On 10/1/24 6:26 PM, Alexei Starovoitov wrote:
On Tue, Oct 1, 2024 at 5:23 PM Kumar Kartikeya Dwivedi <memxor@xxxxxxxxx> wrote:
Makes sense, though will we have cases where hierarchical scheduling
attaches the same prog at different points of the hierarchy?
I'm not sure anyone was asking for such a use case.
Then the
limit of 4 may not be enough (e.g. say with cgroup nested levels > 4).
Well, 4 was the number from TJ.
Anyway the proposed pseudo code:
__bpf_prog_enter_recur_limited()
{
cnt = this_cpu_inc_return(*(prog->active));
if (cnt > 4) {
inc_miss
return 0;
}
// pass cnt into bpf prog somehow, like %rdx ?
// or re-read prog->active from prog
}
then in the prologue emit:
push rbp
mov rbp, rsp
if %rdx == 1
// main prog is called for the first time
mov rsp, pcpu_priv_stack_top
else
// 2+nd time main prog is called or 1+ time subprog
sub rsp, stack_size
if rsp < pcpu_priv_stack_bottom
goto exit // stack is too small, exit
fi
I have tried to implement this approach (not handling
recursion yet) based on the above approach. It works
okay with nested bpf subprogs like
main prog // set rsp = pcpu_priv_stack_top
subprog1 // some stack
subprog2 // some stack
The pcpu_priv_stack is allocated like
priv_stack_ptr = __alloc_percpu_gfp(1024 * 16, 8, GFP_KERNEL);
But whenever the prog called an external function,
e.g. a helper in this case, I will get a double fault.
An example could be
main prog // set rsp = pcpu_priv_stack_top
subprog1 // some stack
subprog2 // some stack
call bpf_seq_printf
(I modified bpf_iter_ipv6_route.c bpf prog for the above
purpose.)
I added some printk statements from the beginning of bpf_seq_printf and
nothing printed out either and of course traps still happens.
I tried another example without subprog and the mainprog calls
a helper and the same double traps happens below too.
The error log looks like
[ 54.024955] traps: PANIC: double fault, error_code: 0x0
[ 54.024969] Oops: double fault: 0000 [#1] PREEMPT SMP KASAN PTI
[ 54.024977] CPU: 3 UID: 0 PID: 1946 Comm: test_progs Tainted: G OE 6.11.0-10577-gf25c172fd840-dirty #968
[ 54.024982] Tainted: [O]=OOT_MODULE, [E]=UNSIGNED_MODULE
[ 54.024983] Hardware name: QEMU Standard PC (i440FX + PIIX, 1996), BIOS rel-1.14.0-0-g155821a1990b-prebuilt.qemu.org 04/01/2014
[ 54.024986] RIP: 0010:error_entry+0x1e/0x140
[ 54.024996] Code: ff ff 90 90 90 90 90 90 90 90 90 90 56 48 8b 74 24 08 48 89 7c 24 08 52 51 50 41 50 41 51 41 52 41 53 53 55 41 54 41 55 41 56 <41> 57 56 31 f6 31 d1
[ 54.024999] RSP: 0018:ffffe8ffff580000 EFLAGS: 00010806
[ 54.025002] RAX: f3f3f300f1f1f1f1 RBX: fffff91fffeb0044 RCX: ffffffff84201701
[ 54.025005] RDX: fffff91fffeb0044 RSI: ffffffff8420128d RDI: ffffe8ffff580178
[ 54.025007] RBP: ffffe8ffff580140 R08: 0000000000000000 R09: 0000000000000000
[ 54.025009] R10: 0000000000000000 R11: 0000000000000000 R12: dffffc0000000000
[ 54.025010] R13: 1ffffd1fffeb0014 R14: 0000000000000003 R15: ffffe8ffff580178
[ 54.025012] FS: 00007fd076525d00(0000) GS:ffff8881f7180000(0000) knlGS:0000000000000000
[ 54.025015] CS: 0010 DS: 0000 ES: 0000 CR0: 0000000080050033
[ 54.025017] CR2: ffffe8ffff57fff8 CR3: 000000010cd80002 CR4: 0000000000370ef0
[ 54.025021] DR0: 0000000000000000 DR1: 0000000000000000 DR2: 0000000000000000
[ 54.025022] DR3: 0000000000000000 DR6: 00000000fffe0ff0 DR7: 0000000000000400
[ 54.025024] Call Trace:
[ 54.025026] <#DF>
[ 54.025028] ? __die_body+0xaf/0xc0
[ 54.025032] ? die+0x2f/0x50
[ 54.025036] ? exc_double_fault+0x73/0x80
[ 54.025040] ? asm_exc_double_fault+0x23/0x30
[ 54.025044] ? common_interrupt_return+0xb1/0xcc
[ 54.025048] ? asm_exc_page_fault+0xd/0x30
[ 54.025051] ? error_entry+0x1e/0x140
[ 54.025055] </#DF>
[ 54.025056] Modules linked in: bpf_testmod(OE)
[ 54.025061] ---[ end trace 0000000000000000 ]---
Maybe somebody could give a hint why I got a double fault
when calling external functions (outside of bpf programs)
with allocated stack?
Since stack bottom/top are known at JIT time we can
generate reliable stack overflow checks.
Much better than guard pages and -fstack-protector.
The prog can alloc percpu
(stack size of main prog + subprogs + extra) * 4
and it likely will be enough.
If not, the stack protection will gently exit the prog
when the stack is too deep.
kfunc won't have such a check, so we need a buffer zone.
Can have a guard page too, but feels like overkill.
