On Mon, Apr 16, 2018 at 9:57 PM, Michal Hocko <mhocko@xxxxxxxxxx> wrote:
> On Mon 16-04-18 21:30:09, Jann Horn wrote:
>> On Mon, Apr 16, 2018 at 9:18 PM, Michal Hocko <mhocko@xxxxxxxxxx> wrote:
> [...]
>> > Yes, reasonably well written application will not have this problem.
>> > That, however, requires an external synchronization and that's why
>> > called it error prone and racy. I guess that was the main motivation for
>> > that part of the man page.
>>
>> What requires external synchronization? I still don't understand at
>> all what you're talking about.
>>
>> The following code:
>>
>> void *try_to_alloc_addr(void *hint, size_t len) {
>> char *x = mmap(hint, len, ...);
>> if (x == MAP_FAILED) return NULL;
>> if (x == hint) return x;
>
> Any other thread can modify the address space at this moment.
But not parts of the address space that were returned by this mmap() call.
> Just
> consider that another thread would does mmap(x, MAP_FIXED) (or any other
> address overlapping [x, x+len] range)
If the other thread does that without previously having created a
mapping covering the area in question, that would be a bug in the
other thread. MAP_FIXED on an unmapped address is almost always a bug
(excluding single-threaded cases with no library code, and even then
it's quite weird) - for example, any malloc() call could also cause
libc to start using the memory range you're trying to map with
MAP_FIXED.
> becaus it is seemingly safe as x
> != hint.
I don't understand this part. Are you talking about a hypothetical
scenario in which a programmer attempts to segment the virtual memory
space into areas that are exclusively used by threads without creating
memory mappings for those areas?
> This will succeed and ...
>> munmap(x, len);
> ... now you are munmaping somebody's else memory range
>
>> return NULL;
>
> Do code _is_ buggy but it is not obvious at all.
>
>> }
>>
>> has no need for any form of external synchronization.
>
> If the above mmap/munmap section was protected by a lock and _all_ other
> mmaps (direct or indirect) would use the same lock then you are safe
> against that.
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