On Fri, Sep 05, 2025 at 11:07:01AM +0200, Florian Weimer wrote:
> > +int x = 0;
> > +
> > +int
> > +here_b_ub ()
> > +{
> > + // missing return triggers UB (we must ignore the warning for this
> > test).
> > +}
> > +
> > +int
> > +main ()
> > +{
> > + __builtin_printf (" start \n");
> > + x += 42;
> > + // Without this checkpoint the addition above is elided (along with the
> > rest
> > + // of main).
> > + __builtin_observable_checkpoint ();
> > + here_b_ub ();
> > +}
>
> What happens if you have this?
>
> static int x = 0;
>
> Does the linkage matter?
It does. If it is static, then it is just fine to optimize it away. You
can see all the code that accesses it.
BTW, I think this testcase should show why expanding it into RTL as nothing
is not a good idea. here_b_ub(); should be turned into
__builtin_unreachable(); call and there is no reason why to increment the
memory when it just invokes UB right after it without anything that could
observe the change.
Jakub
