blocks|key|2477177|text|首先，让我明确声明，|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2477178|.+They+are+supported+as+GCC+extension.**|blockquote|offset|length|2477179|好的，现在，在您的代码中，m()是main()中的一个嵌套函数。它只有main()的块范围。在main()之外，其他函数不能看到m()的存在，也不能直接调用m()。m()只能在main()中调用。|style|CODE|2477180|在您的示例中，调用m()+In+n()是导致问题的原因。即使您将前向声明提供为void+m();，链接器也无法找到m()的定义并抛出错误。|2477181|解决方案：将m()的定义移到main()之外，然后可以从任何其他函数中使用它。|BOLD|2477182|还请注意，推荐的main()签名是int+main(void)。|2477183|entityMap|0|LINK|mutability|MUTABLE|url|https://gcc.gnu.org/onlinedocs/gcc/Nested-Functions.html^0|0|0|14|0|0|D|3|H|6|Z|6|1B|6|1S|3|26|3|2A|3|2G|6|0|9|3|G|3|13|9|1L|3|0|0|5|6|3|E|6|0|8|6|H|E|0^^$0|@$1|2|3|4|5|6|7|Z|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|10|8|@]|9|@$E|11|F|12|1|13]]|A|$]]|$1|G|3|H|5|6|7|14|8|@$E|15|F|16|I|J]|$E|17|F|18|I|J]|$E|19|F|1A|I|J]|$E|1B|F|1C|I|J]|$E|1D|F|1E|I|J]|$E|1F|F|1G|I|J]|$E|1H|F|1I|I|J]|$E|1J|F|1K|I|J]]|9|@]|A|$]]|$1|K|3|L|5|6|7|1L|8|@$E|1M|F|1N|I|J]|$E|1O|F|1P|I|J]|$E|1Q|F|1R|I|J]|$E|1S|F|1T|I|J]]|9|@]|A|$]]|$1|M|3|N|5|6|7|1U|8|@$E|1V|F|1W|I|O]|$E|1X|F|1Y|I|J]|$E|1Z|F|20|I|J]]|9|@]|A|$]]|$1|P|3|Q|5|6|7|21|8|@$E|22|F|23|I|J]|$E|24|F|25|I|J]]|9|@]|A|$]]|$1|R|3|-4|5|6|7|26|8|@]|9|@]|A|$]]]|S|$T|$5|U|V|W|A|$X|Y]]]]

First, let me declare clearly, 

<blockquote>
 <a href="https://gcc.gnu.org/onlinedocs/gcc/Nested-Functions.html" rel="nofollow">Nested functions are not standard <code>C</code>. They are supported as GCC extension.</a>
</blockquote>

OK, now, in your code, <code>m()</code> is a nested function inside <code>main()</code>. It is having block scope for <code>main()</code> only. Outside <code>main()</code> other functions cannot see the existence of <code>m()</code> ,neither can call <code>m()</code> directly. <code>m()</code> can be called only inside <code>main()</code>.

In your case, the call to <code>m()</code> inside <code>n()</code> is causing the issue. Even if you provided the forward declaration as <code>void m();</code>, linker won't be able to find the definition of <code>m()</code> and throw error.

Solution: Move the definition of <code>m()</code> outside <code>main()</code>, then you can use it from any other function.

Also note, the recommended signature of <code>main()</code> is <code>int main(void)</code>.

blocks|key|2139950|text|正如其他地方所解释的，C不支持嵌套函数(gcc这样做是作为扩展，但据我所知，几乎没有其他编译器)。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2139951|您需要将m的定义移出main之外。最好在m被n使用之前定义它|offset|length|style|CODE|2139952|#include+<stdio.h>

void+m()
{
++printf("hi\n");
}

void+n()
{
++m();
}

int+main(+void+)+//+void+main()+is+not+a+valid+signature+for+main
{
++n();+++++++//+call+n,+which+calls+m,+which+prints+"hi"
++return+0;
}|code-block|syntax|javascript|2139953|entityMap^0|0|4|1|A|4|K|1|M|1|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|P|8|@$D|Q|E|R|F|G]|$D|S|E|T|F|G]|$D|U|E|V|F|G]|$D|W|E|X|F|G]]|9|@]|A|$]]|$1|H|3|I|5|J|7|Y|8|@]|9|@]|A|$K|L]]|$1|M|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|N|$]]

As has been explained elsewhere, C doesn't support nested functions as a rule (gcc does as an extension, but almost no other compiler that I know of does).

You need to move the definition of <code>m</code> outside of <code>main</code>. Preferably you should define <code>m</code> before it is used by <code>n</code>:

<pre><code>#include &lt;stdio.h&gt;

void m()
{
 printf("hi\n");
}

void n()
{
 m();
}

int main( void ) // void main() is not a valid signature for main
{
 n(); // call n, which calls m, which prints "hi"
 return 0;
}
</code></pre>

<pre><code>#include &lt;stdio.h&gt;

void m();

void n() {
 m();
}

void main() {
 void m() {
 printf("hi");
 }
}
</code></pre>

On compiling, an error 

<blockquote>
 "undefined reference to m" 
</blockquote>

is shown. Which <code>m</code> is being referred to?

Why does the following code show an error?

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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#include <stdio.h>void m();void n() { m();}void main() { void m() { printf("hi"); }}关于编译，错误“对m的未定义的引用”会显示出来。指的是哪个m？

问为什么下面的代码会显示错误？
EN

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问为什么下面的代码会显示错误？EN