[Haskell-cafe] Re: 'par' - why has it the type a -> b -> b ?
barsoap at web.de
Mon Sep 29 10:53:02 EDT 2008
Henning Thielemann <lemming at henning-thielemann.de> 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.
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