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

par2 :: (a -> b -> c) -> a -> b -> c

...

...

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

Here is the dual: 'par' implemented in terms of parallel application: a `par` b = par2 (\x y-> y) a b

($!) :: (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.

It's not so difficult. (And you may see some similarity with the above defn of par.) a `seq` b = (\x y-> y) $! a b Regards, Malcolm