> applySkip i f ls = (take i) ls ++ f (drop i ls) > > But the following doesn't: > > applySkip i f ls = (take i) ls ++ f $ drop i ls
The issue is with operator precedence. The above is equivalent to: > applySkip i f ls = ((take i) ls ++ f) (drop i ls) (++) binds more strongly than ($). _______________________________________________ Haskell-Cafe mailing list [email protected] http://www.haskell.org/mailman/listinfo/haskell-cafe
