Re: [dm-devel] [PATCH v3 9/9] crypto: shash: Remove VLA usage in unaligned hashing

Kees Cook
2018-07-01T17:04:59Z

On Sat, Jun 30, 2018 at 12:03 AM, Eric Biggers <ebiggers3@gmail.com> wrote:
> On Thu, Jun 28, 2018 at 05:28:43PM -0700, Kees Cook wrote:
>> @@ -88,11 +81,13 @@ static int shash_update_unaligned(struct shash_desc *desc, const u8 *data,
>> unsigned long alignmask = crypto_shash_alignmask(tfm);
>> unsigned int unaligned_len = alignmask + 1 -
>> ((unsigned long)data & alignmask);
>> - u8 ubuf[shash_align_buffer_size(unaligned_len, alignmask)]
>> - __aligned_largest;
>> + u8 ubuf[MAX_ALGAPI_ALIGNMASK + 1];
>> u8 *buf = PTR_ALIGN(&ubuf[0], alignmask + 1);
>> int err;
>>
>> + if (WARN_ON(buf + unaligned_len > ubuf + sizeof(ubuf)))
>> + return -EINVAL;
>> +
>
> How is 'ubuf' guaranteed to be large enough? You removed the __aligned
> attribute, so 'ubuf' can have any alignment. So the aligned pointer 'buf' may
> be as high as '&ubuf[alignmask]'. Then, up to 'alignmask' bytes of data will be
> copied into 'buf'... resulting in up to '2 * alignmask' bytes needed in 'ubuf'.
> But you've only guaranteed 'alignmask + 1' bytes.

Hm, good point. Adding __aligned(MAX_ALGAPI_ALIGNMASK + 1) looks to
fix this, yes?

Also, if __aligned() is used here, can't PTR_ALIGN() be dropped? (I
think you pointed this out earlier.)

Also, is "unaligned_len" being calculated correctly? Let's say
alignmask is 63. If data is binary ...111111, then unaligned_len will
be 64 - 63 == 1, which is fine: we copy 1 byte out, bump the address
by 1, and we're happily aligned to ...000000. If data is ...000000,
then unaligned_len will be 64. But it should be 0. Shouldn't this be:

unsigned int unaligned_len;

unaligned_len = (unsigned long)data & alignmask;
if (unaligned_len)
unaligned_len = alignmask + 1 - unaligned_len;

And then ubuf only needs to be MAX_ALGAPI_ALIGNMASK, without the +1?

-Kees

--
Kees Cook
Pixel Security