Re: Calling through function pointer--syntax question

[Jeffrey Carter <jrcarter@acm.org>]

Adam Beneschan wrote:
> 
> Sorry, I'm not following . . . the default parameter in my example is on the
> function "func", not on the function *pointer*.  In the call
>     P (func (5));
> since an implicit .all is provided prior to "(", why couldn't this be
> interpreted as
>     P (func.all (5));
> [here it's clear that the default parameter is passed to "func"]?

The implicit ".all" also means the "(" starts the parameter list of the
function being pointed to. You never get an implicit ".all" if there are
no explicit parameters.

-- 
Jeff Carter
"Now go away or I shall taunt you a second time."
Monty Python & the Holy Grail