Vijay,
I don't see why this test case is an issue. The outcome you give can be
obtained from an SC execution, so it's certainly permitted.
r1 = 1
b = 1
r2 = 0 (from y)
r3 = 1 (from b)
a = 1
Is the outcome guaranteed? No.
Sylvia.
At 06:06 PM 8/03/2004 -0500, Vijay Saraswat wrote:
>Uh sorry, it looks like my formatting got mangled.. here it is again:
>----------------------------------------------------------
>Test 14a
>Initially: a=b=y=0
>
>Thread 1
>r1 =1
>if (r1 == 0)
> y =1
>else
> b =1
>
>Thread 2
>r2=y
>r3=b
>if (r2+r3 !=0)
>a=1
>
>Question: r1=r3=1, r2=0?
>-----------------------------------------------------------
>
>
>Vijay Saraswat wrote:
>
>>I wonder whether this test case has been discussed before. If so, please
>>point me to the discussion. If not, how should this be dealt with?
>>-----------------------------------------------------------
>>Test 14a
>>Initially: a=b=y=0
>>
>>Thread 1 Thread 2
>>r1=a if (r1==0) r2=y
>> y=1 r3=b
>>else if (r2+r3 != 0)
>> b=1 a=1
>>
>>Question: r1=r3=1, r2=0?
>>-----------------------------------------------------------
>>
>>The case is obtained from Test 14, without making the volatility assumption.
>>
>>Best,
>>Vijay
>>
>>-------------------------------
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>>
>
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