> On 5 Sep 2025, at 10:15, Jakub Jelinek <ja...@redhat.com> wrote:
>
> 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.
Well the intent of the testcase was to demonstate that we correctly prevent
time-traveling UB from doing so. We cannot use any output calls since those
already behave as checkpoints - manipulating a global seemed a reasonable
solution.
Iain
>
> Jakub
>
