On Mon, May 03, 2004 at 08:19:19PM +0200, Holger Schulz wrote:
> I've following declarations
> 
> ---cut---
> type rat = int * int;;
> 
> exception Denominator_Zero;;
> 
> let rec ggT : (int * int) -> int = fun (x,y) -> if y = 0 then x else ggT 
> (y, x mod y)
> 
> let kgV : int * int -> int = fun (x,y) -> (x / ggT (x,y)) * y 
> 
> let kuerze : rat -> rat = function (m,n)  -> if n = 0 then raise 
> Denominator_Zero
> 				else let g=ggT (m,n) in (m div g,n div g)
> ---cut---
> 
> Loading it with #use give the following error message, which I do not 
> understand. Could anyone help me?

> This expression is not a function, it cannot be applied

Sure. Syntactically speaking, the expression 'm div g' appears to be a
call to function 'm' with arguments 'div' and 'g'. Of course, this
doesn't work because 'm' is an integer.

But actually I think your confusion is elsewhere. It looks like you were
trying to use 'div' as a binary operator; as far as I know, there is no
such operator. Try 'm / g' instead.

By the way, when you get error messages like this, it is helpful to pay
close attention to the *exact* position of the error--e.g., in this
case, the message gave a range of characters pointing to 'm' (and maybe
that character was underlined in the toplevel, too), so you know that
the toplevel or compiler expected 'm' to be a function.

-- 
Matt Gushee                 When a nation follows the Way,
Englewood, Colorado, USA    Horses bear manure through
mgushee@havenrock.com           its fields;
http://www.havenrock.com/   When a nation ignores the Way,
                            Horses bear soldiers through
                                its streets.
                                
                            --Lao Tzu (Peter Merel, trans.)

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