convert an integer number into an array

Question

Hi,

I am trying to convert an integer number in C into an array containing each of that number's digits

i.e. if I have

int number = 5400

how can I get to

int numberArray[4]

where

numberArray[0] = 0;
numberArray[1] = 0;
numberArray[2] = 4;
numberArray[3] = 5;

Any suggestions gratefully received

--dave

Answer 1

Hint: Take a look at this earlier question "Sum of digits in C#". It explains how to extract the digits in the number using several methods, some relevant in C.

From Greg Hewgill's answer:

/* count number of digits */
int c = 0; /* digit position */
int n = number;

while (n != 0)
{
    n /= 10;
    c++;
}

int numberArray[c];

c = 0;    
n = number;

/* extract each digit */
while (n != 0)
{
    numberArray[c] = n % 10;
    n /= 10;
    c++;
}

Answer 2

Figuring out the number of digits is tricky, but assuming it's always 4 digits you can do:

for (i = 0; i < 4; i++) {
  numberArray[i] = number%10;
  number = number div 10;
}

Answer 3

This would work for numbers >= 0

#include <math.h>

char * convertNumberIntoArray(unsigned int number) {
    int length = (int)floor(log10((float)number)) + 1;
    char * arr = new char[length];
    int i = 0;
    do {
     arr[i] = number % 10;
     number /= 10;
     i++;
    } while (number != 0);
    return arr;
}

EDIT: Just a little bit more C style but more cryptic.

#include <math.h>

char * convertNumberIntoArray(unsigned int number) {
    unsigned int length = (int)(log10((float)number)) + 1;
    char * arr = (char *) malloc(length * sizeof(char)), * curr = arr;
    do {
     *curr++ = number % 10;
     number /= 10;
    } while (number != 0);
    return arr;
}

Answer 4

C code:

/* one decimal digit takes a few more than 3 bits. (2^3=8, 2^4=16) */
int digits[(sizeof (int) * CHAR_BIT) / 3 + 1], 
    *digitsp = digits;
do {
    *digitsp++ = number % 10;
    number /= 10;
} while(number > 0);

You will see how many digits you converted by taking the difference

digitsp - digits

If you want to put it into a function:

#define MIN_DIGITS_IN_INT ((sizeof (int) * CHAR_BIT) / 3 + 1)

int to_array(int number, int *digits) {
    int *digitsp = digits;
    do {
        *digitsp++ = number % 10;
        number /= 10;
    } while(number > 0);
    return digitsp - digits;
}

int main() {
    int number = rand();
    int digits[MIN_DIGITS_IN_INT];
    int n = to_array(number, digits);

    /* test whether we're right */
    while(n-- > 0) 
        printf("%d", digits[n]);
    }
    printf(" = %d\n", number);
}

I prefer automatic arrays to dynamic memory allocation in this case, since it's easier to do it right and not leak accidentally.

Answer 5

If you need to take negative numbers into account, you might need some extra logic. In fact, when playing around with arrays you don't know the size of upfront, you may want to do some more safety checking, and adding an API for handling the structure of the data is quite handy too.

// returns the number of digits converted
// stores the digits in reverse order (smalles digit first)
// precondition: outputdigits is big enough to store all digits.
//
int convert( int number, int* outputdigits, int* signdigit ) {

  int* workingdigits = outputdigits;

  int sign = 1;
  if( number < 0 ) { *signdigit = -1; number *= -1; }
  ++workingdigits;

  for ( ; number > 0; ++ workingdigits ) {
    *workingdigits = number % 10;
    number = number / 10;
  }

  return workingdigits - outputdigits;
}

void printdigits( int* digits, int size, int signdigit ) {
  if( signdigit < 0 ) printf( "-" );

  for( int* digit = digits+size-1; digit >= digits; --digit ){
    printf( "%d", *digit );
  }
}

int main() {
   int digits[10];
   int signdigit;
   printdigits( digits, convert( 10, digits, &signdigit ), signdigit );
   printdigits( digits, convert( -10, digits, &signdigit ), signdigit );
   printdigits( digits, convert( 1005, digits, &signdigit ), signdigit );

}

Answer 6

You could calculate the number of digits in an integer with logarithm rather than a loop. Thus,

int * toArray(int number)
{
    int n = log10(number) + 1;
    int i;
    int *numberArray = calloc(n, sizeof(int));
    for ( i = 0; i < n; ++i, number /= 10 )
    {
        numberArray[i] = number % 10;
    }
    return numberArray;
}

Answer 7

using vadim's code, I came up with this test program:

#include <stdio.h>
#include <stdlib.h>
#include <math.h>

char * convertNumberIntoArray(unsigned int number) {
    unsigned int length = (int)(log10((float)number)) + 1;
    char * arr = (char *) malloc(length * sizeof(char)), * curr = arr;
    do {
        *curr++ = number % 10;
        number /= 10;
    } while (number != 0);
    return arr;
}



int main(void)
{
    int InputNumber;
    int arr[5];

    printf("enter number: \n");
    scanf("%d", &InputNumber);

    convertNumberIntoArray(InputNumber);

    printf("The number components are: %d %d %d\n", arr[0],arr[1],arr[2]);

    system("PAUSE");    
    return 0;
}

but the output is garbage. Can anyone advise if I have done something stupid here?

/***** output *****/
enter number:
501
The number components are: 2009291924 2009145456 -1
Press any key to continue . . .

--dave