query about precedence: "$", the lazy function application operator

[Mark Phillips]

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

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