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该链接提供了允许合法执行的示例.请参阅第26页第4.8.1
http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.79.629&rep=rep1&type=pdf节.
这个例子是
Initially,x = y = 0
Thread 1 Thread 2
r1 = x; r2 = y;
y = 1; x = r2;
r1 == r2 == 1 is a legal behavior
从纸上,
We wish the action
r2 = y
to see the value 1.
C
1
cannot contain this action
seeing this value: neither write to
y
had been committed.
C
2
may contain this
action; however,the read of
y
must return 0 in
E
2
,because of Rule 6. Execution
E
2
is therefore identical to
E
1
In
E
3
,by Rule 7,
r2 = y
can see any conflicting write that occurs in
C
2
(as long
as that write is happens-before consistent). This action can now see the write of 1
to
y
in Thread 1,which was committed in
C
1
. We commit one additional action in
C
3
: a write of 1 to
x
by
x = r2
.
我们不能提交r2 = y,其中y的值是1,因为y的读取将根据规则6返回0.但是根据E3中下一个提交周期中的文章,r2 = y可以看到值y = 1在Thread1中提交.
我怀疑是执行E3提交r2 = y还应该满足规则6并且应该只看到它之前发生的值,即x = 0.为什么在E2中由于规则6而提交r2 = y,y只能将y值读为0但在E3中提交r2 = y时,y可以读取在Thread1中写入的y = 1？
解决方法
最重要的是要注意的是规则7允许在Ci中执行的读取在Ei中看到的不同写入比在E中看到的那样.但是,薛定谔的读取必须尽早解决.规则5确保这样的读取必须在Ei 1中看到与E中相同的写入.
因此,如果您希望读取看到赛车写入,则不能在提交这两者的同一执行中进行.您需要等待下一次执行,并且您需要确保赛车写入不会违反发生 – 在排序之前(在我们的示例中,主要是在y = 0和r2 = y的排序之前发生).
所以你在C2中提交读取r2 = y,在E2中它只看到写入y = 0,因为写入发生在它之前,但在E3中,r2 = y可以看到它无法建立的赛车写入 – 之前.
附：这篇推理在论文中解释为读r1 = x以下几句：
C4,as part of E4,contains the read r1 = x; it still sees 0,because of Rule 6. In
our final execution E = E5,however,Rule 7 allows r1 = x to see the write of 1 to x that was committed in C3.
（编辑：李大同）
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