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

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?

Altruistic

If your int started life as char data (input from user or text file ...) do not convert to int in the first place (no scanf, no atoi ...) and use the characters to separate the digits. – Rodroda

Answer 1

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

Albina

Answer 2

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;
}

Pion

Answer 3

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.

Sphenic

Answer 4

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;
}

Scherzo

this is not nth digits, this is nth digits starting from the least significant – Sphenic

You don't consider the least significant digit to be the zeroth? – Scherzo

well, I consider first or zeroth digits of 123 to be 1, not 3. – Sphenic

Answer 5

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;
}

Attainment

Answer 6

//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

Pharyngology

Answer 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 [email protected]

    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!

Deviate

Answer 8

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.

Pak

Answer 9

        //This is c#, but you can see how easy it would be to convert
        //to another language.

        int myNumber = 123456;
        string myWord = Convert.ToString(myNumber);

        string partOne = myWord.Substring(0, 2);
        string partTwo = myWord.Substring(2, 2);
        string partThree = myWord.Substring(4, 2);

        int num01 = Convert.ToInt32(partOne);
        int num02 = Convert.ToInt32(partTwo);
        int num03 = Convert.ToInt32(partThree);

        Console.WriteLine(num01);
        Console.WriteLine(num02);
        Console.WriteLine(num03);

        Output:
        12
        34
        56

Bridging

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Answer 10

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;}

Vadim

Never use pow for this – Metropolis

Answer 11

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

Alexi