[Cocci] Not pointer expression

Julia Lawall julia.lawall at lip6.fr
Thu Feb 8 15:44:52 CET 2018



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 at p

@ok@
position p != bad.p;
type T;
T e;
@@

*e at p

If you are actually interested in structure types, you could also say

struct e;

julia



> --
> Francois-Xavier
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