
Also, with -XImpredicativeTypes:
> (seq :: (forall x. Show x => x) -> b -> b) undefined ()
()
On Wed, Apr 2, 2014 at 9:18 PM, Dan Doel
On Wed, Apr 2, 2014 at 8:40 PM, Niklas Haas
wrote: Oh, fair point; this particular F apparently doesn't break it but if you remove the Show x constraint, it does.
Actually, is that a bug in GHC?
λ newtype F = F (forall x. Show x => x -> String) λ F undefined `seq` () () λ undefined `seq` () *** Exception: Prelude.undefined
I'm not sure how to interpret this output.
This is pretty weird, but here's what I think is going on:
F requires an argument of type forall x. Show x => x -> String. This requires abstracting over a dictionary for Show x. So at the core level, this gets expanded to something like:
\showd -> undefined
which is non-bottom.
Even if you annotate undefined with the above type, You'll probably end up with:
(\showd -> undefined) defaultd `seq` ()
after GHC reinstantiates the polymorphic type and then defaults, which will be undefined. So you can only observe this by wrapping the polymorphic expression in a newtype, to keep it abstracted.
I don't know that this qualifies as a bug, but it's definitely pretty subtle behavior.
-- Dan