Re: Self-XORing BPF registers is undefined behavior

On 12/13/2018 02:18 PM, Daniel Borkmann wrote:

On 12/13/2018 01:24 PM, Alexander Potapenko wrote:

quoted

On Thu, Dec 13, 2018 at 1:20 PM Michal Kubecek [off-list ref] wrote:

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On Thu, Dec 13, 2018 at 12:59:36PM +0100, Michal Kubecek wrote:

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On Thu, Dec 13, 2018 at 12:00:59PM +0100, Alexander Potapenko wrote:

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Hi BPF maintainers,

some time ago KMSAN found an issue in BPF code which we decided to
suppress at that point, but now I'd like to bring it to your
attention.
Namely, some BPF programs may contain instructions that XOR a register
with itself.
This effectively results in the following C code:
  regs[BPF_REG_A] = regs[BPF_REG_A] ^ regs[BPF_REG_A];
or
  regs[BPF_REG_X] = regs[BPF_REG_X] ^ regs[BPF_REG_X];
being executed.

According to the C11 standard this is undefined behavior, so KMSAN
reports an error in this case.

Can you elaborate / quote the exact bit from C11 that KMSAN is referring
to? (The below that Michal was quoting or something else?)

Does that only refer to C11 standard? Note that kernel's Makefile +430
explicitly states 'std=gnu89' and not 'std=c11' [0].

  [0] https://git.kernel.org/pub/scm/linux/kernel/git/torvalds/linux.git/commit/?id=51b97e354ba9fce1890cf38ecc754aa49677fc89

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Can you quote the part of the standard saying this is undefined
behavior? I couldn't find anything else than

  If the value being stored in an object is read from another object
  that overlaps in any way the storage of the first object, then the
  overlap shall be exact and the two objects shall have qualified or
  unqualified versions of a compatible type; otherwise, the behavior
  is undefined.

(but I only have a draft for obvious reasons). I'm not sure what exactly
they mean by "exact overlap" and the standard doesn't seem to define
the term but if the two objects are actually the same, they certainly
have compatible types.

Here is an example for the overlap quoted above; I don't think this
applies to our case since it would be "exact". Quote [1]:

  struct S { int x; int y; };
  struct T { int z; struct S s; };
  union U { struct S f ; struct T g; } u;

  main(){
    u.f = u.g.s;
    return 0;
  }

  [1] https://bts.frama-c.com/print_bug_page.php?bug_id=945

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I think I understand now. You didn't want to say that the statement

  regs[BPF_REG_A] = regs[BPF_REG_A] ^ regs[BPF_REG_A];

as such is undefined behavior but that it's UB when regs[BPF_REG_A] is
uninitialized. Right?

Yes. Sorry for being unclear.
By default regs[] is uninitialized, so we need to initialize it before
using the register values.
I am also wondering if it's possible to simply copy the uninitialized
register values from regs[] to the userspace via maps.

Nope, not possible. And to elaborate on cBPF / eBPF cases:

# cat /proc/sys/net/core/bpf_jit_enable
0

For the eBPF case, lets take (test_verifier.c):

{
        "xor test",
        .insns = {
                BPF_ALU64_REG(BPF_XOR, BPF_REG_2, BPF_REG_2),
                BPF_MOV64_REG(BPF_REG_0, BPF_REG_2),
                BPF_EXIT_INSN(),
        },
        .result = ACCEPT,
        .retval = 0,
},

# ./test_verifier 0
#0/u xor test FAIL
Failed to load prog 'Permission denied'!
0: (af) r2 ^= r2
R2 !read_ok
#0/p xor test FAIL
Failed to load prog 'Permission denied'!
0: (af) r2 ^= r2
R2 !read_ok
Summary: 0 PASSED, 0 SKIPPED, 2 FAILED

And for the initialized case:

{
        "xor test",
        .insns = {
                BPF_MOV64_IMM(BPF_REG_2, 42),
                BPF_ALU64_REG(BPF_XOR, BPF_REG_2, BPF_REG_2),
                BPF_MOV64_REG(BPF_REG_0, BPF_REG_2),
                BPF_EXIT_INSN(),
        },
        .result = ACCEPT,
        .retval = 0,
},

# ./test_verifier 0
#0/u xor test OK
#0/p xor test OK
Summary: 2 PASSED, 0 SKIPPED, 0 FAILED

And cBPF (test_bpf.c):

{
        "xor test",
        .u.insns = {
                BPF_STMT(BPF_RET | BPF_A, 0)
        },
        CLASSIC,
        { },
        { { 0, 0 }, },
},

[...]
[88819.276142] test_bpf: #0 xor test jited:0 PASS
[...]

Given latter two use the same underlying interpreter implementation for
the 'good' and 'bad' case, I don't see how compiler would have any degree
of freedom to judge about the runtime dependent BPF insn input stream
here. If it would assume UB, then also 'good' xor case would be broken
and miscompiled which it is clearly not (even for some very obvious case
below). While KMSAN is analyzing runtime memory access, compiler cannot..

# cat foo.c
#include <stdio.h>
int main(void)
{
	int foo = foo ^ foo;
	printf("%u\n", foo);
	return 0;
}

# clang -Wall -O2 foo.c
foo.c:4:12: warning: variable 'foo' is uninitialized when used within its own initialization [-Wuninitialized]
        int foo = foo ^ foo;
            ~~~   ^~~
# ./a.out
0

disasm:

00000000004004d0 <main>:
  4004d0:	50                   	push   %rax
  4004d1:	bf 74 05 40 00       	mov    $0x400574,%edi
  4004d6:	31 f6                	xor    %esi,%esi
  4004d8:	31 c0                	xor    %eax,%eax
  4004da:	e8 f1 fe ff ff       	callq  4003d0 <printf@plt>
  4004df:	31 c0                	xor    %eax,%eax
  4004e1:	59                   	pop    %rcx
  4004e2:	c3                   	retq
  4004e3:	66 2e 0f 1f 84 00 00 	nopw   %cs:0x0(%rax,%rax,1)
  4004ea:	00 00 00
  4004ed:	0f 1f 00             	nopl   (%rax)

# gcc -Wall -O2 foo.c   (neither complains with using -Wextra)
# ./a.out
0

disasm:

0000000000000560 <main>:
 560:	48 8d 35 ad 01 00 00 	lea    0x1ad(%rip),%rsi        # 714 <_IO_stdin_used+0x4>
 567:	48 83 ec 08          	sub    $0x8,%rsp
 56b:	31 d2                	xor    %edx,%edx
 56d:	bf 01 00 00 00       	mov    $0x1,%edi
 572:	31 c0                	xor    %eax,%eax
 574:	e8 c7 ff ff ff       	callq  540 <__printf_chk@plt>
 579:	31 c0                	xor    %eax,%eax
 57b:	48 83 c4 08          	add    $0x8,%rsp
 57f:	c3                   	retq

Cheers,
Daniel