blocks|key|4765324|text|使用同一个标量对象的值进行值计算。x[i]使用i的值。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4765325|自C11以来，赋值中存在一个序列关系。|4765326|4765327|在左操作数和右操作数的值计算之后，对更新左操作数的存储值的副作用进行排序。|blockquote|4765328|4765329|4765330|(C11+6.5.16/3)|4765331|在此之前，标准讨论表达式的方式比较宽松。它没有描述关系之前的顺序。相反，我们有：|4765332|在前一个序列点和下一个序列点之间的|4765333|对象应该通过对表达式的计算来修改其存储值最多一次。此外，应只读取先验值以确定要存储的值。|4765334|4765335|4765336|您没有读取“除了确定要存储的值”以外的值，因此定义了行为。|4765337|(C99+6.5/2)|4765338|entityMap^0|H|4|N|1|0|0|0|0|0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|13|8|@$9|14|A|15|B|C]|$9|16|A|17|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|18|8|@]|D|@]|E|$]]|$1|H|3|-4|5|6|7|19|8|@]|D|@]|E|$]]|$1|I|3|J|5|K|7|1A|8|@]|D|@]|E|$]]|$1|L|3|-4|5|6|7|1B|8|@]|D|@]|E|$]]|$1|M|3|-4|5|6|7|1C|8|@]|D|@]|E|$]]|$1|N|3|O|5|6|7|1D|8|@]|D|@]|E|$]]|$1|P|3|Q|5|6|7|1E|8|@]|D|@]|E|$]]|$1|R|3|S|5|6|7|1F|8|@]|D|@]|E|$]]|$1|T|3|U|5|K|7|1G|8|@]|D|@]|E|$]]|$1|V|3|-4|5|6|7|1H|8|@]|D|@]|E|$]]|$1|W|3|-4|5|6|7|1I|8|@]|D|@]|E|$]]|$1|X|3|Y|5|6|7|1J|8|@]|D|@]|E|$]]|$1|Z|3|10|5|6|7|1K|8|@]|D|@]|E|$]]|$1|11|3|-4|5|6|7|1L|8|@]|D|@]|E|$]]]|12|$]]

There is a value computation using the value of the same scalar object. <code>x[i]</code> uses the value of <code>i</code>.
Since C11, there is a sequence relation in assignment.
<blockquote>
The side effect of updating the stored value of the left operand is
sequenced after the value computations of the left and right operands.
</blockquote>
(C11 6.5.16/3)
Prior to then the way the standard discussed expressions was looser. It did not describe a sequenced before relation. Instead we have:
<blockquote>
Between the previous and next sequence point an object shall have its stored value modified at most once by the evaluation of an expression. Furthermore, the prior value shall be read only to determine the value to be stored.
</blockquote>
You aren't reading the value &quot;other than to determine the value to be stored&quot;, so the behaviour is defined.
(C99 6.5/2)

blocks|key|4765366|text|想象一下：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4765367|i+=+3;
x[]+=+{1,+1,+1,+1,+1};|code-block|syntax|javascript|4765368|因此，x[i]+=+1，x[i]%2B%2B+=2，x变为{1,+1,+2,+1,+1}，i变为1。|offset|length|style|CODE|4765369|为什么会有不明确的行为呢？|4765370|entityMap^0|0|0|3|4|C|6|M|1|P|F|15|1|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@$I|T|J|U|K|L]|$I|V|J|W|K|L]|$I|X|J|Y|K|L]|$I|Z|J|10|K|L]|$I|11|J|12|K|L]]|9|@]|A|$]]|$1|M|3|N|5|6|7|13|8|@]|9|@]|A|$]]|$1|O|3|-4|5|6|7|14|8|@]|9|@]|A|$]]]|P|$]]

Imagine:
<pre><code>i = 3;
x[] = {1, 1, 1, 1, 1};
</code></pre>
So, <code>x[i]</code> equals 1, <code>x[i]++</code> equals 2 and <code>x</code> becomes <code>{1, 1, 2, 1, 1}</code>, and <code>i</code> becomes 1.
Why would there be any undefined behaviour?

blocks|key|4765412|text|如果这是真的|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4765413|4765414|blockquote|4765415|4765416|4765417|没有“对同一个标量对象的不同副作用”|unordered-list-item|4765418|没有“使用同一个标量对象的值计算”|4765419|4765420|(在每个允许排序的子表达式中)，那么您所引用的条款就不会出现特定的问题。也就是说，它的"if“的前因不成立，因此"if”(未定义的行为)的结果不会被断言。|4765421|但是，这对i和使用i值的值计算都有副作用。前者是分配的副作用，后者是x[i]%2B%2B的值计算。但是，这并不是一个问题，因为对于所有形式的任务，|offset|length|style|CODE|4765422|4765423|在左操作数和右操作数的值计算之后，对更新左操作数的存储值的副作用进行排序。|4765424|4765425|4765426|(C17+6.5.16/3)|4765427|同时，为了完整起见，|4765428|4765429|操作符的操作数的值计算是在运算符结果的值计算之前进行排序的。|4765430|4765431|4765432|(C17+6.5/1)|4765433|因此，分配对i的副作用在x[i]%2B%2B的值计算之后被排序，在i的评估后被排序。|4765434|entityMap^0|0|0|0|0|0|0|0|0|0|5|1|9|1|Y|6|0|0|0|0|0|0|0|0|0|0|0|0|6|1|C|6|T|1|0^^$0|@$1|2|3|4|5|6|7|1E|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|1F|8|@]|9|@]|A|$]]|$1|C|3|-4|5|D|7|1G|8|@]|9|@]|A|$]]|$1|E|3|-4|5|6|7|1H|8|@]|9|@]|A|$]]|$1|F|3|-4|5|6|7|1I|8|@]|9|@]|A|$]]|$1|G|3|H|5|I|7|1J|8|@]|9|@]|A|$]]|$1|J|3|K|5|I|7|1K|8|@]|9|@]|A|$]]|$1|L|3|-4|5|6|7|1L|8|@]|9|@]|A|$]]|$1|M|3|N|5|6|7|1M|8|@]|9|@]|A|$]]|$1|O|3|P|5|6|7|1N|8|@$Q|1O|R|1P|S|T]|$Q|1Q|R|1R|S|T]|$Q|1S|R|1T|S|T]]|9|@]|A|$]]|$1|U|3|-4|5|6|7|1U|8|@]|9|@]|A|$]]|$1|V|3|W|5|D|7|1V|8|@]|9|@]|A|$]]|$1|X|3|-4|5|6|7|1W|8|@]|9|@]|A|$]]|$1|Y|3|-4|5|6|7|1X|8|@]|9|@]|A|$]]|$1|Z|3|10|5|6|7|1Y|8|@]|9|@]|A|$]]|$1|11|3|12|5|6|7|1Z|8|@]|9|@]|A|$]]|$1|13|3|-4|5|6|7|20|8|@]|9|@]|A|$]]|$1|14|3|15|5|D|7|21|8|@]|9|@]|A|$]]|$1|16|3|-4|5|6|7|22|8|@]|9|@]|A|$]]|$1|17|3|-4|5|6|7|23|8|@]|9|@]|A|$]]|$1|18|3|19|5|6|7|24|8|@]|9|@]|A|$]]|$1|1A|3|1B|5|6|7|25|8|@$Q|26|R|27|S|T]|$Q|28|R|29|S|T]|$Q|2A|R|2B|S|T]]|9|@]|A|$]]|$1|1C|3|-4|5|6|7|2C|8|@]|9|@]|A|$]]]|1D|$]]

If it were true that
<blockquote>
<ul>
<li>there is no &quot;different side effect on the same scalar object&quot;</li>
<li>there is no &quot;value computation using the value of the same scalar object&quot;</li>
</ul>
</blockquote>
(in every allowed ordering of the subexpressions), then the provision you cite would present no particular issue. That is, the antecedent of its &quot;if&quot; would not hold, so the consequence of that &quot;if&quot; (undefined behavior) would not be asserted.
However, there is both a side effect on <code>i</code> and a value computation using the value of <code>i</code>. The former is the side effect of the assignment, and the latter is the value computation of <code>x[i]++</code>. This is not a problem, however, because, for all forms of assignment,
<blockquote>
The side effect of updating the stored value of the left operand is sequenced after the value computations of the left and right operands.
</blockquote>
(C17 6.5.16/3)
Also, for completeness,
<blockquote>
The value computations of the operands of an operator are sequenced before the value computation of the result of the operator.
</blockquote>
(C17 6.5/1)
Thus, the assignment's side effect on <code>i</code> is sequenced after the value computation of <code>x[i]++</code>, which is sequenced after the evaluation of <code>i</code>.

Can someone please explain whether <code>i = x[i]++;</code> lead to undefined behavior?
Note: <code>x[i]</code> and <code>i</code> are not both volatile and <code>x[i]</code> does not overlap <code>i</code>.
There is C11, 6.5 Expressions, 2 (emphasis added):
<blockquote>
If a side effect on a scalar object is unsequenced relative to either a different side effect
on the same scalar object or a value computation using the value of the same scalar
object, the behavior is undefined. If there are multiple allowable orderings of the
subexpressions of an expression, the behavior is undefined if such an unsequenced side
effect occurs in any of the orderings. 84)
</blockquote>
As I understand:
<ul>
<li>there is no &quot;different side effect on the same scalar object&quot;</li>
<li>there is no &quot;value computation using the value of the same scalar object&quot;</li>
</ul>
Are there &quot;multiple allowable orderings&quot;?
Overall: how can the <code>i = x[i]++;</code> be interpreted w.r.t. sequence points, side effects, and undefined behavior (if any)?
<hr />
UPD. Conclusion: the <code>i = x[i]++;</code> leads to 2 side effects:
<ol>
<li>&quot;the value of the operand object is incremented&quot; (Postfix increment)</li>
<li>&quot;updating the stored value of the left operand&quot; (Assignment operators)</li>
</ol>
The Standard does not define the order in which the side effects take place.
Hence, per C11, 4. Conformance, 2:
<blockquote>
Undefined behavior is otherwise indicated in this International Standard by the words ‘‘undefined behavior’’ or by the omission of any explicit definition of behavior.
</blockquote>
Experiments show that GCC/LLVM/ICC have order <code>1-2</code>, while MSVC (and some others) have order <code>2-1</code>.
<hr />
Extra (speculating): why not making it unspecified behavior? Example: &quot;an example of unspecified behavior is the order in which the side effects take place&quot;?

Does i = x[i]++; lead to undefined behavior?

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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请有人解释一下i = x[i]++;是否会导致未定义的行为？注意：x[i]和i都不是易失性的，x[i]不重叠i。这里有C11，6.5个表达式，2(重点是添加的)：如果对标量对象的副作用相对于同一标量对象上的不同副作用或使用相同标量对象值计算的值而言没有排序，则行为是未定义的。如果表达式的子表达式有多个允许的命令，则如果...

问I= x[i]++；会导致未定义的行为吗？
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问I= x[i]++；会导致未定义的行为吗？EN