On 7/24/24 10:00 AM, Amery Hung wrote:
On Tue, Jul 23, 2024 at 5:32 PM Martin KaFai Lau <martin.lau@xxxxxxxxx> wrote:
On 7/14/24 10:51 AM, Amery Hung wrote:
@@ -21004,6 +21025,13 @@ static int do_check_common(struct bpf_verifier_env *env, int subprog)
mark_reg_known_zero(env, regs, BPF_REG_1);
}
+ if (env->prog->type == BPF_PROG_TYPE_STRUCT_OPS) {
+ ctx_arg_info = (struct bpf_ctx_arg_aux *)env->prog->aux->ctx_arg_info;
+ for (i = 0; i < env->prog->aux->ctx_arg_info_size; i++)
+ if (ctx_arg_info[i].refcounted)
+ ctx_arg_info[i].ref_obj_id = acquire_reference_state(env, 0);
+ }
+
I think this will miss a case when passing the struct_ops prog ctx (i.e. "__u64
*ctx") to a global subprog. Something like this:
__noinline int subprog_release(__u64 *ctx __arg_ctx)
{
struct task_struct *task = (struct task_struct *)ctx[1];
int dummy = (int)ctx[0];
bpf_task_release(task);
return dummy + 1;
}
SEC("struct_ops/subprog_ref")
__failure
int test_subprog_ref(__u64 *ctx)
{
struct task_struct *task = (struct task_struct *)ctx[1];
bpf_task_release(task);
return subprog_release(ctx);;
}
SEC(".struct_ops.link")
struct bpf_testmod_ops subprog_ref = {
.test_refcounted = (void *)test_subprog_ref,
};
Thanks for pointing this out. The test did failed.
A quick thought is, I think tracking the ctx's ref id in the env->cur_state may
not be the correct place.
I think it is a bit tricky because subprogs are checked independently
and their state is folded (i.e., there can be multiple edges from the
main program to a subprog).
Maybe the verifier can rewrite the program: set the refcounted ctx to
NULL when releasing reference. Then, in do_check_common(), if it is a
global subprog, we mark refcounted ctx as PTR_MAYBE_NULL to force a
runtime check. How does it sound?
don't know how to get the ctx pointer to patch the code. It is not always in r1.
A case like this should still break even with the PTR_MAYBE_NULL marking in all
main and subprog (I haven't tried this one myself):
SEC("struct_ops/subprog_ref")
int test_subprog_ref(__u64 *ctx)
{
struct task_struct *task = (struct task_struct *)ctx[1];
if (task) {
subprog_release(ctx);
bpf_task_release(task);
}
return;
}
afaik, the global subprog is checked independently from the main prog and it
does not know the state of the main prog. Take a look at the subprog_is_global()
case in the check_func_call().
How about only acquire_reference_state() for the main prog? Yes, the global
subprog cannot do the bpf_kptr_xchg() and bpf_qdisc_skb_drop() but it can still
read the skb. The non-global subprog (static) should work though (please test).
I don't have other better idea. May be Alexei can provide some guidance here?
