blocks|key|203040|text|是的，代码是有效的，没有未定义的行为。在计算S的构造函数之前，初始化器列表中的表达式从左到右进行计算和排序。因此，您的程序应该始终如一地为变量i赋值2。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|203041|引用C%2B%2B标准的§8.5.4：|203042|在带括号的初始化列表的初始化器列表中，初始化器子句(包括任何由包扩展(14.5.3)产生的子句)将按照它们出现的顺序进行评估。也就是说，与给定的初始化器子句关联的每个值计算和副作用将在初始化程序列表的逗号分隔列表中与其后面的任何初始化器子句关联的每个值计算和副作用之前排序。|BOLD|203043|因此，发生的情况是：|203044|对|203045|%2B%2Bi求值，产生constructor);|ordered-list-item|203046|%2B%2Bi+(计算S的i+=+1的第一个参数，执行S的constructor);|203047|S's构造函数的第二个参数是2;|203048|value转换运算符，返回值2被赋值给i+(它已经具有值2).|203049|203050|本标准的另一个相关段落是§1.9/15，其中也提到了具有未定义行为的类似示例：|203051|i+=+v[i%2B%2B];+//+the+behavior+is+undefined
i+=+i%2B%2B+%2B+1;+//+the+behavior+is+undefined|code-block|syntax|javascript|203052|然而，同一段话说：|203053|除非另有说明，否则对单个运算符的操作数和子表达式的计算是不排序的。...当调用函数(无论函数是否为内联函数)时，在执行被调用函数体中的每个表达式或语句之前，与任何参数表达式或指定被调用函数的后缀表达式相关联的每个值计算和副作用都会被排序。|203054|由于1)初始化器列表中表达式的求值是从左到右排序的，2)+S的构造函数的执行是在初始化器列表中的所有表达式求值之后排序的，3)对i的赋值是在S的构造函数(及其转换运算符)执行之后排序的，所以行为是定义良好的。|203055|entityMap^0|M|1|1Z|1|22|1|0|0|26|1N|0|0|A|0|0|1|0|0|L|0|3|0|0|12|0|3|7|1|9|5|N|1|0|0|G|0|1|E|1|0|0|V|E|1|J|1|S|1|0|0|0|13|0|0|0|9|0|0|F|13|2|0|T|1|1S|1|1Y|1|0^^$0|@$1|2|3|4|5|6|7|1D|8|@$9|1E|A|1F|B|C]|$9|1G|A|1H|B|C]|$9|1I|A|1J|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|1K|8|@]|D|@]|E|$]]|$1|H|3|I|5|6|7|1L|8|@$9|1M|A|1N|B|J]]|D|@]|E|$]]|$1|K|3|L|5|6|7|1O|8|@$9|1P|A|1Q|B|J]]|D|@]|E|$]]|$1|M|3|N|5|6|7|1R|8|@$9|1S|A|1T|B|J]]|D|@]|E|$]]|$1|O|3|P|5|Q|7|1U|8|@$9|1V|A|1W|B|J]|$9|1X|A|1Y|B|C]]|D|@]|E|$]]|$1|R|3|S|5|Q|7|1Z|8|@$9|20|A|21|B|J]|$9|22|A|23|B|C]|$9|24|A|25|B|C]|$9|26|A|27|B|C]|$9|28|A|29|B|C]]|D|@]|E|$]]|$1|T|3|U|5|Q|7|2A|8|@$9|2B|A|2C|B|J]|$9|2D|A|2E|B|C]|$9|2F|A|2G|B|C]]|D|@]|E|$]]|$1|V|3|W|5|Q|7|2H|8|@$9|2I|A|2J|B|J]|$9|2K|A|2L|B|C]|$9|2M|A|2N|B|C]|$9|2O|A|2P|B|C]]|D|@]|E|$]]|$1|X|3|-4|5|6|7|2Q|8|@]|D|@]|E|$]]|$1|Y|3|Z|5|6|7|2R|8|@$9|2S|A|2T|B|J]]|D|@]|E|$]]|$1|10|3|11|5|12|7|2U|8|@]|D|@]|E|$13|14]]|$1|15|3|16|5|6|7|2V|8|@$9|2W|A|2X|B|J]]|D|@]|E|$]]|$1|17|3|18|5|6|7|2Y|8|@$9|2Z|A|30|B|J]|$9|31|A|32|B|J]]|D|@]|E|$]]|$1|19|3|1A|5|6|7|33|8|@$9|34|A|35|B|C]|$9|36|A|37|B|C]|$9|38|A|39|B|C]]|D|@]|E|$]]|$1|1B|3|-4|5|6|7|3A|8|@]|D|@]|E|$]]]|1C|$]]

Yes, the code is valid and does not have undefined behavior. Expressions in an initizalizer list are evaluated left-to-right and sequenced before the evaluation of the constructor of <code>S</code>. Therefore, your program should consistently assign value <code>2</code> to variable <code>i</code>.

Quoting § 8.5.4 of the C++ Standard:

"Within the initializer-list of a braced-init-list, the initializer-clauses, including any that result from pack expansions (14.5.3), are evaluated in the order in which they appear. That is, every value computation and side effect associated with a given initializer-clause is sequenced before every value computation and side effect associated with any initializer-clause that follows it in the comma-separated list of the initializer-list."

Thus, what happens is:

<ol>
<li><code>++i</code> gets evaluated, yielding <code>i = 1</code> (first argument of <code>S</code>'s constructor);</li>
<li><code>++i</code> gets evaluated, yielding <code>i = 2</code> (second argument of <code>S</code>'s constructor);</li>
<li><code>S</code>'s constructor is executed;</li>
<li><code>S</code>'s conversion operator is executed, returning value <code>2</code>;</li>
<li>value <code>2</code> is assigned to <code>i</code> (which already had value <code>2</code>).</li>
</ol>

Another relevant paragraph of the Standard is § 1.9/15, which also mentions similar examples that do have undefined behavior:

<pre><code>i = v[i++]; // the behavior is undefined
i = i++ + 1; // the behavior is undefined
</code></pre>

However, the same paragraph says:

"Except where noted, evaluations of operands of individual operators and of subexpressions of individual expressions are unsequenced. [...] When calling a function (whether or not the function is inline), every value computation and side effect associated with any argument expression, or with the postfix expression designating the called function, is sequenced before execution of every expression or statement in the body of the called function."

Since 1) the evaluation of the expressions in the initializer list is sequenced left-to-right, 2) the execution of the constructor of <code>S</code> is sequenced after the evaluation of all expressions in the initializer list, and 3) the assignment to <code>i</code> is sequenced after the execution of the constructor of <code>S</code> (and its conversion operator), the behavior is well-defined.

blocks|key|198441|text|是的，确实存在未定义行为的情况。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|198442|以下是导致未定义行为的情况的示例：|198443|在一个序列点内，|198444|A变量被多次更改。作为一个典型的例子，经常引用i=i%2B%2B表达式，其中i变量的赋值和它的递增是同时执行的。要了解有关此类错误的更多信息，请在初始化变量之前阅读"sequence+points".|unordered-list-item|offset|length|style|BOLD|198445|Using+a+variable一节。尝试使用new+[]运算符使用variable.|198446|Memory分配并使用delete运算符进行后续释放时，会发生未定义的行为。例如:t+*p+=新建T10；删除p；。正确的代码是:t+*p+=新建；T10+[]+p；。|198447|198448|编辑的|198449|此外，您的代码S{%2B%2Bi，%2B%2Bi}；未针对VS2012进行编译。你的意思可能是S(%2B%2Bi，%2B%2Bi);？如果使用"()“，则存在未定义的行为。在其他情况下，源代码是不正确的。|198450|entityMap|0|LINK|mutability|MUTABLE|url|http://www.viva64.com/en/t/0065/^0|0|0|0|0|2O|26|H|0|0|0|17|0|0|2C|0|0|0|3|0|0|2E|0^^$0|@$1|2|3|4|5|6|7|13|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|14|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|15|8|@]|9|@]|A|$]]|$1|F|3|G|5|H|7|16|8|@$I|17|J|18|K|L]]|9|@$I|19|J|1A|1|1B]]|A|$]]|$1|M|3|N|5|H|7|1C|8|@$I|1D|J|1E|K|L]]|9|@]|A|$]]|$1|O|3|P|5|H|7|1F|8|@$I|1G|J|1H|K|L]]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|1I|8|@]|9|@]|A|$]]|$1|R|3|S|5|6|7|1J|8|@$I|1K|J|1L|K|L]]|9|@]|A|$]]|$1|T|3|U|5|6|7|1M|8|@$I|1N|J|1O|K|L]]|9|@]|A|$]]|$1|V|3|-4|5|6|7|1P|8|@]|9|@]|A|$]]]|W|$X|$5|Y|Z|10|A|$11|12]]]]

Yes indeed you have case of undefined behavior.

Below are examples of situations causing undefined behavior:

<ul>
<li>A variable is changed several times within one sequence point. As a
canonical example, the i=i++ expression is often cited where
assignment of the i variable and its increment are performed at the
same time. To learn more about this kind of errors, read the section
<a href="http://www.viva64.com/en/t/0065/" rel="nofollow">"sequence points"</a>.</li>
<li>Using a variable before initializing it. Undefined behavior occurs
when trying to use the variable.</li>
<li>Memory allocation using the new [] operator and subsequent release
using the delete operator. For example: T *p = new T[10]; delete p;.
The correct code is: T *p = new T[10]; delete [] p;.</li>
</ul>

EDITED

Also your code S{++i, ++i}; is not compiled for VS2012. May be you mean S(++i, ++i);?
If you use "()" then undefined behavour is present. In other case your source code is incorrect.

I am curious about initializer lists and sequence points. I read a while ago that the order of evaluation in initializer lists is left to right. If that is so, then there must be some kind of sequence point between the points of evaluation, am I wrong? So with that said is the following valid code? Is there anything that causes undefined behavior in it?

<pre><code>int i = 0;

struct S {
 S(...) {} 
 operator int() { return i; }
};

int main() {
 i = S{++i, ++i};
}
</code></pre>

Any and all responses are appreciated.

Are multiple mutations within initializer lists undefined behavior?

我对初始化器列表和序列点很感兴趣。不久前我读到初始化器列表中的求值顺序是从左到右。如果是这样，那么在评估点之间一定有某种序列点，我错了吗？那么，下面的代码是有效的吗？其中是否存在导致未定义行为的原因？int i = 0;struct S {    S(...) {}     operator int() { retur...

问初始化器列表中的多个突变是否有未定义的行为？
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问初始化器列表中的多个突变是否有未定义的行为？
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