[Haskell-cafe] Re: 'par' - why has it the type a -> b -> b ?

Henning Thielemann wrote: par2 :: (a -> b -> c) -> a -> b -> c

...

par2 f x y = f x (par x y)

($!) :: (a -> b) -> a -> b f $! x = x `seq` f x It's terseness vs. maximum composability. I don't even want to think about implementing seq in terms of $!, makes my brain twist. -- (c) this sig last receiving data processing entity. Inspect headers for copyright history. All rights reserved. Copying, hiring, renting, performance and/or broadcasting of this signature prohibited.