blocks|key|2856844|text|这里的问题是您没有静态地初始化s。所以每次你调用f()的时候，它总是从0开始，f()一次又一次地调用它自己，但是从来没有遇到过if(s+==+10)。调用堆栈最终会溢出，并突然出现分段错误。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2856845|entityMap^0|F|1|O|3|13|3|1R|B|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]|$9|K|A|L|B|C]|$9|M|A|N|B|C]|$9|O|A|P|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|G|$]]

The problem here is that you're not statically initializing <code>s</code>. So it'll always start off as 0 every time you call <code>f()</code>, and <code>f()</code> keeps calling itself over and over again but <code>if(s == 10)</code> is never met. The call stack eventually overflows and boom, segmentation fault.

blocks|key|946113|text|因为s是一个局部变量，所以每次对f()的递归调用都会获得它自己的副本。所以每次s都会是1，你会得到一个堆栈溢出异常。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|946114|entityMap^0|2|1|G|3|13|1|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]|$9|K|A|L|B|C]|$9|M|A|N|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|O|8|@]|D|@]|E|$]]]|G|$]]

Since <code>s</code> is a local variable, each recursive call to <code>f()</code> gets its own copy of it. So every time <code>s</code> will be 1 and the you get a stack overflow exception.

blocks|key|948346|text|问题是您递归地调用f，而不是每次都更改任何内容。你可能会得到一个堆栈溢出。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|948347|每次调用f时，都会将s初始化为0并递增。它永远不会达到10。|948348|您应该做的是要么将s设置为f的参数，要么将s设置为静态。|948349|entityMap^0|9|1|0|4|1|A|1|F|1|0|9|1|D|1|L|1|0^^$0|@$1|2|3|4|5|6|7|L|8|@$9|M|A|N|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|O|8|@$9|P|A|Q|B|C]|$9|R|A|S|B|C]|$9|T|A|U|B|C]]|D|@]|E|$]]|$1|H|3|I|5|6|7|V|8|@$9|W|A|X|B|C]|$9|Y|A|Z|B|C]|$9|10|A|11|B|C]]|D|@]|E|$]]|$1|J|3|-4|5|6|7|12|8|@]|D|@]|E|$]]]|K|$]]

The problem is that you're calling <code>f</code> recursively, without changing anything each time. You're probably getting a stack overflow.

Each time you call <code>f</code>, you initialize <code>s</code> to <code>0</code> and increment it. It will never reach 10.

What you should do is either make <code>s</code> a parameter to <code>f</code> or else make <code>s</code> static.

blocks|key|2856865|text|你的程序是一个无限循环；你得到了分段错误，因为你最终溢出了堆栈。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2856866|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|$]]

Your program is an infinite loop; you get the segmentation fault because you eventually overflow your stack.

blocks|key|2856971|text|程序进入无限循环。原因是:s对于f()来说是局部的，所以只要f()被调用，s就会被初始化为0。然后它加1，&+f()再次被调用。这一次，f()使用它的本地副本，将s初始化为0，并再次递归调用f()。这会生成一个无限循环&程序会生成一个分段错误。seg错误的原因是递归无限调用导致的堆栈溢出。要解决这个问题，要么将s设置为静态，要么将其设置为全局的，而不是局部的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2856972|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 program goes into an infinite loop.
The reason for that is :
s is local to f(), so whenever f() is called, s is initialized to 0. Then it is incremented by 1, &amp; f() is again called. This time as well, f() uses its local copy &amp; initializes s to 0 &amp; again calls f() recursively. This generates an infinite loop &amp; the program generates a segmentation fault. The reason for seg fault is stack overflow due to recursive infinite calls.
To remedy this, either make s static or make it global, rather than local.

blocks|key|1694551|text|把你的变量设为静态的怎么样？|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1694552|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|$]]

How about making your variable s static?

blocks|key|2857005|text|这是一个调用自身的递归函数。但是，它是一个无限循环，因为int+s+=+0;在每次调用它时都会重置计数器。它将打印无休止的0行。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2857006|示例解决方案：|2857007|int+main+(void)+{
+++f(0);
}

void+f+(int+s)+{
+++if+(s+==+10)
++++++return;
+++printf("%25d+",+s);
+++f(s%2B%2B);
}|code-block|syntax|javascript|2857008|我对C的语法不太在行，但是我的想法很好。|2857009|entityMap^0|S|A|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|T|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|U|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|V|8|@]|D|@]|E|$]]|$1|O|3|-4|5|6|7|W|8|@]|D|@]|E|$]]]|P|$]]

This is a recursive function that calls itself. However, it is an infinite loop because <code>int s = 0;</code> resets the counter every time it is called. It will print an endless line of 0s.

Sample solution:

<pre><code>int main (void) {
 f(0);
}

void f (int s) {
 if (s == 10)
 return;
 printf("%d ", s);
 f(s++);
}
</code></pre>

I'm not good with C syntax, but the idea is sound.

blocks|key|2857017|text|无限循环。这可能是你想要的东西|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2857018|#include<stdio.h>

void+f(int+s)
{
++++s%2B%2B;
++++if(s+==+10)
++++++++return;
++++f(s);
++++printf("%25d+",+s);
}
int+main(void)
{
++++f(0);
}|code-block|syntax|javascript|2857019|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Infinite loop. Here is probably something similar to what you want

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

void f(int s)
{
 s++;
 if(s == 10)
 return;
 f(s);
 printf("%d ", s);
}
int main(void)
{
 f(0);
}
</code></pre>

blocks|key|1694600|text|这可能是一个更有趣的程序：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1694601|void+f(int+s)
{
++s%2B%2B;
++if(s+==+10)
++++return;
++f(s);
++printf("%25d+",+s);
}

int+main(void)+
{
++f(0);
}|code-block|syntax|javascript|1694602|这将打印：|1694603|9+8+7+6+5+4+3+2+1|offset|length|style|CODE|1694604|entityMap^0|0|0|0|0|H|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]]|$1|G|3|H|5|6|7|S|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|T|8|@$K|U|L|V|M|N]]|9|@]|A|$]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

This might be a more interesting program:

<pre><code>void f(int s)
{
 s++;
 if(s == 10)
 return;
 f(s);
 printf("%d ", s);
}

int main(void) 
{
 f(0);
}
</code></pre>

That will print:

<code>9 8 7 6 5 4 3 2 1</code>

blocks|key|2857079|text|这个程序的输出是这个站点的名称。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2857080|正如其他人建议的那样，您应该将s声明为static+int+s+=+0|offset|length|style|CODE|2857081|entityMap^0|0|F|1|J|G|0^^$0|@$1|2|3|4|5|6|7|J|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|K|8|@$D|L|E|M|F|G]|$D|N|E|O|F|G]]|9|@]|A|$]]|$1|H|3|-4|5|6|7|P|8|@]|9|@]|A|$]]]|I|$]]

The output of this program is the name of this site.

As others suggested you should declare <code>s</code> as <code>static int s = 0</code>

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

void f(void)
{
 int s = 0;
 s++;
 if(s == 10)
 return;
 f();
 printf("%d ", s);
}
int main(void)
{
 f();
}
</code></pre>

what is the output of the programme!??
i m segmentation fault ...what is it?

output of the programme

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 f(void){ int s = 0; s++; if(s == 10) return; f(); printf("%d ", s);}int main(void){ f();}程序的输出是什么！？？我是分段错误...what，是吗？

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