blocks|key|1838608|text|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1838609|根据运算符的优先级，迭代器应先递增，然后赋值给自身|blockquote|1838610|1838611|1838612|运算符优先级不决定计算顺序。|1838613|如果表达式d+=+(a%2Bb)*c、a、b、c和d可以按任意顺序求值。a%2Bb必须在a和b之后求值，(a%2Bb)*c在a%2Bb和c之后求值，赋值结果在d和乘法结果之后求值，但仅此而已。|offset|length|style|CODE|1838614|赋值的情况也是如此。|1838615|之前赋值的左侧和右侧可以按任何顺序进行计算。具体地说，可以实现以下顺序：|1838616|对|1838617|iter求值，结果是我们称为temp|unordered-list-item|1838618|%2B%2Biter的值被求值(这不会影响temp)+|1838619|这会导致iter递增，并且该值是新递增的iter+temp|1838620|%2B%2Biter|1838621|1838622|1838623|​|1838624|将|1838625|temp赋值给在|1838626|之上求值的iter+r值，最终结果是iter+|1838627|不变|1838628|另一种评估顺序过去也是可能的。|1838629|1838630|%2B%2Biter进行求值，这会导致iter递增，该值是新递增的iter+|1838631|1838632|1838633|1838634|再次对|1838635|iter+(赋值的右侧)求值，结果是新增加的lvalue|1838636|转换为|1838637|1838638|1838639|1838640|1838641|分配发生，使iter按预期递增。|1838642|1838643|然而，C%2B%2B17+has+changed+this。赋值现在从右到左严格计算。这意味着现在保证了第一个求值顺序，而不允许第二个求值顺序。|1838644|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/questions/38501587/what-are-the-evaluation-order-guarantees-introduced-by-c17^0|0|0|0|0|0|5|B|H|1|J|1|L|1|N|1|Y|3|14|1|16|1|1C|7|1K|3|1O|1|1Z|1|0|0|0|0|0|4|E|4|0|0|6|H|4|1|4|4|K|4|P|4|1|0|6|1|0|0|0|0|0|4|0|5|4|I|4|0|0|0|0|0|6|F|4|T|4|0|0|0|0|0|0|4|0|0|0|0|0|0|6|4|0|0|9|G|0|0^^$0|@$1|2|3|-4|4|5|6|29|7|@]|8|@]|9|$]]|$1|A|3|B|4|C|6|2A|7|@]|8|@]|9|$]]|$1|D|3|-4|4|5|6|2B|7|@]|8|@]|9|$]]|$1|E|3|-4|4|5|6|2C|7|@]|8|@]|9|$]]|$1|F|3|G|4|5|6|2D|7|@]|8|@]|9|$]]|$1|H|3|I|4|5|6|2E|7|@$J|2F|K|2G|L|M]|$J|2H|K|2I|L|M]|$J|2J|K|2K|L|M]|$J|2L|K|2M|L|M]|$J|2N|K|2O|L|M]|$J|2P|K|2Q|L|M]|$J|2R|K|2S|L|M]|$J|2T|K|2U|L|M]|$J|2V|K|2W|L|M]|$J|2X|K|2Y|L|M]|$J|2Z|K|30|L|M]|$J|31|K|32|L|M]]|8|@]|9|$]]|$1|N|3|O|4|5|6|33|7|@]|8|@]|9|$]]|$1|P|3|Q|4|5|6|34|7|@]|8|@]|9|$]]|$1|R|3|S|4|5|6|35|7|@]|8|@]|9|$]]|$1|T|3|U|4|V|6|36|7|@$J|37|K|38|L|M]|$J|39|K|3A|L|M]]|8|@]|9|$]]|$1|W|3|X|4|V|6|3B|7|@$J|3C|K|3D|L|M]|$J|3E|K|3F|L|M]]|8|@]|9|$]]|$1|Y|3|Z|4|V|6|3G|7|@$J|3H|K|3I|L|M]|$J|3J|K|3K|L|M]|$J|3L|K|3M|L|M]]|8|@]|9|$]]|$1|10|3|11|4|V|6|3N|7|@$J|3O|K|3P|L|M]]|8|@]|9|$]]|$1|12|3|-4|4|5|6|3Q|7|@]|8|@]|9|$]]|$1|13|3|-4|4|5|6|3R|7|@]|8|@]|9|$]]|$1|14|3|15|4|5|6|3S|7|@]|8|@]|9|$]]|$1|16|3|17|4|5|6|3T|7|@]|8|@]|9|$]]|$1|18|3|19|4|V|6|3U|7|@$J|3V|K|3W|L|M]]|8|@]|9|$]]|$1|1A|3|1B|4|V|6|3X|7|@$J|3Y|K|3Z|L|M]|$J|40|K|41|L|M]]|8|@]|9|$]]|$1|1C|3|1D|4|5|6|42|7|@]|8|@]|9|$]]|$1|1E|3|1F|4|5|6|43|7|@]|8|@]|9|$]]|$1|1G|3|S|4|5|6|44|7|@]|8|@]|9|$]]|$1|1H|3|1I|4|V|6|45|7|@$J|46|K|47|L|M]|$J|48|K|49|L|M]|$J|4A|K|4B|L|M]]|8|@]|9|$]]|$1|1J|3|-4|4|V|6|4C|7|@]|8|@]|9|$]]|$1|1K|3|-4|4|5|6|4D|7|@]|8|@]|9|$]]|$1|1L|3|15|4|5|6|4E|7|@]|8|@]|9|$]]|$1|1M|3|1N|4|5|6|4F|7|@]|8|@]|9|$]]|$1|1O|3|1P|4|V|6|4G|7|@$J|4H|K|4I|L|M]]|8|@]|9|$]]|$1|1Q|3|1R|4|5|6|4J|7|@]|8|@]|9|$]]|$1|1S|3|-4|4|V|6|4K|7|@]|8|@]|9|$]]|$1|1T|3|-4|4|5|6|4L|7|@]|8|@]|9|$]]|$1|1U|3|15|4|5|6|4M|7|@]|8|@]|9|$]]|$1|1V|3|-4|4|5|6|4N|7|@]|8|@]|9|$]]|$1|1W|3|1X|4|V|6|4O|7|@$J|4P|K|4Q|L|M]]|8|@]|9|$]]|$1|1Y|3|-4|4|5|6|4R|7|@]|8|@]|9|$]]|$1|1Z|3|20|4|5|6|4S|7|@]|8|@$J|4T|K|4U|1|4V]]|9|$]]|$1|21|3|-4|4|5|6|4W|7|@]|8|@]|9|$]]]|22|$23|$4|24|25|26|9|$27|28]]]]

<blockquote>
 according to operator precedence, iterator should first be incremented, then assigned to self
</blockquote>

Operator precedence does not determine evaluation order. 

If the expression <code>d = (a+b)*c</code>, <code>a</code>, <code>b</code>, <code>c</code> and <code>d</code> can be evaluated in any order. <code>a+b</code> has to be evaluated after <code>a</code> and <code>b</code>, <code>(a+b)*c</code> after both <code>a+b</code> and <code>c</code>, and the result of the assignment after both <code>d</code> and the result of the multiplication, but that's about it.

The same used to be the case with the assignment.

The left-hand side and the right-hand side of the assignment previously could be evaluated in any order. In particular, the following order could be realised:

<ul>
<li><code>iter</code> is evaluated, resulting in a value we call <code>temp</code></li>
<li><code>++iter</code> is evaluated (this doesn't affect <code>temp</code>)

<ul>
<li>this results in incrementing of <code>iter</code>, and the value is the newly-incremented <code>iter</code> rvalue</li>
</ul></li>
<li><code>temp</code> is assigned to the <code>iter</code> rvalue evaluated above</li>
<li>the net result is <code>iter</code> being unchanged </li>
</ul>

Another evaluation order also used to be possible.

<ul>
<li><code>++iter</code> is evaluated 

<ul>
<li>this results in incrementing of <code>iter</code>, and the value is the newly-incremented <code>iter</code> rvalue</li>
</ul></li>
<li><code>iter</code> (the right hand side of the assignment) is evaluated again

<ul>
<li>the result is the newly-incremented <code>iter</code> converted to lvalue</li>
</ul></li>
<li>the assignment takes place, leaving <code>iter</code> incremented as expected.</li>
</ul>

However C++17 <a href="https://stackoverflow.com/questions/38501587/what-are-the-evaluation-order-guarantees-introduced-by-c17">has changed this</a>. Assignment is now evaluated strictly from right to left. This means the first evaluation order is now guaranteed, and the second one is not allowed.

I had a typo in code recently, that was causing an infinite loop in some cases (but not in all environments, it seems). 
It looked like this:

<pre><code>for (std::vector&lt;myString&gt;::iterator iter = myVector.begin(); iter != myVector.end(); ++iter = iter)
{
 ...
}
</code></pre>

The typo is the assignment of the iterator to itself. Removing it fixes the issues. 

I am wondering what exactly happens during the statement ++iter = iter? 
I thought that according to operator precedence, iterator should first be incremented, then assigned to self, but I seem to be missing some steps (otherwise there would not be infinite loops). 
My feeling is that it also involves dereferencing of the variable, but I am not sure to completely understand what was happening with this bug. 
Also, why did it seem to not cause infinite loops on some platforms?

What happens during ++iter = iter?

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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我最近在代码中有一个打字错误，这在某些情况下导致了无限循环(但似乎并不是在所有环境中)。它看起来是这样的：for (std::vector<myString>::iterator iter = myVector.begin(); iter != myVector.end(); ++iter = iter){ ...}拼写错误是迭代器对自身的赋值。删除它可以解决问题。我想知道在语句++iter

问在++iter = iter期间发生了什么？
EN

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问在++iter = iter期间发生了什么？EN