[Haskell-cafe] Re: Re-order type (flip map)

Max Rabkin max.rabkin at gmail.com
Sun Oct 10 19:44:20 EDT 2010


On Mon, Oct 11, 2010 at 00:51, Ozgur Akgun <ozgurakgun at gmail.com> wrote:
> My point was: you need to find/define two operators, not just one. That
> still holds :)

No it doesn't.

f $ g $ h $ x == f (g (h x)) == f . g . h $ x == x $$ h $$ g $$ f

if you have the correct associativity for ($$)

--Max


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