14 Feb
2005
14 Feb
'05
11 a.m.
Lennart Augustsson wrote (on Mon, 14 Feb 2005 at 14:55): > Any definition can be made point free if you have a > complete combinator base at your disposal, e.g., S and K.
> Haskell has K (called const), but lacks S. S could be > defined as > spread f g x = f x (g x) > Given that large set of Haskell prelude functions I would > not be surprised if spread could already be defined point > free in Haskell. :) It sometimes surprises me the prelude doesn't have diag f x = f x x (aka W. It already has B, C, K and I: (.), flip, const and id.) Peter Hancock