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

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?

Answer 1

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

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;
}

Answer 3

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;
}

Answer 4

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.

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;
}

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

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 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!

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.

Answer 9

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;}

Answer 10

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