Simon Peyton-Jones wrote: [snip]
Now you are proposing that
data Bar a = MkBar (!a -> a) MkBar f = :MkBar (\x. x `seq` f x)
I suppose you can combine the two notations:
data Baz a = MkBaz !(!a -> a) MkBaz f = f `seq` :MkBaz (\x. x `seq` f x)
Interesting. Is that what you meant? An undesirable consequence would be that case (MkBar bot) of MkBar f -> f `seq` 0 would return 0, because the MkBar constructor puts a lambda inside. This seems bad. Maybe you can only put a ! inside the function type if you have a bang at the top (like MkBaz).
Another possible fix could be defining data Bar a = MkBar (!a -> a) to mean MkBar f = :MkBar (f `seq` (\x -> x `seq` f x)) In that case, case (MkBar bot) of MkBar f -> f `seq` 0 would diverge. Roberto.