blocks|key|434724|text|作为提示，获得数字中的第n位非常简单；除以10n次，然后mod+10，或C：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|434725|int+nthdig(int+n,+int+k){
+++++while(n--)
+++++++++k/=10;
+++++return+k%2510;
}|code-block|syntax|javascript|434726|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|$]]

As a hint, getting the nth digit in the number is pretty easy; divide by 10 n times, then mod 10, or in C:

<pre><code>int nthdig(int n, int k){
 while(n--)
 k/=10;
 return k%10;
}
</code></pre>

blocks|key|649164|text|int+value+=+123;
while+(value+>+0)+{
+int+digit+=+value+%25+10;
+//+do+something+with+digit
+value+/=+10;
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|649165|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>int value = 123;
while (value &gt; 0) {
 int digit = value % 10;
 // do something with digit
 value /= 10;
}
</code></pre>

blocks|key|434851|text|123的最后一个数字是123+%25+10。你可以通过做123/10来去掉123的最后一个数字--使用整数除法会得到12。要回答你关于“我怎么知道你有多少个数字”的问题--试着按照上面描述的那样做，你就会知道什么时候停止。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|434852|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 last digits of 123 is 123 % 10.
You can drop the last digit of 123 by doing 123/10 -- using integer division this will give you 12.
To answer your question about "how do I know how many digits you have" -- 
try doing it as described above and you will see how to know when to stop.

blocks|key|649363|text|首先，计算数字：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|649364|unsigned+int+count(unsigned+int+i)+{
+unsigned+int+ret=1;
+while+(i/=10)+ret%2B%2B;
+return+ret;
}|code-block|syntax|javascript|649365|然后，您可以将它们存储在一个数组中：|649366|unsigned+int+num=123;+//for+example
unsigned+int+dig=count(num);
char+arr[dig];
while+(dig--)+{
+arr[dig]=num%2510;
+num/=10;
}|649367|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

First, count the digits:

<pre><code>unsigned int count(unsigned int i) {
 unsigned int ret=1;
 while (i/=10) ret++;
 return ret;
}
</code></pre>

Then, you can store them in an array:

<pre><code>unsigned int num=123; //for example
unsigned int dig=count(num);
char arr[dig];
while (dig--) {
 arr[dig]=num%10;
 num/=10;
}
</code></pre>

blocks|key|434961|text|我认为下面这段代码会有所帮助……|type|unstyled|depth|inlineStyleRanges|entityRanges|data|434962|temp+=+num;
while(temp)
{
++++temp=temp/10;
++++factor+=+factor*10;
}

printf("\n%25d\n",+factor);
printf("Each+digits+of+given+number+are:\n");

while(factor>1)
{
++++factor+=+factor/10;
++++printf("%25d\t",num/factor);
++++i%2B%2B;
++++num+=+num+%25+factor;
}|code-block|syntax|javascript|434963|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|$]]

I think below piece of code will help....

<pre><code>temp = num;
while(temp)
{
 temp=temp/10;
 factor = factor*10;
}

printf("\n%d\n", factor);
printf("Each digits of given number are:\n");

while(factor&gt;1)
{
 factor = factor/10;
 printf("%d\t",num/factor);
 i++;
 num = num % factor;
}
</code></pre>

blocks|key|649539|text|我们可以使用这个程序作为一个函数，在"while(a%2B%2B<2)“中使用3+arguments.Here，2是你需要的位数(可以给出一个参数)用你需要的位数替换2。这里我们可以使用"z/=pow(10,6)“，如果我们不需要最后一个特定的数字，用你不需要的数字数替换6(可以作为另一个参数)，第三个参数是你需要换算的数字。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|649540|int+main(){
long+signed+c=0,z,a=0,b=1,d=1;
scanf("%25ld",&z);
while(a%2B%2B<2){
+++++++if(d%2B%2B==1)+
+++++++z/=pow(10,6);
+++++++c%2B=(z%2510)*b;+
+++++++z/=10;
+++++++b*=10;}
++++++++return+c;}|code-block|syntax|javascript|649541|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|$]]

we can use this program as a function with 3 arguments.Here in "while(a++&lt;2)", 2 is the number of digits you need(can give as one argument)replace 2 with no of digits you need. Here we can use "z/=pow(10,6)" if we don't need last certain digits ,replace 6 by the no of digits you don't need(can give as another argument),and the third argument is the number you need to break.

<pre><code>int main(){
long signed c=0,z,a=0,b=1,d=1;
scanf("%ld",&amp;z);
while(a++&lt;2){
 if(d++==1) 
 z/=pow(10,6);
 c+=(z%10)*b; 
 z/=10;
 b*=10;}
 return c;}
</code></pre>

blocks|key|435197|text|//Based+on+Tony's+answer
#include+<stdio.h>+
int+nthdig(int+n,+int+k){
++++while(n--)
++++++++k/=10;
++++return+k%2510;
}

int+main()+{
++++int+numberToSplit+=+987;
++++printf("Hundreds+=+%25i\n",nthdig(2,+numberToSplit));
++++printf("Tens+++++=+%25i\n",nthdig(1,+numberToSplit));
++++printf("Units++++=+%25i\n",nthdig(0,+numberToSplit));
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|435198|这将导致以下打印输出：|unstyled|435199|数百=9|435200|Tens+=8|435201|单位=7|435202|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|P|8|@]|9|@]|A|$]]|$1|G|3|H|5|F|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|F|7|R|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|S|8|@]|9|@]|A|$]]|$1|M|3|-4|5|F|7|T|8|@]|9|@]|A|$]]]|N|$]]

<pre><code>//Based on Tony's answer
#include &lt;stdio.h&gt; 
int nthdig(int n, int k){
 while(n--)
 k/=10;
 return k%10;
}

int main() {
 int numberToSplit = 987;
 printf("Hundreds = %i\n",nthdig(2, numberToSplit));
 printf("Tens = %i\n",nthdig(1, numberToSplit));
 printf("Units = %i\n",nthdig(0, numberToSplit));
}
</code></pre>

This results in the following printout:

Hundreds = 9

Tens = 8

Units = 7

blocks|key|649694|text|我基于@asaelr的代码实现了这一点：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|649695|typedef+struct+digitsArrayPlusNumber+{
++++uint32_t+*baseAddress;
++++uint32_t+number;
}+digitsArrayPlusNumber;

digitsArrayPlusNumber+*splitDigits+(uint32_t+inValue)+{
++++//+based+on+code+from+asaelr@stackoverflow.com

++++uint32_t+inputValue+=+inValue;

++++//Count+digits

++++uint32_t+theCount+=+1;
++++while+(inputValue+/=+10)
++++++++theCount%2B%2B;

++++//+put+in+array
++++uint32_t+*arr+=+malloc(sizeof(uint32_t)+*+theCount);
++++uint32_t+dig+=+theCount;
++++while+(dig--)+{
++++++++arr[dig]=inValue+%25+10;
++++++++inValue+/=+10;
++++++++//++printf+("%25d\n",+arr[dig]);
++++}

++++digitsArrayPlusNumber+*dandn+=+malloc+(sizeof(digitsArrayPlusNumber));

++++dandn->baseAddress+=+arr;
++++dandn->number+=+theCount;

++++return+dandn;

}

int+main(int+argc,+const+char+*+argv[])+{


++++for+(int+d+=+0;+d+<+splitDigits(12345678)->number;+d%2B%2B)
++++++++printf+("%25u\n",+(splitDigits(12345678)->baseAddress)[d]);

}|code-block|syntax|javascript|649696|它工作得很好，谢谢！|649697|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 made this based on the code from @asaelr:

<pre><code>typedef struct digitsArrayPlusNumber {
 uint32_t *baseAddress;
 uint32_t number;
} digitsArrayPlusNumber;

digitsArrayPlusNumber *splitDigits (uint32_t inValue) {
 // based on code from asaelr@stackoverflow.com

 uint32_t inputValue = inValue;

 //Count digits

 uint32_t theCount = 1;
 while (inputValue /= 10)
 theCount++;

 // put in array
 uint32_t *arr = malloc(sizeof(uint32_t) * theCount);
 uint32_t dig = theCount;
 while (dig--) {
 arr[dig]=inValue % 10;
 inValue /= 10;
 // printf ("%d\n", arr[dig]);
 }

 digitsArrayPlusNumber *dandn = malloc (sizeof(digitsArrayPlusNumber));

 dandn-&gt;baseAddress = arr;
 dandn-&gt;number = theCount;

 return dandn;

}

int main(int argc, const char * argv[]) {


 for (int d = 0; d &lt; splitDigits(12345678)-&gt;number; d++)
 printf ("%u\n", (splitDigits(12345678)-&gt;baseAddress)[d]);

}
</code></pre>

It works quite well, thanks!

blocks|key|649754|text|你可以分而治之，但是你已经重写了所有的算术库。我建议使用多精确库https://gmplib.org，但这当然是一个很好的实践|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|649755|entityMap|0|LINK|mutability|MUTABLE|url|https://gmplib.org/^0|W|I|0|0^^$0|@$1|2|3|4|5|6|7|L|8|@]|9|@$A|M|B|N|1|O]]|C|$]]|$1|D|3|-4|5|6|7|P|8|@]|9|@]|C|$]]]|E|$F|$5|G|H|I|C|$J|K]]]]

You can divide and conquer but you have rewrite all of arithmetic libraries. I suggest using a multi-precision library <a href="https://gmplib.org" rel="nofollow noreferrer">https://gmplib.org</a> But of course it is good practice

blocks|key|435377|text|您可以使用%2510，这意味着在除以后的数字的余数。所以123+%25+10是3，因为余数是3，从123中减去3，然后是120，然后用10除以120，也就是12。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|435378|entityMap^0|Q|8|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 use %10, which means the remainder if the number after you divided it. So <code>123 % 10</code> is 3, because the remainder is 3, substract the 3 from 123, then it is 120, then divide 120 with 10 which is 12. And do the same process.

Say I have a multi-digit integer in C. I want to break it up into single-digit integers.

<code>123</code> would turn into <code>1</code>, <code>2</code>, and <code>3</code>.

How can I do this, especially if I don't know how many digits the integer has?

C: how to break apart a multi digit number into separate variables?

假设我有一个用C写的多位整数，我想把它分成几个一位的整数。123会变成1、2和3。我如何做到这一点，特别是当我不知道整数有多少位的时候？

问C:如何将一个多位数分解成独立的变量？
EN

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