Re: drivers: random: Shift out-of-bounds in _mix_pool_bytes

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On 10/24/2014 06:01 AM, Peter Zijlstra wrote:
> On Mon, Oct 20, 2014 at 10:16:35AM -0400, Theodore Ts'o wrote:
>> > On Mon, Oct 20, 2014 at 04:09:30PM +0200, Daniel Borkmann wrote:
>>>> > > >
>>>> > > >It's triggering when input_rotate == 0, so UBSan complains about right shift in rol32()
>>>> > > >
>>>> > > >static inline __u32 rol32(__u32 word, unsigned int shift)
>>>> > > >{
>>>> > > >	return (word << shift) | (word >> (32 - shift));
>>>> > > >}
>>> > > 
>>> > > So that would be the case when the entropy store's input_rotate calls
>>> > > _mix_pool_bytes() for the very first time ... I don't think it's an
>>> > > issue though.
>> > 
>> > I'm sure it's not an issue, but it's still true that 
>> > 
>> > 	return (word << 0) | (word >> 32);
>> > 
>> > is technically not undefined, and while it would be unfortunate (and
>> > highly unlikely) if gcc were to say, start nethack, it's technically
>> > allowed by the C spec.  :-)
> In fact, n >> 32 == n.
> 
> #include <stdio.h>
> 
> int main(int argc, char **argv)
> {
> 	int i = atoi(argv[1]);
> 	int shift = atoi(argv[2]);
> 	printf("%x\n", i >> shift);
> 	return 0;
> }
> 
> $ ./shift 5 32
> 5
> 
> On x86 at least the shift ops simply mask out the upper bits and
> therefore the 32 == 0.
> 
> So you end up OR-ing the same value twice, which is harmless.
> 
> So no misbehaviour on the rol32() function.
> 
> I think I've ran into this before, in that case I did get fail because I
> did indeed expect the 0 and things didn't work out.

i >> 32 may happen to be "i", but is there anything that prevents the compiler
from returning, let's say, 42?


Thanks,
Sasha
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