Re: [Haskell-cafe] ContT and ST stack

On Thursday 10 March 2011 17:15:29, Bas van Dijk wrote:

On 10 March 2011 14:47, Daniel Fischer 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:

foo :: (forall s. ST s a) -> a foo st = runST $ st

Because GHC contains a special rule for infix $. Also see:

http://article.gmane.org/gmane.comp.lang.haskell.glasgow.user/19152

Bas

Bedankt. I sort of knew there was a special case for ($) because runST $ do ... works, but not exactly what was handled. So until I forget, I now know it's infix ($) and only that. Cheers, Daniel