On Mon, Feb 3, 2014 at 10:25 AM, Martin Frb <laza...@mfriebe.de> wrote:
> It does not state if or if not it is an operator. And also give no
> indication on its precedence, or if it is applied before or after operators.
>
> This is not so much about he -a^ (which is a (very) constructed case), but
> about @a^ which is possible (not doing much so)
>
> So if {$T-}
> a : pinteger;
> @a^ gives an untyped pointer (same as "pointer(a)" ? / but shorter)
> But only because it is @(a^) and not (@a)^ . The second would only compile
> wit {$T+}
>
Both still both expressions are parsed the same (not if $T is enabled or
not).
If you treat ^ as a part of identified, you'll be able to parse the
expression.
thanks,
Dmitry
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