blocks|key|4441890|text|预处理器在编译器之前运行，这意味着预处理器对编译器定义的符号一无所知，因此不能依赖这些符号执行操作。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4441891|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|$]]

The preprocessor runs before the compiler, which means that the preprocessor doesn't know anything about symbols defined by the compiler and therefore it can't act depending on them.

blocks|key|4442239|text|这是因为预处理器在编译前工作。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4442240|由于枚举定义发生在编译时，因此A和B在预处理时都被定义为空(pp-+0)，因此相等，因此输出语句包含在编译后的代码中。|offset|length|style|CODE|4442241|当您使用#define时，它们在预处理时的定义不同，因此语句的计算结果为false。|4442242|关于你想要做什么的评论，你不需要使用预处理器#if来做这件事。您可以只使用标准if，因为MODE和MODE_GREY+(或MODE_RGB或MODE_CMYK)都仍然是已定义的：|4442243|#include+<iostream>

enum+T+{+MODE_RGB+=+1,+MODE_GREY+=+2,+MODE_CMYK+=+3+};

#define+MODE+MODE_GREY

int+main()
{
++++if(+MODE+==+MODE_GREY+)
++++++++std::cout+<<+"Grey+mode"+<<+std::endl;
++++else+if(+MODE+==+MODE_RGB+)
++++++++std::cout+<<+"RGB+mode"+<<+std::endl;
++++else+if(+MODE+==+MODE_CMYK+)
++++++++std::cout+<<+"CMYK+mode"+<<+std::endl;

++++return+0;
}|code-block|syntax|javascript|4442244|只使用预处理器的另一个选项是@TripeHound+correctly+answered+below。|4442245|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/a/34678859/3933895^0|0|Y|1|0|4|7|0|M|3|13|2|18|4|1D|9|1P|8|1Y|9|0|0|Q|O|0|0^^$0|@$1|2|3|4|5|6|7|10|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|11|8|@$D|12|E|13|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|14|8|@$D|15|E|16|F|G]]|9|@]|A|$]]|$1|J|3|K|5|6|7|17|8|@$D|18|E|19|F|G]|$D|1A|E|1B|F|G]|$D|1C|E|1D|F|G]|$D|1E|E|1F|F|G]|$D|1G|E|1H|F|G]|$D|1I|E|1J|F|G]]|9|@]|A|$]]|$1|L|3|M|5|N|7|1K|8|@]|9|@]|A|$O|P]]|$1|Q|3|R|5|6|7|1L|8|@]|9|@$D|1M|E|1N|1|1O]]|A|$]]|$1|S|3|-4|5|6|7|1P|8|@]|9|@]|A|$]]]|T|$U|$5|V|W|X|A|$Y|Z]]]]

This is because the preprocessor works before compile time.

As the enum definitions occur at compile time, A and B will both be defined as empty (pp-number <code>0</code>) - and thus equal - at pre-processing time, and thus the output statement is included in the compiled code.

When you use <code>#define</code> they are defined differently at pre-processing time and thus the statement evaluates to false.

In relation to your comment about what you want to do, you don't need to use pre-processor <code>#if</code> to do this. You can just use the standard <code>if</code> as both <code>MODE</code> and <code>MODE_GREY</code> (or <code>MODE_RGB</code> or <code>MODE_CMYK</code>) are all still defined:

<pre><code>#include &lt;iostream&gt;

enum T { MODE_RGB = 1, MODE_GREY = 2, MODE_CMYK = 3 };

#define MODE MODE_GREY

int main()
{
 if( MODE == MODE_GREY )
 std::cout &lt;&lt; "Grey mode" &lt;&lt; std::endl;
 else if( MODE == MODE_RGB )
 std::cout &lt;&lt; "RGB mode" &lt;&lt; std::endl;
 else if( MODE == MODE_CMYK )
 std::cout &lt;&lt; "CMYK mode" &lt;&lt; std::endl;

 return 0;
}
</code></pre>

The other option using only the pre-processor is to do this as @TripeHound <a href="https://stackoverflow.com/a/34678859/3933895">correctly answered below</a>.

blocks|key|4442315|text|在条件预处理器指令中，未定义宏的标识符被解释为值0。因此，由于您没有定义宏A和B，因此它们都被认为是0，并且两个0彼此相等。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4442316|未定义(对于预处理器)标识符被视为0的原因是因为它允许在条件中使用未定义的宏，而不使用#ifdef。|4442317|entityMap^0|11|1|13|1|0|17|6|0^^$0|@$1|2|3|4|5|6|7|J|8|@$9|K|A|L|B|C]|$9|M|A|N|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|O|8|@$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|H|3|-4|5|6|7|R|8|@]|D|@]|E|$]]]|I|$]]

Identifiers that are not defined macros are interpreted as value 0 in conditional preprocessor directives. Therefore, since you hadn't defined macros <code>A</code> and <code>B</code>, they are both considered 0 and two 0 are equal to each other.

The reason why undefined (to the pre-processor) identifiers are considered 0 is because it allows using undefined macros in the conditional without using <code>#ifdef</code>.

blocks|key|4442123|text|没有名为A或B的宏，因此在您的#if行中，A和B将被0替换，因此您实际上拥有：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4442124|enum+T+{+A+=+1,+B+=+2+};

int+main()+{
#if+(0+==+0)
++++std::cout+<<+A+<<+B;
#endif
}|code-block|syntax|javascript|4442125|在编译器了解您的enum之前，预处理器就会运行。预处理器只知道宏(#define)。|4442126|entityMap^0|4|1|6|1|F|3|L|1|N|1|Q|1|0|0|8|4|X|7|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]|$9|R|A|S|B|C]|$9|T|A|U|B|C]|$9|V|A|W|B|C]|$9|X|A|Y|B|C]|$9|Z|A|10|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|11|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|12|8|@$9|13|A|14|B|C]|$9|15|A|16|B|C]]|D|@]|E|$]]|$1|M|3|-4|5|6|7|17|8|@]|D|@]|E|$]]]|N|$]]

There are no macros called <code>A</code> or <code>B</code>, so on your <code>#if</code> line, <code>A</code> and <code>B</code> get replaced by <code>0</code>, so you actually have:

<pre><code>enum T { A = 1, B = 2 };

int main() {
#if (0 == 0)
 std::cout &lt;&lt; A &lt;&lt; B;
#endif
}
</code></pre>

The preprocessor runs before the compiler knows anything about your <code>enum</code>. The preprocessor only knows about macros (<code>#define</code>).

blocks|key|4442387|text|其他答案解释了为什么你正在尝试的东西不起作用；如果是另一种选择，我可能会选择：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4442388|#define+RGB+1
#define+GREY+2
#define+CMYK+3
#define+MODE+RGB

#if+MODE+==+RGB
++++//RGB-mode+code
#elif+MODE+==+GREY
++++//Greyscale+code
#elif+MODE+==+CMYK
++++//CMYK+code
#else
#++++error+Undefined+MODE
#endif|code-block|syntax|javascript|4442389|如果存在与“真实”源代码冲突的危险，您可能需要在RGB/GREY/CMYK上添加前缀。|4442390|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

Other answers explain why what you're trying doesn't work; for an alternative, I'd probably go with:

<pre><code>#define RGB 1
#define GREY 2
#define CMYK 3
#define MODE RGB

#if MODE == RGB
 //RGB-mode code
#elif MODE == GREY
 //Greyscale code
#elif MODE == CMYK
 //CMYK code
#else
# error Undefined MODE
#endif
</code></pre>

You might want prefixes on the RGB/GREY/CMYK if there's a danger of clashes with "real" source code.

blocks|key|4442399|text|文章已经解释了原因，但是一个保持可读性的可能的解决方案可能是这样的|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4442400|#define+MODE_RGB

int+main()
{++++++++
++++#ifdef+MODE_RGB
++++++++std::cout+<<+"RGB+mode"+<<+std::endl;
++++#elif+defined+MODE_GREY
++++++++std::cout+<<+"Grey+mode"+<<+std::endl;
++++#elif+defined+MODE_CMYK
++++++++std::cout+<<+"CMYK+mode"+<<+std::endl;
++++#endif
}|code-block|syntax|javascript|4442401|然后，您只需将顶部的宏更改为仅定义您感兴趣的宏。您还可以包含一个检查，以确保定义了一个且仅定义了一个，如果未定义，则执行#error+"You+must+define+MODE_RGB,+MODE_GREY+or+MODE_CMYK|offset|length|style|CODE|4442402|entityMap^0|0|0|1O|1K|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@$I|R|J|S|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

The posts have explained why, but a possible solution for you that keeps readability might be like this

<pre><code>#define MODE_RGB

int main()
{ 
 #ifdef MODE_RGB
 std::cout &lt;&lt; "RGB mode" &lt;&lt; std::endl;
 #elif defined MODE_GREY
 std::cout &lt;&lt; "Grey mode" &lt;&lt; std::endl;
 #elif defined MODE_CMYK
 std::cout &lt;&lt; "CMYK mode" &lt;&lt; std::endl;
 #endif
}
</code></pre>

You just then need to change the macro at the top, to only the macro you are interested in is defined. You could also include a check to make sure that one and only one is defined and if not then and do <code>#error "You must define MODE_RGB, MODE_GREY or MODE_CMYK</code>

Why does the <code>std::cout</code> line in the following code run even though <code>A</code> and <code>B</code> are different?

<pre><code>#include &lt;iostream&gt;

enum T { A = 1, B = 2 };
// #define A 1
// #define B 2

int main() {
#if (A == B)
 std::cout &lt;&lt; A &lt;&lt; B;
#endif
}
</code></pre>

If I use <code>#define</code> instead (as commented out), I get no output as I expect.

Reason for the question:

I want to have a mode selector for some test code in which I can easily change modes by commenting/uncommenting lines on top:

<pre><code>enum T { MODE_RGB = 1, MODE_GREY = 2, MODE_CMYK = 3 };
// #define MODE MODE_RGB
#define MODE MODE_GREY
// #define MODE MODE_CMYK

int main() {
#if (MODE == MODE_RGB)
 // do RGB stuff
#elif (MODE == MODE_GREY)
 // do greyscale stuff
#else
 // do CMYK stuff
#endif

 // some common code

 some_function(arg1, arg2,
#if (MODE == MODE_RGB)
 // RGB calculation for arg3,
#elif (MODE == MODE_GREY)
 // greyscale calculation for arg3,
#else
 // CMYK calculation for arg3,
#endif
 arg4, arg5);
}
</code></pre>

I know I can use numeric values e.g.

<pre><code>#define MODE 1 // RGB
...
#if (MODE == 1) // RGB
</code></pre>

but it makes the code less readable.

Is there an elegant solution for this?

Why does the C preprocessor consider enum values as equal?

为什么在A和B不同的情况下，以下代码中的std::cout行仍然可以运行？#include <iostream>enum T { A = 1, B = 2 };// #define A 1// #define B 2int main() {#if (A == B) std::cout << A << B;#end...

问为什么C预处理器认为枚举值相等？
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