On 22/02/2011, at 7:46 PM, z_axis wrote: > > I want to split "2,15,33,0,8,1,16,18" to ([2,15,33],[8,1,16,18]). the "0" > will by discarded. > > However, it seems that there isnot any standard function to do it.
let (front,(_:back)) = List.span (/= 0) xs in (front,back) What do you plan to do if there is no 0 there? [Yes, I know this can be done in point free style. I hereby announce defeat in code golf.] _______________________________________________ Haskell-Cafe mailing list [email protected] http://www.haskell.org/mailman/listinfo/haskell-cafe
