Hi

Isn't _|_ = \x -> _|_?

_|_ `seq` () = _|_
(\x -> _|_) `seq` () = ()

Whether this is the fault of seq or not is your call...

Thanks

Neil
_______________________________________________
Haskell-Cafe mailing list
Haskell-Cafe@haskell.org
http://www.haskell.org/mailman/listinfo/haskell-cafe

Reply via email to