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.
> +{
> + /*
> + * 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?
> + cpumask_and(local_cpus, cpus, p->cpus_ptr))
> + return local_cpus;
> +
> + return NULL;
and return values need some explanation too.
Thanks.
--
tejun
