* Iain Sandoe:
>> On 5 Sep 2025, at 10:07, Florian Weimer <fwei...@redhat.com> wrote:
>>
>> * Iain Sandoe:
>>
>>> diff --git a/gcc/testsuite/g++.dg/cpp26/observable-checkpoint.C
>>> b/gcc/testsuite/g++.dg/cpp26/observable-checkpoint.C
>>> new file mode 100644
>>> index 00000000000..886cda7ae33
>>> --- /dev/null
>>> +++ b/gcc/testsuite/g++.dg/cpp26/observable-checkpoint.C
>>> @@ -0,0 +1,24 @@
>>> +// P1494R5+P3641R0 - Partial program correctness.
>>> +// { dg-do compile { target c++11 } }
>>> +// { dg-additional-options "-fdump-tree-optimized -Wno-return-type -O" }
>>> +// { dg-final { scan-tree-dump {\+\s42} "optimized" } }
>>> +// { dg-final { scan-tree-dump {__builtin_observable_checkpoint}
>>> "optimized" } }
>>> +
>>> +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?
>
> I don’t think it does.
Jakub and the compiler disagree. 8-)
I don't think the test case (without the static) is valid: the write to
x is not considered observable behavior by the standard. Even without a
closed-world assumption, there is no synchronization with other threads,
so if they produce observable behavior in relation to x, it does not
become part of the observable prefix.
Thanks,
Florian
