On Mon, Mar 17, 2025 at 08:22:35AM -1000, Tejun Heo wrote:
> Hello,
>
> On Mon, Mar 17, 2025 at 06:53:24PM +0100, Andrea Righi wrote:
> > +/*
> > + * Return the subset of @cpus that task @p can use or NULL if none of the
> > + * CPUs in the @cpus cpumask can be used.
> > + */
> > +static const struct cpumask *task_cpumask(const struct task_struct *p, const struct cpumask *cpus,
> > + struct cpumask *local_cpus)
>
> task_cpus_allowed_and()? It also would help to add comment explaining the
> parameters as the function is a bit unusual.
Ack.
>
> > +{
> > + /*
> > + * If the task is allowed to run on all CPUs, simply use the
> > + * architecture's cpumask directly. Otherwise, compute the
> > + * intersection of the architecture's cpumask and the task's
> > + * allowed cpumask.
> > + */
> > + if (!cpus || p->nr_cpus_allowed >= num_possible_cpus() ||
> > + cpumask_subset(cpus, p->cpus_ptr))
> > + return cpus;
> > +
> > + if (!cpumask_equal(cpus, p->cpus_ptr) &&
>
> Hmm... isn't this covered by the preceding cpumask_subset() test? Here, cpus
> is not a subset of p->cpus_ptr, so how can it be the same as p->cpus_ptr?
Oh that's right, I missed that between all the refactoring, thanks for
catching it. Will remove it.
>
> > + cpumask_and(local_cpus, cpus, p->cpus_ptr))
> > + return local_cpus;
> > +
> > + return NULL;
>
> and return values need some explanation too.
Ok.
Thanks,
-Andrea
