blocks|key|4435982|text|更改为：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4435983|printf("Please+enter+an+integer+");
scanf("+%25d\n",&choosen_number);|code-block|syntax|javascript|4435984|按照您的方式，choosen_number将读取为0。|4435985|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|$]]

Change to:

<pre><code>printf("Please enter an integer ");
scanf(" %d\n",&amp;choosen_number);
</code></pre>

as the way you have it, choosen_number will read as 0.

blocks|key|5904088|text|你的代码有一些问题。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5904089|除了在stdin中用于忽略空白的前缀空格之外，|offset|length|style|CODE|5904090|scanf应该只包含您要扫描的格式说明符。|ordered-list-item|5904091|5904092|printf(“请输入整数：")；scanf(”%25d\n"，&choosen_number);|5904093|In您的第二个while循环，当您逐位处理数字时，仅在当前处理的数字等于1时递减count。这意味着对于任何不完全由1组成的数字，您将有一个无限循环。|unordered-list-item|5904094|5904095|entityMap^0|0|3|5|0|0|5|0|0|0|7|5|14|5|0|0^^$0|@$1|2|3|4|5|6|7|T|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|U|8|@$D|V|E|W|F|G]]|9|@]|A|$]]|$1|H|3|I|5|J|7|X|8|@$D|Y|E|Z|F|G]]|9|@]|A|$]]|$1|K|3|-4|5|6|7|10|8|@]|9|@]|A|$]]|$1|L|3|M|5|6|7|11|8|@]|9|@]|A|$]]|$1|N|3|O|5|P|7|12|8|@$D|13|E|14|F|G]|$D|15|E|16|F|G]]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|17|8|@]|9|@]|A|$]]|$1|R|3|-4|5|6|7|18|8|@]|9|@]|A|$]]]|S|$]]

There are a few problems with your code.

<ol>
<li><code>scanf</code> should just contain the format specifier you're scanning for, in addition to a prepended space for ignoring whitespace in <code>stdin</code>. 

<pre><code>printf("Please enter an integer: ");
scanf(" %d\n", &amp;choosen_number);
</code></pre></li>
<li>In your second <code>while</code> loop where you process the number digit-by-digit, you only decrement <code>count</code> if the digit you're currently processing is equal to 1. This means that for any number that isn't composed entirely of 1s, you will have an infinite loop.</li>
</ol>

blocks|key|2119055|text|1这条线|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2119056|scanf("Please+enter+an+integer+%25d\n",&choosen_number);|code-block|syntax|javascript|2119057|应该是|2119058|printf("Please+enter+an+integer:+");
scanf("%25d",+&choosen_number);|2119059|2你可以这样做：|2119060|#include+<stdio.h>

int+main(){
++int+choosen_number;
++int+count+=+0;

++printf("Please+enter+an+integer:+");
++scanf("%25d",+&choosen_number);
++while+(choosen_number+!=+0)+{
++++if+(choosen_number%2510+==+1)
++++++%2B%2Bcount;
++++choosen_number+/=+10;
++}
++printf("the+number+of+ones+is+%25d\n",+count);
++return+0;++
}|2119061|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|$]]

1 The line 

<pre><code>scanf("Please enter an integer %d\n",&amp;choosen_number);
</code></pre>

should be

<pre><code>printf("Please enter an integer: ");
scanf("%d", &amp;choosen_number);
</code></pre>

2 You can do like this: 

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

int main(){
 int choosen_number;
 int count = 0;

 printf("Please enter an integer: ");
 scanf("%d", &amp;choosen_number);
 while (choosen_number != 0) {
 if (choosen_number%10 == 1)
 ++count;
 choosen_number /= 10;
 }
 printf("the number of ones is %d\n", count);
 return 0; 
}
</code></pre>

blocks|key|4436086|text|您甚至可以在不使用"if“语句的情况下完成此操作|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4436087|#include<stdio.h>
int+main()
{
++++unsigned+int+choosen_number,digit,onesnumber+=0;++++++
++++printf("Please+gine+a+positive+integer\n");
++++scanf("+%25d",&choosen_number);+
++++//+Do+some+error+handling+here++
++++while+(choosen_number>0)
++++{
++++++++digit=choosen_number%2510;+
++++++++onesnumber+=onesnumber+%2B(digit==1);//(digit==1)+will+return+0+or+1
++++++++choosen_number=choosen_number/10;+++++
++++}
++++printf("the+number+of+ones+is+%25d\n",onesnumber+);
++++return+0;
}|code-block|syntax|javascript|4436088|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|$]]

You can even do it without using "if" statement

<pre><code>#include&lt;stdio.h&gt;
int main()
{
 unsigned int choosen_number,digit,onesnumber =0; 
 printf("Please gine a positive integer\n");
 scanf(" %d",&amp;choosen_number); 
 // Do some error handling here 
 while (choosen_number&gt;0)
 {
 digit=choosen_number%10; 
 onesnumber =onesnumber +(digit==1);//(digit==1) will return 0 or 1
 choosen_number=choosen_number/10; 
 }
 printf("the number of ones is %d\n",onesnumber );
 return 0;
}
</code></pre>

my mission was to take integer from a user and calculate how many times the number 1 is in it , here is my code:

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


int main(){
int choosen_number;
int count = 0;
int onesnumber = 0;
 scanf("Please enter an integer %d\n",&amp;choosen_number);
 int indicator = choosen_number;
 while (indicator &gt; 0){
 indicator = indicator/10;
 count = count + 1;
 }
 while (count!=0){

 if (choosen_number%10 ==1){
 count = count - 1;
 onesnumber = onesnumber + 1;
 }
 choosen_number = choosen_number /10;

 }


 printf("the number of ones is %d\n",onesnumber );
 }
</code></pre>

As i am trying to execute it with a number like 12312 it says the number of ones are 0 everytime i try something like that.

C beginner execise , taking integer from user and calculate how many time the number of 1 is in it

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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我的任务是从用户那里获取整数，并计算数字1在其中的次数，以下是我的代码：#include <stdio.h>int main(){int choosen_number;int count = 0;int onesnumber = 0; scanf("Please enter an integer %d\n",...

问C初学者练习，从用户那里取整数，计算1在其中的次数
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问C初学者练习，从用户那里取整数，计算1在其中的次数EN