Re: [Haskell-cafe] question regarding the $ apply operator

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 ($).