dear list,
I have the following (simplified) piece of code:
find :: Int - [Int]
find i = runST . (`runContT` return) $
callCC $ \escape - do
return []
which used to compile correctly under GHC 6.12.3.
Now that I've switched to 7.0.2 it gets rejected with the following error:
On Thursday 10 March 2011 14:18:24, Anakim Border wrote:
Dear list,
I have the following (simplified) piece of code:
find :: Int - [Int]
find i = runST . (`runContT` return) $
callCC $ \escape - do
return []
which used to compile correctly under GHC 6.12.3.
Now that I've
On 10 March 2011 14:47, Daniel Fischer daniel.is.fisc...@googlemail.com wrote:
If memory serves correctly, it's impredicative polymorphism.
Indeed. For example the following also doesn't type check in GHC-7:
foo :: (forall s. ST s a) - a
foo st = ($) runST st
Surprisingly the following does:
On Thursday 10 March 2011 17:15:29, Bas van Dijk wrote:
On 10 March 2011 14:47, Daniel Fischer daniel.is.fisc...@googlemail.com
wrote:
If memory serves correctly, it's impredicative polymorphism.
Indeed. For example the following also doesn't type check in GHC-7:
foo :: (forall s. ST s
Why has the operator (.) troubles with a type like (forall s. ST s a)?
Why can't it match the type 'b' in (.) definition?
2011/3/10 Daniel Fischer daniel.is.fisc...@googlemail.com
On Thursday 10 March 2011 14:18:24, Anakim Border wrote:
Dear list,
I have the following (simplified) piece
On 10 March 2011 18:24, Yves Parès limestr...@gmail.com wrote:
Why has the operator (.) troubles with a type like (forall s. ST s a)?
Why can't it match the type 'b' in (.) definition?
As explained by the email from SPJ that I linked to, instantiating a
type variable (like 'b') with a
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