No, wrong. I am speaking nonsense here. Of course one also needs to define a *forward* function composition operator to get the effect you originally wanted.
My point was: you need to find/define two operators, not just one. That still holds :) Best, On 10 October 2010 23:47, Ozgur Akgun <ozgurak...@gmail.com> wrote: > > On 10 October 2010 22:32, Johannes Waldmann > <waldm...@imn.htwk-leipzig.de>wrote: > >> Oh, and while we're at it - are there standard notations >> for "forward" function composition and application? >> >> I mean instead of h . g . f $ x >> I'd sometimes prefer x ? f ? g ? h >> but what are the "?" >> > > While asking you use the same symbol for function composition, and > something like inverse function application. I don't think there exists an > operator ?, such that h . g . f $ x is equivalent to x ? f ? g ? h. > > But you can simply define an inverse function application like the > following and have a close enough alternative, > > ($$) :: a -> (a -> b) -> b > ($$) = flip ($) > infixl 5 $$ > > Now the following two expression are identical, I suppose: > > h . g . f $ x > x $$ f . g . h > > Cheers, > Ozgur > -- Ozgur Akgun
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