Listing digits in an int, without losing 0 - c

Question

I'm writing a program that list the digits of integer input.
example:
Please enter an integer number: 5021
The digits are:
5
0
2
1
The program works fine except when it comes to numbers ending in 0 since I have reversed the number first using a while loop (while num>0)in order to print the numbers in the order shown above. I hope someone can point me in the right direction as I really can't seem to think of another way just now :)
sorry couldn't figure out how to add code without doing 4 spaces in front of each line (is there another(easier) way to add code?).
int main() {
int userInt; /* integer input by user */
int revInt; /* reversed integer */
printf("Please enter an integer number: ");
scanf("%d", &userInt);
/* reverse int */
revInt = reverseInteger(userInt);
/* print digits */
printDigits(revInt);
return 0;
}
int reverseInteger(int num) {
long sum = 0;
int remainder;
while(num) {
remainder = num%10; /* get last digit of number */
sum = sum*10 + remainder; /* move digits by one place in sum and add remainder of number */
num = num/10; /* remove last digit of number */
}
return sum;
}
void printDigits(int num) {
int remainder;
printf("The digits are:\n");
while(num) {
remainder = num%10; /* get last digit of number */
printf("%d \n", remainder); /* print last digit */
num = num/10; /* remove last digit of number */
}
}

Answer 1

The maximum number of digits in an int is small so you could safely use recursion in this case:
#include <stdio.h>
#include <stdlib.h>
void
print_digits(unsigned n) {
div_t q = div(n, 10);
if (q.quot > 0) print_digits(q.quot);
printf("%d\n", q.rem);
}
int main() {
print_digits(50210);
return 0;
}
Output
5
0
2
1
0

Answer 2

In your code for ReverseInteger, the variable remainder always keeps the last digit of the digit of the number you've entered (even when there are zeros at the end). So your function is almost ready, we can just tweak it a little bit:
void reverseIntegerAndPrint(int num) {
int remainder;
while(num) {
remainder = num%10; /* get last digit of number */
printf("%d \n", remainder); /* prints last digit!! */
num = num/10; /* remove last digit of number */
}
return;
}

Answer 3

You need to store the number of times you passed in the loop in reverseInteger and reuse this information when you loop in printDigits function.
Instead of reversing your integer in an int, you also can just store all the digits in the reverse order in an array and store in another integer variable the number of digits. Then you can print the elements of array (the digits) element by element starting from the end.

Answer 4

Before reverting the number,do something like this:
while(number%10 == 0)
{
number /= 10;
}
...//invering code here

Answer 5

The problem is that reverseInteger(1) == reverseInteger(10) == reverseInteger(100) == reverseInteger(1000) == ... and hence, once you use this function (which returns an int) you've already lost the information you need. One solution is to count the trailing zeros first, leading to something like this (assuming userInt > 0):
int numTrailingZeros = 0;
while(!(userInt % 10)) {
++numTrailingZeros;
userInt /= 10;
}
revInt = reverseInteger(userInt);
printDigits(revInt);
while(numTrailingZeros) {
printf("0 \n");
--numTrailingZeros;
}

Answer 6

Also is possible to convert it to string and print reserved:
#include "stdio.h"
#include "string.h"
int main(void)
{
int i,j;
char s[20];
scanf("%i", &i);
sprintf(s, "%i", i);
for(j=strlen(s); j>0; j--) {
printf("%c \n", s[j-1]);
}
}