Messages in this thread | | | Date | Wed, 14 Mar 2018 11:54:52 -0700 | From | Ram Pai <> | Subject | Re: [PATCH 1/1 v2] x86: pkey-mprotect must allow pkey-0 |
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On Wed, Mar 14, 2018 at 10:51:26AM -0700, Dave Hansen wrote: > On 03/14/2018 10:14 AM, Ram Pai wrote: > > I look at key-0 as 'the key'. It has special status. > > (a) It always exist. > > Do you mean "is always allocated"?
always allocated and cannot be freed. Hence always exists.
If we let it freed, than yes 'it is always allocated', but may not 'always exist'.
> > > (b) it cannot be freed. > > This is the one I'm questioning.
this is a philosophical question. Should we allow the application shoot-its-own-feet or help it from doing so. I tend to gravitate towards the later.
> > > (c) it is assigned by default. > > I agree on this totally. :)
good. we have some common ground :)
> > > (d) its permissions cannot be modified. > > Why not? You could pretty easily get a thread going that had its stack > covered with another pkey and that was being very careful what it > accesses. It could pretty easily set pkey-0's access or write-disable bits.
ok. I see your point. Will not argue against it.
> > > (e) it bypasses key-permission checks when assigned. > > I don't think this is necessary. I think the only rule we *need* is: > > pkey-0 is allocated implicitly at execve() time. You do not > need to call pkey_alloc() to allocate it.
And can be explicitly associated with any address range ?
> > > An arch need not necessarily map 'the key-0' to its key-0. It could > > internally map it to any of its internal key of its choice, transparent > > to the application. > > I don't understand what you are saying here.
I was trying to disassociate the notion that "application's key-0 means hardware's key-0". Nevermind. its not important for this discussion.
-- Ram Pai
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