Thomas Wang wrote:
>Initially, x = y = v = 0. v is a volatile variable.
>
>Thread 1:
>r1 = x
>v = 0
>y = 1
>
>Thread 2:
>r3 = y
>v = 0
>x = 1
>
>Is the behavior r1 == r3 == 1 possible?
>
Yes.
Because there are no reads to the variable, it doesnt matter whether the
variable is volatile (and therefore strong vs weak interpretations dont
matter). The program is equivalent to:
================================================
Initially, x = y = v = 0. v is a volatile variable.
Thread 1:
r1 = x
y = 1
Thread 2:
r3 = y
x = 1
======================================================
-------------------------------
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