>> On Nov 25, 2018, at 6:53 AM, Jarkko Sakkinen 
>> <jarkko.sakki...@linux.intel.com> wrote:
>> 
>> On Sat, Nov 24, 2018 at 09:21:14AM -0800, Jarkko Sakkinen wrote:
>> On Thu, Nov 22, 2018 at 07:21:08AM -0800, Andy Lutomirski wrote:
>>>> At a high level, addressing these issues is straight forward.  First,
>>>> the driver needs to support authorization equivalent to that which is
>>>> implemented in the current Intel Launch Enclave, ie. control over the
>>>> SGX_FLAGS_PROVISION_KEY attribute.
>>> 
>>> I agree, hence my email :)
>> 
>> Started to scratch my head that is it really an issue that any enclave
>> can provision in the end?
>> 
>> Direct quote from your first response:
>> 
>> "In particular, the ability to run enclaves with the provisioning bit set
>> is somewhat sensitive, since it effectively allows access to a stable
>> fingerprint of the system."
>> 
>> As can be seen from the key derivation table this does not exactly hold
>> so you should refine your original argument before we can consider any
>> type of change.
>> 
>> I just don't see what it is so wrong for any enclave to be able to tell
>> that it really is an enclave.
> 
> I mean I can understand why Greg wants LE although I don't understand
> what benefit does it bring to anyone to lock in for enclave to allow
> to identify itself.
> 
> What you are proposing does not really bring any additional security if
> we consider a threat model where the kernel is an adversary but it makes
> the software stack more clanky to use.

Agreed. What I’m proposing adds additional security if the kernel is *not* 
compromised.

There are other ways to accomplish it that might be better in some respects.  
For example, there could be /dev/sgx and /dev/sgx_rights/provision.  The former 
exposes the whole sgx API, except that it doesn’t allow provisioning by 
default. The latter does nothing by itself. To run a provisioning enclave, you 
open both nodes, then do something like:

ioctl(sgx, SGX_IOC_ADD_RIGHT, sgx_provisioning);

This requires extra syscalls, but it doesn’t have the combinatorial explosion 
problem.

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