Secret registers can be leaked under certain circumstances

[HopeDevote]

Malicious modification of the IR does not cause changes in the program results, leading to the leakage of secret register (s9) values

High-level source code is:

a = sint(1)
b = sint(2)
c = sint(3)

a.update(c)
b.update(a > sfix(8))

checksum = a + b + c
print_ln('checksum: %s', checksum.reveal())

The source IR is as follows:

...
subs s9, s4, s7 # 8
LTZ (s8, s9, 32) # 9
....

Full source IR:
source IR.txt

The maliciously modified IR is as follows：

....
subml s14, s4, c2 # 10
LTZ (s8, s9, 32) # 11
...

Full modified IR:
modified IR.txt

When running VMs based on these two IRs, the results are consistent.
However, we can see from the modified IR that s9 is not initialized. Based on previous experience, we know that VM generally assigns uninitialized registers to 0. When the program result does not change, we can infer that the value of register s9 in the source IR is not less than 0, which means that the value of s9 in the source IR is leaked.

Is this a bug?

Thanks!

[mkskeller]

What do you mean by running VMs? Which protocol? LTZ is not valid instruction for all proper protocols. More generally, changing the instructions would imply a corrupted party in MPC security, so one would expect only protocols with malicious security to protect against that.

[HopeDevote]

What do you mean by running VMs? Which protocol? LTZ is not valid instruction for all proper protocols. More generally, changing the instructions would imply a corrupted party in MPC security, so one would expect only protocols with malicious security to protect against that.

Thanks for your reply！
I run this: emulate.x

I understand that each party needs to generate consistent IR after compiling the source code. Does that mean that if the compilation process is normal, undefined registers will not appear?

[mkskeller]

Thanks for your reply！ I run this: emulate.x

I don't think that using emulate.x says much about the security posture because there is no protocol being executed.

I understand that each party needs to generate consistent IR after compiling the source code. Does that mean that if the compilation process is normal, undefined registers will not appear?

Yes. The compiler emits a warning if there are undefined registers, and with the argument --stop it aborts.