Andrii Nakryiko <andrii.nakryiko@xxxxxxxxx> writes:
> On Mon, Aug 18, 2025 at 10:06 AM Roman Gushchin
> <roman.gushchin@xxxxxxxxx> wrote:
>>
>> Implement a new bpf_psi_create_trigger() bpf kfunc, which allows
>> to create new psi triggers and attach them to cgroups or be
>> system-wide.
>>
>> Created triggers will exist until the struct ops is loaded and
>> if they are attached to a cgroup until the cgroup exists.
>>
>> Due to a limitation of 5 arguments, the resource type and the "full"
>> bit are squeezed into a single u32.
>>
>> Signed-off-by: Roman Gushchin <roman.gushchin@xxxxxxxxx>
>> ---
>> kernel/sched/bpf_psi.c | 84 ++++++++++++++++++++++++++++++++++++++++++
>> 1 file changed, 84 insertions(+)
>>
>> diff --git a/kernel/sched/bpf_psi.c b/kernel/sched/bpf_psi.c
>> index 2ea9d7276b21..94b684221708 100644
>> --- a/kernel/sched/bpf_psi.c
>> +++ b/kernel/sched/bpf_psi.c
>> @@ -156,6 +156,83 @@ static const struct bpf_verifier_ops bpf_psi_verifier_ops = {
>> .is_valid_access = bpf_psi_ops_is_valid_access,
>> };
>>
>> +__bpf_kfunc_start_defs();
>> +
>> +/**
>> + * bpf_psi_create_trigger - Create a PSI trigger
>> + * @bpf_psi: bpf_psi struct to attach the trigger to
>> + * @cgroup_id: cgroup Id to attach the trigger; 0 for system-wide scope
>> + * @resource: resource to monitor (PSI_MEM, PSI_IO, etc) and the full bit.
>> + * @threshold_us: threshold in us
>> + * @window_us: window in us
>> + *
>> + * Creates a PSI trigger and attached is to bpf_psi. The trigger will be
>> + * active unless bpf struct ops is unloaded or the corresponding cgroup
>> + * is deleted.
>> + *
>> + * Resource's most significant bit encodes whether "some" or "full"
>> + * PSI state should be tracked.
>> + *
>> + * Returns 0 on success and the error code on failure.
>> + */
>> +__bpf_kfunc int bpf_psi_create_trigger(struct bpf_psi *bpf_psi,
>> + u64 cgroup_id, u32 resource,
>> + u32 threshold_us, u32 window_us)
>> +{
>> + enum psi_res res = resource & ~BPF_PSI_FULL;
>> + bool full = resource & BPF_PSI_FULL;
>> + struct psi_trigger_params params;
>> + struct cgroup *cgroup __maybe_unused = NULL;
>> + struct psi_group *group;
>> + struct psi_trigger *t;
>> + int ret = 0;
>> +
>> + if (res >= NR_PSI_RESOURCES)
>> + return -EINVAL;
>> +
>> +#ifdef CONFIG_CGROUPS
>> + if (cgroup_id) {
>> + cgroup = cgroup_get_from_id(cgroup_id);
>> + if (IS_ERR_OR_NULL(cgroup))
>> + return PTR_ERR(cgroup);
>> +
>> + group = cgroup_psi(cgroup);
>> + } else
>> +#endif
>> + group = &psi_system;
>
> just a drive-by comment while skimming through the patch set: can't
> you use IS_ENABLED(CONFIG_CGROUPS) and have a proper if/else with
> proper {} ?
Fixed.
It required defining cgroup_get_from_id() and cgroup_psi()
for !CONFIG_CGROUPS, but I agree, it's much better.
Thanks
>
>> +
>> + params.type = PSI_BPF;
>> + params.bpf_psi = bpf_psi;
>> + params.privileged = capable(CAP_SYS_RESOURCE);
>> + params.res = res;
>> + params.full = full;
>> + params.threshold_us = threshold_us;
>> + params.window_us = window_us;
>> +
>> + t = psi_trigger_create(group, ¶ms);
>> + if (IS_ERR(t))
>> + ret = PTR_ERR(t);
>> + else
>> + t->cgroup_id = cgroup_id;
>> +
>> +#ifdef CONFIG_CGROUPS
>> + if (cgroup)
>> + cgroup_put(cgroup);
>> +#endif
>> +
>> + return ret;
>> +}
>> +__bpf_kfunc_end_defs();
>> +
>> +BTF_KFUNCS_START(bpf_psi_kfuncs)
>> +BTF_ID_FLAGS(func, bpf_psi_create_trigger, KF_TRUSTED_ARGS)
>> +BTF_KFUNCS_END(bpf_psi_kfuncs)
>> +
>> +static const struct btf_kfunc_id_set bpf_psi_kfunc_set = {
>> + .owner = THIS_MODULE,
>> + .set = &bpf_psi_kfuncs,
>> +};
>> +
>> static int bpf_psi_ops_reg(void *kdata, struct bpf_link *link)
>> {
>> struct bpf_psi_ops *ops = kdata;
>> @@ -238,6 +315,13 @@ static int __init bpf_psi_struct_ops_init(void)
>> if (!bpf_psi_wq)
>> return -ENOMEM;
>>
>> + err = register_btf_kfunc_id_set(BPF_PROG_TYPE_STRUCT_OPS,
>> + &bpf_psi_kfunc_set);
>
> would this make kfunc callable from any struct_ops, not just this psi
> one?
It will. Idk how big of a problem it is, given that the caller needs
a trusted reference to bpf_psi. Also, is there a simple way to constrain
it? Wdyt?
