blocks|key|813339|text|用粗体表示的段落指的是#if表达式，因此可以使用运算符：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|813340|#if+VARIABLE+>+3|code-block|syntax|javascript|813341|等等，而不是你定义的宏中的代码。|813342|entityMap^0|B|3|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|S|8|@]|D|@]|E|$]]|$1|M|3|-4|5|6|7|T|8|@]|D|@]|E|$]]]|N|$]]

The paragraph you put in bold refers to <code>#if</code> expressions, so you can use operators:

<pre><code>#if VARIABLE &gt; 3
</code></pre>

etc., not the code that is inside the macros you define.

blocks|key|813421|text|条件表达式可以包含除赋值运算符、增量运算符和递减运算符以外的任何C运算符。|type|blockquote|depth|inlineStyleRanges|entityRanges|data|813422|上面说，你不能写这些，|unstyled|813423|#if+X+=+4+//wrong

#if+X%2B%2B++//this+too+wrong

#if+X--+//this+too|code-block|syntax|javascript|813424|编辑：|813425|我刚读到你对迭戈·塞维利亚职位的评论，|offset|length|813426|同意，但为什么不允许这样的事情(ideone.com/ahXDA)？|813427|这段代码实际上相当于您在全局范围中编写的代码，|813428|//global+code
int+b=7;+
b%2B%2B;++//ill-formed!|813429|这在C(和C%2B%2B)中是不允许的。不过，这在C中是允许的，|813430|//legal+in+C+Only
int+b+=+7;
int+b;+//well-formed!|813431|这里，第二个定义称为暂定定义。C标准在$6.9.2/2中，|style|BOLD|813432|对于没有初始化器的具有文件作用域的对象的标识符的声明，以及没有存储类说明符或具有存储类说明符静态的的声明构成暂定的definition.If翻译单元包含一个或多个标识符的暂定定义，并且翻译单元不包含该标识符的外部定义，然后该行为就像翻译单元包含该标识符的文件范围声明一样，转换单元的复合类型作为翻译单元的末尾，其初始化器等于0。|813433|这也意味着，不允许这样做，|813434|//illegal+in+both+C+and+C%2B%2B
int+b=7;
int+b=8;+//error+|813435|这不再是一个暂时的定义了。这是重新定义，这是不允许的！|813436|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/questions/4554313/doubt-in-conditional-inclusion/4554353#4554353^0|0|0|0|0|6|9|0|0|0|0|0|0|0|A|4|0|8|8|1D|37|0|0|D|0|0|0|R|0^^$0|@$1|2|3|4|5|6|7|1J|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|1K|8|@]|9|@]|A|$]]|$1|E|3|F|5|G|7|1L|8|@]|9|@]|A|$H|I]]|$1|J|3|K|5|D|7|1M|8|@]|9|@]|A|$]]|$1|L|3|M|5|D|7|1N|8|@]|9|@$N|1O|O|1P|1|1Q]]|A|$]]|$1|P|3|Q|5|6|7|1R|8|@]|9|@]|A|$]]|$1|R|3|S|5|D|7|1S|8|@]|9|@]|A|$]]|$1|T|3|U|5|G|7|1T|8|@]|9|@]|A|$H|I]]|$1|V|3|W|5|D|7|1U|8|@]|9|@]|A|$]]|$1|X|3|Y|5|G|7|1V|8|@]|9|@]|A|$H|I]]|$1|Z|3|10|5|D|7|1W|8|@$N|1X|O|1Y|11|12]]|9|@]|A|$]]|$1|13|3|14|5|6|7|1Z|8|@$N|20|O|21|11|12]|$N|22|O|23|11|12]]|9|@]|A|$]]|$1|15|3|16|5|D|7|24|8|@$N|25|O|26|11|12]]|9|@]|A|$]]|$1|17|3|18|5|G|7|27|8|@]|9|@]|A|$H|I]]|$1|19|3|1A|5|D|7|28|8|@$N|29|O|2A|11|12]]|9|@]|A|$]]|$1|1B|3|-4|5|D|7|2B|8|@]|9|@]|A|$]]]|1C|$1D|$5|1E|1F|1G|A|$1H|1I]]]]

<blockquote>
 The conditional expression could
 contain any C operator except for the
 assignment operators,increment, and
 decrement operators.
</blockquote>

This says, you cannot write these,

<pre><code>#if X = 4 //wrong

#if X++ //this too wrong

#if X-- //this too
</code></pre>

<hr>

EDIT:

I just read your comment on <a href="https://stackoverflow.com/questions/4554313/doubt-in-conditional-inclusion/4554353#4554353">Diego Sevilla's post</a>,

<blockquote>
 Agreed but why something like this
 (ideone.com/ahXDA) is not allowed ? –
 Philando Gullible
</blockquote>

This code actually equivalent to this which you write in the global scope,

<pre><code>//global code
int b=7; 
b++; //ill-formed!
</code></pre>

Which is not allowed in C (and C++). This however is allowed in C ONLY,

<pre><code>//legal in C Only
int b = 7;
int b; //well-formed!
</code></pre>

Here the second definition is called tentative definition. C Standard says in $6.9.2/2,

<blockquote>
 A declaration of an identifier for an
 object that has file scope without an
 initializer, and without a
 storage-class specifier or with the
 storage-class specifier static,
 constitutes a tentative definition.If a translation unit contains one or more tentative definitions for an identifier, and the translation unit contains no external definition for that identifier, then the behavior is exactly as if the translation unit contains a file scope declaration of that identifier, with the composite type as of the end of the translation unit, with an initializer equal to 0.
</blockquote>

Which again implies, the following is not allowed,

<pre><code>//illegal in both C and C++
int b=7;
int b=8; //error 
</code></pre>

It's not a tentative definition anymore. It's redefinition, which is not allowed!

blocks|key|2010990|text|你似乎误解了条件表达式指的是什么。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2010991|在这个片段里|2010992|#if+defined+TEST
++int+a+=+0;
#endif|code-block|syntax|javascript|2010993|条件表达式是由#if测试的部分，这意味着它是defined+TEST部件。|offset|length|style|CODE|2010994|不允许赋值、增量和递减的原因是，这些运算符想要更改变量，这在预处理器的上下文中是不合理的。|2010995|预处理程序完全基于文本替换和结果常量表达式的计算。|2010996|如果你有这段代码|2010997|#define+X+a%2B%2B
#define+Y+42

#if+X+==+Y
#endif|2010998|然后在测试#if+X+==+Y中，首先将X和Y替换为它们的宏展开(分别为a%2B%2B和42)，从而导致|2010999|#if+a%2B%2B+==+42|2011000|接下来，a被其宏扩展所取代。由于没有宏a，因此将替换定义为生成0。|2011001|#if+0%2B%2B+==+42|2011002|现在，没有可能的宏名称需要展开，因此预处理程序尝试计算条件。由于试图增加常量0，此计算失败时出错。|2011003|entityMap^0|0|0|0|M|C|0|0|0|0|0|5|A|K|1|M|1|10|3|14|2|0|0|4|1|J|1|V|1|0|0|12|1|0^^$0|@$1|2|3|4|5|6|7|18|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|19|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|1A|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|1B|8|@$K|1C|L|1D|M|N]]|9|@]|A|$]]|$1|O|3|P|5|6|7|1E|8|@]|9|@]|A|$]]|$1|Q|3|R|5|6|7|1F|8|@]|9|@]|A|$]]|$1|S|3|T|5|6|7|1G|8|@]|9|@]|A|$]]|$1|U|3|V|5|F|7|1H|8|@]|9|@]|A|$G|H]]|$1|W|3|X|5|6|7|1I|8|@$K|1J|L|1K|M|N]|$K|1L|L|1M|M|N]|$K|1N|L|1O|M|N]|$K|1P|L|1Q|M|N]|$K|1R|L|1S|M|N]]|9|@]|A|$]]|$1|Y|3|Z|5|F|7|1T|8|@]|9|@]|A|$G|H]]|$1|10|3|11|5|6|7|1U|8|@$K|1V|L|1W|M|N]|$K|1X|L|1Y|M|N]|$K|1Z|L|20|M|N]]|9|@]|A|$]]|$1|12|3|13|5|F|7|21|8|@]|9|@]|A|$G|H]]|$1|14|3|15|5|6|7|22|8|@$K|23|L|24|M|N]]|9|@]|A|$]]|$1|16|3|-4|5|6|7|25|8|@]|9|@]|A|$]]]|17|$]]

You seem to misunderstand what the phrase conditional expression refers to. 
In this snippet

<pre><code>#if defined TEST
 int a = 0;
#endif
</code></pre>

The conditional expression is the part being tested by the #if, meaning it is the <code>defined TEST</code> part.

The reason that assignment, increment and decrement are not allowed is because those operators want to change a variable, which is nonsensical in the context of the preprocessor. 
The preprocessor works entirely based on textual substitution and evaluation of the resulting constant expressions.

If you have this code

<pre><code>#define X a++
#define Y 42

#if X == Y
#endif
</code></pre>

Then in the test <code>#if X == Y</code>, first <code>X</code> and <code>Y</code> are replaced by their macro expansion (respectively <code>a++</code> and <code>42</code>), resulting in

<pre><code>#if a++ == 42
</code></pre>

Next, <code>a</code> is replaced by its macro expansion. As there is no macro <code>a</code>, the replacement is defined to result in <code>0</code>:

<pre><code>#if 0++ == 42
</code></pre>

Now there are no possible macro names left to expand, so the preprocessor tries to evaluate the condition. As there is an attempt to increment the constant <code>0</code>, this evaluation fails with an error.

This is actually extracted from my module (Pre-processor in C)

The conditional expression could contain any C operator except for the assignment operators,increment, and decrement operators.

I am not sure if I am getting this statement or not since I tried using <a href="http://ideone.com/r5Spc" rel="nofollow">this</a> and it worked.Also for other manipulation a probable work around would be to simply declare macro or function inside the conditional expression,<a href="http://ideone.com/83Vm0" rel="nofollow">something like this to be precise</a>.

Also I don't understand what is the rationale behind this rule. Could somebody explain?

Thanks

Doubt in Conditional inclusion

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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这实际上是从我的模块(C中的预处理器)中提取的。条件表达式可以包含任何C运算符，但赋值运算符、增量运算符和递减运算符除外。我不确定我是否得到了这条语句，因为我尝试使用和worked.Also进行其他操作，可能的工作是简单地在条件表达式中声明宏或函数。而且，我也不明白这条规则背后的理由是什么。有人能解释一下吗？谢谢

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