Probably self-application, λx. x x.
In Haskell
bar :: (forall x. x) -> c bar = \ x -> x xIf we leave out the type signature, type inference fails, e.g.
$ ghc -fglasgow-exts ... Occurs check: cannot construct the infinite type: t = t -> t1 Expected type: t Inferred type: t -> t1 In the first argument of `x', namely `x' In a lambda abstraction: x xWe can also type self-application with the recursive type T = T → T.
Andreas Abel,
http://www.tcs.informatik.uni-muenchen.de/~abel/fomega/
Last modified: Fri Feb 6 15:21:07 CET 2004 |