Hi, Newbie question. Given the inferred type for square, the inferred types for quad1, quad2 and quad3 are what I would expect. Is there a straightforward explanation (i.e. one that a newbie would understand) as to why the inferred type for quad4 is less general? Regards, dl -- GHC Interactive, version 6.4, for Haskell 98. Prelude> let square x = x * x Prelude> :t square square :: (Num a) => a -> a Prelude> let quad1 x = square (square x) Prelude> :t quad1 quad1 :: (Num a) => a -> a Prelude> let quad2 x = square $ square x Prelude> :t quad2 quad2 :: (Num a) => a -> a Prelude> let quad3 x = (square . square) x Prelude> :t quad3 quad3 :: (Num a) => a -> a Prelude> let quad4 = square . square Prelude> :t quad4 quad4 :: Integer -> Integer ___________________________________________________________ Yahoo! Photos NEW, now offering a quality print service from just 8p a photo http://uk.photos.yahoo.com