On 08/02/2018 15:44, Julia Lawall wrote:
>
>
> On Thu, 8 Feb 2018, Francois-Xavier Le Bail wrote:
>
>> Hi,
>>
>> expression *e;
>> define 'e' is a pointer expression.
>>
>> Is there a way to define 'e' is *not* a pointer expression?
>
> @@
> expression *e;
> expression e1;
> @@
>
> (
> e
> |
> *e1
> )
>
> The above will highlight all expressions that have not been inferred to
> have a pointer type. But note that that doesn't mean that it has been
> inferred that the expression as a non-pointer type. It could be that not
> enough information is available.
>
> Another option would be:
>
> @@
> expression *e;
> type T;
> T e1;
> @@
>
> (
> e
> |
> *e1
> )
>
> Now it should require that e1 actually has a known type and it is not a
> pointer type.
>
> Another approach would be:
>
> @bad@
> position p;
> expression *e;
> @@
>
> e@p
>
> @ok@
> position p != bad.p;
> type T;
> T e;
> @@
>
> *e@p
>
> If you are actually interested in structure types, you could also say
>
> struct e;
Thanks, I will do some tests with these different approaches.
--
Francois-Xavier
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