The theory behind vdso randomization is that it's mapped at a random
offset above the top of the stack.  To avoid wasting a page of
memory for an extra page table, the vdso isn't supposed to extend
past the lowest PMD into which it can fit.  Other than that, the
address should be a uniformly distributed address that meets all of
the alignment requirements.

The current algorithm is buggy: the vdso has about a 50% probability
of being at the very end of a PMD.  The current algorithm also has a
decent chance of failing outright due to incorrect handling of the
case where the top of the stack is near the top of its PMD.

This fixes the implementation.  The paxtest estimate of vdso
"randomisation" improves from 11 bits to 18 bits.  (Disclaimer: I
don't know what the paxtest code is actually calculating.)

It's worth noting that this algorithm is inherently biased: the vdso
is more likely to end up near the end of its PMD than near the
beginning.  Ideally we would either nix the PMD sharing requirement
or jointly randomize the vdso and the stack to reduce the bias.

In the mean time, this is a considerable improvement with basically
no risk of compatibility issues, since the allowed outputs of the
algorithm are unchanged.

As an easy test, doing this:

for i in `seq 10000`
  do grep -P vdso /proc/self/maps |cut -d- -f1
done |sort |uniq -d

used to produce lots of output (1445 lines on my most recent run).
A tiny subset looks like this:

7fffdfffe000
7fffe01fe000
7fffe05fe000
7fffe07fe000
7fffe09fe000
7fffe0bfe000
7fffe0dfe000

Note the suspicious fe000 endings.  With the fix, I get a much more
palatable 76 repeated addresses.
Reviewed-by: Kees Cook <keescook@chromium.org>
Cc: stable@vger.kernel.org
Signed-off-by: Andy Lutomirski <luto@amacapital.net>