Hi, As I understand it, function application has highest precedence (10 even!) whereas $, the operator which does the same thing, has lowest precedence (0 even!). But there's something that doesn't make sense to me. Suppose I have functions f :: Int -> Int f x -> x * x g :: Int -> Int g x -> x + 1 The lazy application operator "$" allows me to do: f $ g x instead of f (g x) But I don't understand why it works this way! Let me explain. f is a function, and application has highest precedence, so unless it sees a bracket, it should take the next thing it sees as an argument. Lo and behold the next thing it sees is "$", which is not a bracket! So it should try and apply f to argument $. But oh dear, $ is a function (operator actually); it is not an Int at all! So an error should be reported! So what am I not understanding properly? Thanks, Mark. -- Dr Mark H Phillips Research Analyst (Mathematician) AUSTRICS - Smarter Scheduling Solutions - www.austrics.com Level 2, 50 Pirie Street, Adelaide SA 5000, Australia Phone +61 8 8226 9850 Fax +61 8 8231 4821 Email mark@austrics.com.au