blocks|key|998960|text|表达式count%2B%2B返回count的原始值，然后递增该值。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|998961|因此，您每次都会用相同的值覆盖count。只需执行以下操作：|998962|count%2B%2B;|code-block|syntax|javascript|998963|对于好奇的人，这里有一个指向Eric+Lippert的文章的链接，该文章区分了运算符优先级和计算顺序--这是一个有趣的阅读：|998964|http://blogs.msdn.com/b/ericlippert/archive/2009/08/10/precedence-vs-order-redux.aspx|998965|entityMap|0|LINK|mutability|MUTABLE|url^0|3|7|0|0|0|0|0|2D|0|0^^$0|@$1|2|3|4|5|6|7|X|8|@$9|Y|A|Z|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|10|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|11|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|12|8|@]|D|@]|E|$]]|$1|O|3|P|5|6|7|13|8|@]|D|@$9|14|A|15|1|16]]|E|$]]|$1|Q|3|-4|5|6|7|17|8|@]|D|@]|E|$]]]|R|$S|$5|T|U|V|E|$W|P]]]]

The expression <code>count++</code> returns the original value of count, then increments the value afterwards.

So you are overwriting count with the same value every time. Just do this:

<pre><code>count++;
</code></pre>

For the curious, here's a link to Eric Lippert's article which distinguishes between operator precedence and the order of evaluation -- it's an interesting read:

<a href="http://blogs.msdn.com/b/ericlippert/archive/2009/08/10/precedence-vs-order-redux.aspx" rel="noreferrer">http://blogs.msdn.com/b/ericlippert/archive/2009/08/10/precedence-vs-order-redux.aspx</a>

blocks|key|998981|text|count++=+count%2B%2B;不会将count递增1。x%2B%2B是后增量运算符，这意味着表达式返回的值是旧值。因此，在您的代码中，会发生以下情况：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|998982|int+oldValue+=+count;
count+=+count+%2B+1;
count+=+oldValue;|code-block|syntax|javascript|998983|你可能想要写的是count%2B%2B;+(没有"count+=")。|998984|有关这方面的更多详细信息，请参阅以下SO问题：|998985|998986|What+does+"count%2B%2B"+return+in+C#?|unordered-list-item|998987|998988|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/q/5030653/87698^0|0|H|K|5|T|3|0|0|8|8|L|7|0|0|0|0|X|0|0|0^^$0|@$1|2|3|4|5|6|7|11|8|@$9|12|A|13|B|C]|$9|14|A|15|B|C]|$9|16|A|17|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|18|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|19|8|@$9|1A|A|1B|B|C]|$9|1C|A|1D|B|C]]|D|@]|E|$]]|$1|M|3|N|5|6|7|1E|8|@]|D|@]|E|$]]|$1|O|3|-4|5|6|7|1F|8|@]|D|@]|E|$]]|$1|P|3|Q|5|R|7|1G|8|@]|D|@$9|1H|A|1I|1|1J]]|E|$]]|$1|S|3|-4|5|6|7|1K|8|@]|D|@]|E|$]]|$1|T|3|-4|5|6|7|1L|8|@]|D|@]|E|$]]]|U|$V|$5|W|X|Y|E|$Z|10]]]]

<code>count = count++;</code> does not increment <code>count</code> by one. <code>x++</code> is the post increment operator, which means that the value returned by the expression is the old value. Thus, in your code, the following happens:

<pre><code>int oldValue = count;
count = count + 1;
count = oldValue;
</code></pre>

What you probably meant to write was <code>count++;</code> (without the "<code>count =</code>").

More details about this can be found in the following SO question:

<ul>
<li><a href="https://stackoverflow.com/q/5030653/87698">What does &quot;count++&quot; return in C#?</a></li>
</ul>

blocks|key|999012|text|%2B%2B操作符首先保存当前值，然后递增，最后返回保存的值，因此count永远不会改变。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|999013|Eiter使用%2B%2B运算符或执行赋值。这些都是等效的：|999014|count%2B%2B;
count+%2B=+1;
count+=+count+%2B+1;|code-block|syntax|javascript|999015|entityMap^0|T|5|0|7|2|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|R|8|@$9|S|A|T|B|C]]|D|@]|E|$]]|$1|H|3|I|5|J|7|U|8|@]|D|@]|E|$K|L]]|$1|M|3|-4|5|6|7|V|8|@]|D|@]|E|$]]]|N|$]]

The ++ operator first saves the current value then increments and finally returns the saved value, so <code>count</code> will never change.

Eiter use the <code>++</code> operator or do an assignment. These are all equivalent:

<pre><code>count++;
count += 1;
count = count + 1;
</code></pre>

blocks|key|5437489|text|这将无限循环。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5437490|有两种类型的变量递增：|5437491|在这里，count%2B%2B和%2B%2Bcount是不同的，如果你使用过%2B%2Bcount，它将会工作。|offset|length|style|CODE|5437492|在这里，count+=+count%2B%2B表示计数变量将递增1，然后将较早的值1赋给计数变量本身，因此计数保持不变。|5437493|entityMap^0|0|0|4|7|C|7|U|7|0|4|F|0^^$0|@$1|2|3|4|5|6|7|N|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|O|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|P|8|@$F|Q|G|R|H|I]|$F|S|G|T|H|I]|$F|U|G|V|H|I]]|9|@]|A|$]]|$1|J|3|K|5|6|7|W|8|@$F|X|G|Y|H|I]]|9|@]|A|$]]|$1|L|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|M|$]]

This will loop infinitely.

There are two types of incrementing a variable:

Here <code>count++</code> and <code>++count</code> both are different if you have used <code>++count</code> it will work.

Here <code>count = count++</code> means count variable will be incremented by one then assigns the earlier value 1 to the count variable itself so count remains unchanged.

blocks|key|5437543|text|因为|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5437544|count%2B%2B|code-block|syntax|javascript|5437545|返回count，而不是count+%2B+1|offset|length|style|CODE|5437546|只需使用不带赋值的count%2B%2B或：|5437547|count+=+%2B%2Bcount;|5437548|最后一个只解释，但你不应该使用它…|5437549|来自：%2B%2B+Operator+(C#+Reference)|5437550|5437551|第一种形式是前缀递增操作。运算的结果是操作数递增后的值。|blockquote|5437552|5437553|第二种形式是后缀增量操作。操作的结果是操作数在被递增之前的值。|5437554|数字和枚举类型具有预定义的增量运算符。用户定义的类型可以重载%2B%2B运算符。整数类型的运算通常允许在枚举上进行。|5437555|entityMap|0|LINK|mutability|MUTABLE|url|http://msdn.microsoft.com/en-us/library/36x43w8w.aspx^0|0|0|2|5|B|9|0|0|0|0|3|Q|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|1B|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|1C|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|1D|8|@$I|1E|J|1F|K|L]|$I|1G|J|1H|K|L]]|9|@]|A|$]]|$1|M|3|N|5|6|7|1I|8|@]|9|@]|A|$]]|$1|O|3|P|5|D|7|1J|8|@]|9|@]|A|$E|F]]|$1|Q|3|R|5|6|7|1K|8|@]|9|@]|A|$]]|$1|S|3|T|5|6|7|1L|8|@]|9|@$I|1M|J|1N|1|1O]]|A|$]]|$1|U|3|-4|5|6|7|1P|8|@]|9|@]|A|$]]|$1|V|3|W|5|X|7|1Q|8|@]|9|@]|A|$]]|$1|Y|3|-4|5|6|7|1R|8|@]|9|@]|A|$]]|$1|Z|3|10|5|6|7|1S|8|@]|9|@]|A|$]]|$1|11|3|12|5|6|7|1T|8|@]|9|@]|A|$]]|$1|13|3|-4|5|6|7|1U|8|@]|9|@]|A|$]]]|14|$15|$5|16|17|18|A|$19|1A]]]]

because

<pre><code>count++
</code></pre>

returns <code>count</code>, not <code>count + 1</code>

just have count++ with no assignment or:

<pre><code>count = ++count;
</code></pre>

the last one only to explain but you should not use it...

from: <a href="http://msdn.microsoft.com/en-us/library/36x43w8w.aspx" rel="nofollow">++ Operator (C# Reference)</a>

<blockquote>
 The first form is a prefix increment operation. The result of the
 operation is the value of the operand after it has been incremented.
 
 The second form is a postfix increment operation. The result of the
 operation is the value of the operand before it has been incremented.
 
 Numeric and enumeration types have predefined increment operators.
 User-defined types can overload the ++ operator. Operations on
 integral types are generally allowed on enumeration.
</blockquote>

blocks|key|5437584|text|count++=+count%2B%2B;|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|5437585|这是一个后增量。它执行以下操作。|unstyled|5437586|int+temp+=+count;
count%2B%2B;
count+=+temp;|5437587|所以你不是在增加计数。请改用以下内容：|5437588|while+(count++<+10)
{
++++%2B%2Bcount;
}|5437589|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|P|8|@]|9|@]|A|$]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$B|C]]|$1|I|3|J|5|F|7|R|8|@]|9|@]|A|$]]|$1|K|3|L|5|6|7|S|8|@]|9|@]|A|$B|C]]|$1|M|3|-4|5|F|7|T|8|@]|9|@]|A|$]]]|N|$]]

<pre><code>count = count++;
</code></pre>

This is a post-increment. It does the following.

<pre><code>int temp = count;
count++;
count = temp;
</code></pre>

So you're not incrementing count. Use the following instead:

<pre><code>while (count &lt; 10)
{
 ++count;
}
</code></pre>

blocks|key|1001871|text|它是无限的，因为您实际上并没有递增计数。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1001872|count+=+count%2B%2B；将值1赋给count，然后递增count，但是由于没有分配递增的值，count永远不会递增。|1001873|您需要执行以下任一操作：|1001874|count%2B%2B;|code-block|syntax|javascript|1001875|或|1001876|count+=+%2B%2Bcount;|1001877|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|R|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|S|8|@]|9|@]|A|$]]|$1|F|3|G|5|H|7|T|8|@]|9|@]|A|$I|J]]|$1|K|3|L|5|6|7|U|8|@]|9|@]|A|$]]|$1|M|3|N|5|H|7|V|8|@]|9|@]|A|$I|J]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

It is infinite because you aren't actually incrementing count.

count = count++; assigns the value of 1 to count and then increments count but since you don't assign the incremented value count never increases.

You need to do either:

<pre><code>count++;
</code></pre>

or

<pre><code>count = ++count;
</code></pre>

blocks|key|999133|text|让我问你一个问题，为什么你在一个变量上做两个操作，而一个操作就足够了？你的意图是什么？count%2B%2B本身就足够了，所以为什么还要赋值给count。也许你想做些别的事情。你可以只使用count%2B%2B，或者%2B%2Bcount或者count%2B1。我认为其他的方式会导致两种操作。很抱歉我的写作方式。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|999134|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

Let me ask you a question why do you make two operations on a single variable while one is enough?
what was your intention? count++ itself was enough so why again assign to count. May be you want to do something else.
You could have only count++, or ++count or count+1. I think other ways causes two operations.
Sorry for my way of writing.

This is a simple while loop in C# but it is working infinitely.

<pre><code>int count = 1;
while (count &lt; 10)
{
 count = count++;
}
</code></pre>

Why is this so?

Why is this Loop Working Infinitely

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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这是C#中的一个简单的while循环，但它可以无限地工作。int count = 1;while (count < 10){ count = count++;}为何会这样呢？

问为什么这个循环会无限地工作
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问为什么这个循环会无限地工作EN