Question

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

123 would turn into 1, 2, and 3.

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

Was it helpful?

Solution

int value = 123;
while (value > 0) {
 int digit = value % 10;
 // do something with digit
 value /= 10;
}

OTHER TIPS

First, count the digits:

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

Then, you can store them in an array:

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

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

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

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.

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

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>1)
{
    factor = factor/10;
    printf("%d\t",num/factor);
    i++;
    num = num % factor;
}

we can use this program as a function with 3 arguments.Here in "while(a++<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.

int main(){
long signed c=0,z,a=0,b=1,d=1;
scanf("%ld",&z);
while(a++<2){
       if(d++==1) 
       z/=pow(10,6);
       c+=(z%10)*b; 
       z/=10;
       b*=10;}
        return c;}
//Based on Tony's answer
#include <stdio.h> 
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));
}

This results in the following printout:

Hundreds = 9

Tens = 8

Units = 7

I made this based on the code from @asaelr:

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

    return dandn;

}

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


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

}

It works quite well, thanks!

Try this code if you want to separate digits in the same order without using arrays.

//Separate number digits
#include <stdio.h>
#include <math.h>

void main()
{
    int x, y, n = 0;

    scanf("%d", &x);

    //counting digits
    y = x;
    while (y != 0)
    {
        n += 1;
        y /= 10;
    }

    //printing separated digits
    int i;
    for (i = ceil(pow(10, (n - 1))); i != 0; i /= 10)
        printf("%d  ", (x / i) % 10);
}

You can use %10, which means the remainder if the number after you divided it. So 123 % 10 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.

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

int l1; //123456 for example
scanf("%d",&l1);
char s[sizeof(l1)];
sprintf(s,"%5d",l1);'

//This will give you separate digits of the number in char format inside s[0],s[1] 
//and so on.

//If you want them in int format, declare a int array say int i[sizeof(l1)] and add 
//the following code

for(int c=1;c<=sizeof(l1);c++){
i[c] = s[c] - '0';
} 

//Now i[0], i[1] etc will have the digits in int format
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