Split int into single digits. How to handle zeroes - c

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.

Answer 2

Given the precisely worded constraints of the problem: a function that will take an int and separately return the leftmost digit and the rest of the number it cannot be done, because it is impossible to represent the leading zeros in an integer, so passing back the "rest of the number" is bound to cause permanent loss of information
As the robot in Lost in Space would say, "It does not compute."

Answer 3

Putting an appropriate loop around something like this should work if I understand the question correctly.
//! Given a number return the least significant digit.
//! Return true if all went well. False otherwise. (Clearly we
//! can handle errors better.)
/*!
\param[in/out] n
The number to strip the least significant digit from.
\param[out] d
The least significant digit of n.
*/
bool digitize(int& n, int& d)
{
if ( n == 0 )
return false
else
{
d = n % 10;
n = n / 10;
return true;
}
}