blocks|key|1335300|text|正如前面提到的--什么都没有。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1335301|此外，您的代码中还存在打印错误。|1335302|它应该是#else而不是#elif。如果是#elif，则后跟新条件|offset|length|style|BOLD|1335303|#include+<iostream>+++

#ifdef+NOOP+++++++
++++#define+conditional_noop(x)+do+{}+while(0)
#else+++++++
++++#define+conditional_noop(x)+std::cout+<<+(x)+++
#endif++|code-block|syntax|javascript|1335304|祝你编码愉快！编辑:根据另一个答案中的建议，添加了do结构以实现健壮性。|1335305|entityMap^0|0|0|4|5|C|5|L|5|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|T|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|U|8|@$F|V|G|W|H|I]|$F|X|G|Y|H|I]|$F|Z|G|10|H|I]]|9|@]|A|$]]|$1|J|3|K|5|L|7|11|8|@]|9|@]|A|$M|N]]|$1|O|3|P|5|6|7|12|8|@]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|13|8|@]|9|@]|A|$]]]|R|$]]

As mentioned before - nothing. 
Also, there is a misprint in your code. 
it should be #else not #elif. if it is #elif it is to be followed by the new condition

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

#ifdef NOOP 
 #define conditional_noop(x) do {} while(0)
#else 
 #define conditional_noop(x) std::cout &lt;&lt; (x) 
#endif 
</code></pre>

Have fun coding!
EDIT: added the [do] construct for robustness as suggested in another answer.

blocks|key|231457|text|虽然将其留空是显而易见的选择，但我会选择|type|unstyled|depth|inlineStyleRanges|entityRanges|data|231458|#define+conditional_noop(x)+do+{}+while(0)|code-block|syntax|javascript|231459|这个技巧显然是无操作的，但是会迫使您在conditional_noop(123)后面写一个分号。|offset|length|style|CODE|231460|entityMap^0|0|0|J|L|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|$]]

While leaving it blank is the obvious option, I'd go with

<pre><code>#define conditional_noop(x) do {} while(0)
</code></pre>

This trick is obviously no-op, but forces you to write a semicolon after <code>conditional_noop(123)</code>.

blocks|key|1532885|text|将宏定义为void可以很好地传达您的意图。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1532886|#ifdef+NOOP
++++#define+conditional_noop(x)+(void)0
#else|code-block|syntax|javascript|1532887|entityMap^0|5|4|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

Defining the macro to be <code>void</code> conveys your intent well.

<pre><code>#ifdef NOOP
 #define conditional_noop(x) (void)0
#else
</code></pre>

blocks|key|4551620|text|+#ifdef+NOOP+++++++
+++++#define+conditional_noop(x)+++
+#elif++|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|4551621|没什么!|unstyled|4551622|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|J|8|@]|9|@]|A|$]]|$1|G|3|-4|5|F|7|K|8|@]|9|@]|A|$]]]|H|$]]

<pre><code> #ifdef NOOP 
 #define conditional_noop(x) 
 #elif 
</code></pre>

nothing!

blocks|key|231340|text|#ifdef+NOOP
++++static+inline+void+conditional_noop(int+x)+{+}
#else+
++++static+inline+void+conditional_noop(int+x)+{+std::cout+<<+x;+}
#endif|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|231341|使用内联函数void可以启用类型检查，即使没有定义NOOP。因此，当NOOP未定义时，您仍然无法将结构传递给该函数或未定义的变量。当您打开NOOP标志时，这最终将防止您获得编译器错误。|unstyled|offset|length|style|CODE|231342|entityMap^0|0|P|4|Y|4|1X|4|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|N|8|@$G|O|H|P|I|J]|$G|Q|H|R|I|J]|$G|S|H|T|I|J]]|9|@]|A|$]]|$1|K|3|-4|5|F|7|U|8|@]|9|@]|A|$]]]|L|$]]

<pre><code>#ifdef NOOP
 static inline void conditional_noop(int x) { }
#else 
 static inline void conditional_noop(int x) { std::cout &lt;&lt; x; }
#endif
</code></pre>

Using inline function void enables type checking, even when <code>NOOP</code> isn't defined. So when <code>NOOP</code> isn't defined, you still won't be able to pass a struct to that function, or an undefined variable.
This will eventually prevent you from getting compiler errors when you turn the <code>NOOP</code> flag on.

blocks|key|1530969|text|你可以把它留空。您不需要遵循#define中的任何内容。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1530970|entityMap^0|E|7|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|K|8|@]|D|@]|E|$]]]|G|$]]

You can just leave it blank. You don't need to follow the <code>#define</code> with anything.

blocks|key|4551845|text|就像其他人说的，把它留空。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4551846|您应该使用的一个技巧是将(void)0添加到宏中，强制用户在宏后添加分号：|offset|length|style|CODE|4551847|#ifdef+NOOP+++++++
++++#define+conditional_noop(x)+(void)0
#else+++++++
++++#define+conditional_noop(x)+std::cout+<<+(x);+(void)0
#endif++|code-block|syntax|javascript|4551848|在C%2B%2B中，(void)0什么也不做。This+article+explains其他不太好的选项，以及它们背后的理由。|4551849|entityMap|0|LINK|mutability|MUTABLE|url|http://www.parashift.com/c%2B%2B-faq-lite/misc-technical-issues.html#faq-39.5^0|0|C|7|0|0|6|7|J|L|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|X|8|@$D|Y|E|Z|F|G]]|9|@]|A|$]]|$1|H|3|I|5|J|7|10|8|@]|9|@]|A|$K|L]]|$1|M|3|N|5|6|7|11|8|@$D|12|E|13|F|G]]|9|@$D|14|E|15|1|16]]|A|$]]|$1|O|3|-4|5|6|7|17|8|@]|9|@]|A|$]]]|P|$Q|$5|R|S|T|A|$U|V]]]]

Like others have said, leave it blank.

A trick you should use is to add <code>(void)0</code> to the macro, forcing users to add a semicolon after it:

<pre><code>#ifdef NOOP 
 #define conditional_noop(x) (void)0
#else 
 #define conditional_noop(x) std::cout &lt;&lt; (x); (void)0
#endif 
</code></pre>

In C++, <code>(void)0</code> does nothing. <a href="http://www.parashift.com/c++-faq-lite/misc-technical-issues.html#faq-39.5" rel="nofollow noreferrer">This article explains</a> other not-as-good options, as well as the rationale behind them.

blocks|key|1532950|text|由于这是一个宏，因此您还应该考虑如下情况|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1532951|if+(other_cond)
++++conditional_noop(123);|code-block|syntax|javascript|1532952|为了安全起见，您可以提供一个空语句，如下所示|1532953|#define+conditional_noop(X)+{}|1532954|对于旧的C语言，有时你需要这样定义empty语句(也应该进行优化)：|1532955|#define+conditional_noop(X)+do+{}+while(0)|1532956|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|U|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

As this is a macro, you should also consider a case like

<pre><code>if (other_cond)
 conditional_noop(123);
</code></pre>

to be on the safe side, you can give an empty statement like

<pre><code>#define conditional_noop(X) {}
</code></pre>

for older C sometimes you need to define the empty statment this way (should also get optimized away):

<pre><code>#define conditional_noop(X) do {} while(0)
</code></pre>

blocks|key|4551955|text|我认为结合前面的变种是一个很好的解决方案：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4551956|#ifdef+NOOP
++++static+inline+void+conditional_noop(int+x)+do+{}+while(0)
#else+
++++static+inline+void+conditional_noop(int+x)+do+{+std::cout+<<+x;+}+while(0)
#endif|code-block|syntax|javascript|4551957|好消息是，这两个代码只是在块内部不同，这意味着它们在外部的行为对于解析器来说是完全相同的。|4551958|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|$]]

I think that a combination of the previous variants is a good solution:

<pre><code>#ifdef NOOP
 static inline void conditional_noop(int x) do {} while(0)
#else 
 static inline void conditional_noop(int x) do { std::cout &lt;&lt; x; } while(0)
#endif
</code></pre>

The good thing is that these two codes differ only inside a block, which means that their behaviour for the outside is completely identical for the parser.

How do I implement no-op macro in C++?

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

#ifdef NOOP 
 #define conditional_noop(x) what goes here? 
#else 
 #define conditional_noop(x) std::cout &lt;&lt; (x) 
#endif 
int main() { 
 conditional_noop(123); 
}
</code></pre>

I want this to do nothing when NOOP is defined and print "123", when NOOP is not defined.

How do I implement no-op macro (or template) in C++?

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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如何在C++中实现无操作宏？#include <iostream> #ifdef NOOP #define conditional_noop(x) what goes here? #else #define conditional_noop(x) std::cout << (x) #endif int main() { c

问如何在C++中实现无操作宏(或模板)？
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