Question

I am trying to create a function that will take an int and separately return the leftmost digit and the rest of the number.

int idigitizer(int *number) {
int i = 1;
int head = 0;
int tmp = 0;

tmp = *number;
while (tmp > 9) {
    if ((tmp/i) < 10) {
        head = tmp/i;
        *number = *number - (head*i);
        return head;
    } else {
        i = i*10;
    }
}
number = 0;
return tmp;
} 

idigitizer returns the leftmost part of the number and *number will carry the rest.
I will have a loop in my main that will keep calling idigitizer until all the digits of the number get separated.
The thing is I don’t know how to handle zeroes and how to make this process terminate correctly when it is done with the last digit. Any help is welcome. Thanks in advance.

EDIT : To make it clearer. I don’t want the possible zeroes in the middle of a number to get lost. If i get the number 100047 as input I want idigitizer to return:

return - *number
         100047
1        00047
0        0047
0        047
0        47
4        7
7

Answer 1

I would use something like this instead:

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

int getFirstDigit( int number );
int getRestOfNumber( int number );
int getNumberOfMissingZeros (int number, int restOfNumber);

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

   int x = 500574;
   int firstDigit;
   int restOfNumber;

   int n; /* loop index */

   firstDigit = getFirstDigit(x);
   restOfNumber = getRestOfNumber(x);

   printf("The first digit of %d is %d.\n",x,firstDigit);
   printf("The rest of the number is ");

   for( n = 0; n<getNumberOfMissingZeros(x,restOfNumber); ++n ){
     printf("0");
   }

   printf("%d.",restOfNumber);

   return EXIT_SUCCESS;
  }

int getFirstDigit( int number ){

  return number / (int)floor( pow(10,floor(log10( (double)number ))) );

}

int getRestOfNumber( int number){

   return number % (int)floor( pow(10,floor(log10( (double)number ))) );

}

int getNumberOfMissingZeros (int number, int restOfNumber){

  int digitsOriginally;
  int digitsInRestOfNumber;

  digitsOriginally = floor(log10( (double)number ) );
  digitsInRestOfNumber = floor(log10( (double)restOfNumber ) );

  return digitsOriginally - digitsInRestOfNumber - 1;

}

The magic is in the expression (int)floor( pow(10,floor(log10( (double)number ))) ); This gets the size of the value, like for 52330, it would return 10000, since there are five digits in 52330. This makes it easy to extract the highest digit.