Re: bpf: undefined shift in __bpf_prog_run

[Alexei Starovoitov <alexei.starovoitov@gmail.com>]

On Fri, Dec 04, 2015 at 12:44:09PM -0800, Kostya Serebryany wrote:
> On Fri, Dec 4, 2015 at 12:35 PM, Alexei Starovoitov <
> alexei.starovoitov@gmail.com> wrote:
> 
> > On Fri, Dec 04, 2015 at 08:48:57PM +0100, Dmitry Vyukov wrote:
> > >
> > > For example, a compiler can assume that result of left shift is larger
> > > or equal to first operand, which in turn can allow it to elide some
> > > bounds check in code, which in turn can lead to an exploit. I am not
> > > saying that this particular pattern is present in the code, what I
> > > want to say is that such undefined behaviors can lead to very
> > > unpredictable and unexpected consequences.
> >
> > Within bpf it cannot.
> > shift is not used in any memory or bounds operations.
> > so reg <<= 1234 cannot be exploited.
> >
> 
> I afraid this is not that simple.
> In C, undefined behavior applies to the entire program, not just to a
> single instruction.
> My favorite example:
> http://stackoverflow.com/questions/7682477/why-does-integer-overflow-on-x86-with-gcc-cause-an-infinite-loop
> Here an undefined behavior in one instruction causes *other* instructions
> to misbehave.

that's actually not related example. There compiler takes advantage of undefined
behavior which is very typical for compiler to do.
for(int i = 0; i < 100; i++)
and
for(unsigned int i = 0; i < 100; i++)
are very different loops from compiler point of view.

but that is not applicable in bpf world.
there are no loops in bpf in the first place.