Re: Bug in herd7 [Was: Re: Litmus test for question from Al Viro]

[Akira Yokosawa <akiyks@gmail.com>]

On Sat, 3 Oct 2020 13:13:38 -0400, Alan Stern wrote:
> On Sun, Oct 04, 2020 at 12:16:31AM +0900, Akira Yokosawa wrote:
>> Hi Alan,
>>
>> Just a minor nit in the litmus test.
>>
>> On Sat, 3 Oct 2020 09:22:12 -0400, Alan Stern wrote:
>>> To expand on my statement about the LKMM's weakness regarding control 
>>> constructs, here is a litmus test to illustrate the issue.  You might 
>>> want to add this to one of the archives.
>>>
>>> Alan
>>>
>>> C crypto-control-data
>>> (*
>>>  * LB plus crypto-control-data plus data
>>>  *
>>>  * Expected result: allowed
>>>  *
>>>  * This is an example of OOTA and we would like it to be forbidden.
>>>  * The WRITE_ONCE in P0 is both data-dependent and (at the hardware level)
>>>  * control-dependent on the preceding READ_ONCE.  But the dependencies are
>>>  * hidden by the form of the conditional control construct, hence the 
>>>  * name "crypto-control-data".  The memory model doesn't recognize them.
>>>  *)
>>>
>>> {}
>>>
>>> P0(int *x, int *y)
>>> {
>>> 	int r1;
>>>
>>> 	r1 = 1;
>>> 	if (READ_ONCE(*x) == 0)
>>> 		r1 = 0;
>>> 	WRITE_ONCE(*y, r1);
>>> }
>>>
>>> P1(int *x, int *y)
>>> {
>>> 	WRITE_ONCE(*x, READ_ONCE(*y));
>>
>> Looks like this one-liner doesn't provide data-dependency of y -> x on herd7.
> 
> You're right.  This is definitely a bug in herd7.
> 
> Luc, were you aware of this?
> 
>> When I changed P1 to
>>
>> P1(int *x, int *y)
>> {
>> 	int r1;
>>
>> 	r1 = READ_ONCE(*y);
>> 	WRITE_ONCE(*x, r1);
>> }
>>
>> and replaced the WRITE_ONCE() in P0 with smp_store_release(),
>> I got the result of:
>>
>> -----
>> Test crypto-control-data Allowed
>> States 1
>> 0:r1=0;
>> No
>> Witnesses
>> Positive: 0 Negative: 3
>> Condition exists (0:r1=1)
>> Observation crypto-control-data Never 0 3
>> Time crypto-control-data 0.01
>> Hash=9b9aebbaf945dad8183d2be0ccb88e11
>> -----
>>
>> Restoring the WRITE_ONCE() in P0, I got the result of:
>>
>> -----
>> Test crypto-control-data Allowed
>> States 2
>> 0:r1=0;
>> 0:r1=1;
>> Ok
>> Witnesses
>> Positive: 1 Negative: 4
>> Condition exists (0:r1=1)
>> Observation crypto-control-data Sometimes 1 4
>> Time crypto-control-data 0.01
>> Hash=843eaa4974cec0efae79ce3cb73a1278
>> -----
> 
> What you should have done was put smp_store_release in P0 and left P1 in 
> its original form.  That test should not be allowed, but herd7 says that 
> it is.

Yea, that was what I tried first, expecting the result of "Never".

> 
>> As this is the same as the expected result, I suppose you have missed another
>> limitation of herd7 + LKMM.
> 
> It would be more accurate to say that we all missed it.  :-)  (And it's 
> a bug in herd7, not a limitation of either herd7 or LKMM.)  How did you 
> notice it?

:-) :-) :-)

Well, I thought I had never seen a litmus test with such one-liner.
So I split the READ_ONCE() and WRITE_ONCE() into two lines and
got the expected result.

I don't expect much from herd7's C mode in the first place.
(No offense intended!)

 
>> By the way, I think this weakness on control dependency + data dependency
>> deserves an entry in tools/memory-model/Documentation/litmus-tests.txt.
>>
>> In the LIMITATIONS section, item #1 mentions some situation where
>> LKMM may not recognize possible losses of control-dependencies by
>> compiler optimizations.
>>
>> What this litmus test demonstrates is a different class of mismatch.
> 
> Yes, one in which LKMM does not recognize a genuine dependency because 
> it can't tell that some optimizations are not valid.
> 
> This flaw is fundamental to the way herd7 works.  It examines only one 
> execution at a time, and it doesn't consider the code in a conditional 
> branch while it's examining an execution where that branch wasn't taken.  
> Therefore it has no way to know that the code in the unexecuted branch 
> would prevent a certain optimization.  But the compiler does consider 
> all the code in all branches when deciding what optimizations to apply.

I see.

> 
> Here's another trivial example:
> 
> 	r1 = READ_ONCE(*x);
> 	if (r1 == 0)
> 		smp_mb();
> 	WRITE_ONCE(*y, 1);
> 
> The compiler can't move the WRITE_ONCE before the READ_ONCE or the "if" 
> statement, because it's not allowed to move shared memory accesses past 
> a memory barrier -- even if that memory barrier isn't always executed.  
> Therefore the WRITE_ONCE actually is ordered after the READ_ONCE, but 
> the memory model doesn't realize it.> 
>> Alan, can you come up with an update in this regard?
> 
> I'll write something.

Thanks!

        Akira

> 
> Alan
>