> Hi,
>
> I think a lone ';' will match empty statement case, so something like this
> should work:
>
> @@
> statement S;
> @@
> (
>  ;
> |
>  S
> -;
> )

It's a nice idea, but I don't think it works.  The problem is that the
match of the empty statement will be at a different place in the
control-flow graph than the match of S;, because S; is two separate
statements.  So the disjunction doesn't realize that the two matches
overlap.

julia
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