On Mon, Dec 9, 2013 at 1:13 PM, Jiri Kosina <[email protected]> wrote:
> On Mon, 9 Dec 2013, H.J. Lu wrote:
>
>> >> Normally, a PIE executable has zero virtual address on the first PT_LOAD
>> >> segment and kernel will load such executable at random address when
>> >> randomization is enabled.  If randomization is disabled, kernel will load
>> >> it at a fixed address.  But if a PIE executable has non-zero virtual
>> >> address on the first PT_LOAD segment, kernel will load such executable
>> >> at the non-zero virtual address when randomization is enabled. But when
>> >> randomization is disabled, kernel ignores the non-zero virtual address
>> >> at the non-zero virtual address when randomization is enabled.
>> >
>> > Hmm ... isn't actually this the thing that needs to be fixed instead?
>> >
>> > IOW, when randomization is enabled, is there a reason not to load on
>> > randomized address? (even if the first PT_LOAD segment has non-zero
>> > vaddr?)
>>
>> No, please don't do that.  Normally, PIE has zero load address and kernel
>> can load it anywhere.  There are multiple reasons why PIE has non-zero
>> load address.  Saying you need to load a program above 4GB under x86-64,
>> you can't do that with normal dynamic executable.  PIE with non-zero load
>> address is the only way to do that on x86-64.
>
> Hmm, so if it's because of 4G PT_LOAD limit, how about at least adding

Yes.

> randomized offset to the supplied vaddr?

Yes, people who build PIE with non-zero vaddr can use
 randomized vaddr. But kernel must follow the non-zero vaddr.

> PT_LOAD being non-zero causing randomization to be turned off seems like
> quite unexpected behavior to me, with a great potential to cause a lot of
> confusion.
>

There should be no difference between dynamic executable and PIE
with non-zero vaddr when choosing where to load them.

-- 
H.J.
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