On 15/09/2015 20:42, Caldarale, Charles R wrote:
>> From: Mark Thomas [mailto:ma...@apache.org]
>> Subject: Re: RV-Predict bugs
>
>> Putting it into my own words to check my understanding:
>
>> - The two reads in T2 may be re-ordered because, in T2, there is nothing
>> that requires a happens-before relationship between the two reads
>
> Depends on what was really meant by the ??? notation. If it were another
> reference to st_200, then the example ignores the potential write to st_200.
>
>> - If T2 was executing in isolation the order of the reads wouldn't
>> matter
>
> Correct, assuming there is no write in T2's control flow.
>
>> - However, T1 is writing.
>
>> So if the writes
>
> I'll assume you meant "reads" there.
I'd did. Thanks.
>> in T2 are re-ordered and the write from T1 takes place between them the T2
>> read for the line 'st_200 == null' could return a non-null value for st_200
>> (the value after the write from T1) while the read for the line 'return
>> st_200;' could return null (the value from before the write in T1).
>
> No - that would violate program order (intra-thread semantics). The compiler
> cannot legally move the read for "return st_200" to a point before the
> potential write to st_200. The CPU can speculatively read for the return
> value, but must discard such speculation if a write were to occur (or use
> store forwarding to update the read).
Thanks for sticking with me on this. I'm finding it hugely helpful. I
think I need a few more references to make things clearer so I'm going
to restate the problem.
st_200 is non-volatile
L1 if (st_200 == null ) {
L2 st_200 = sm.getString("sc.200");
L3 }
L4 return st_200;
Ln=Line n
Tx=Thread x
Rn=Read at line n
Wn=Write at line n
So T2 has two reads, T2R1 and T2R4.
Depending on the value read for T2R1 there may be a write at T2W2.
If there is a write there is a happens before relationship between T2R1
and T2R4.
Consider the following sequence
T2R4 (out of order read returns null)
T1R1 (returns null)
T1W2 (writes non-null value)
T1R4 (reads new non-null value)
T2R1 (reads new non-null value)
Because T2R1 reads a non-null value there is no write in T2.
Therefore there is no happens-before relationship between T2R1 and T2R4
because there is no intervening write in that thread (the write happens
in T1).
Therefore the re-ordering is allowed to happen.
And we get the unexpected result.
Or
The write at T1W2 is sufficient to enforce the happens before
relationship between T2R1 and T2R4. In which case what is the point of
volatile? What do I gain by using it?
A still slightly confused,
Mark
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