integer - Java Int(Why returning 0 when number is within range)

Question

Thank you for your time!

1. for value upto 2147483641 code is working fine after that it is returning 0(why).. as per my understanding program should return 0 only when overflow occurs.. (for -2147483648 and 2147483647 ) not for value falling in the range.

2. Also please share any link for leading zero number reversal.. I could not find any online.

    public class ReverseDigit {
        public int reverse(int integer) {
            boolean negflag=false;
            if(integer<0){
                      negflag=true;
                      integer=integer*-1;
                          }
            int rev=0;
            int rem=0;
            while(integer!=0){
                     rem=integer%10;
                     int newrev= rev*10+rem;
                     if((newrev-rem)/10!=rev){
                     return 0;
                                              }
             else{
   
                     rev=newrev;
                 }
                     integer=integer/10;
        }
                      return  rev = negflag?rev*-1:rev;
    }
   
    public static void main(String[] args) {
                ReverseDigit rd = new ReverseDigit();
                System.out.println(rd.reverse(**2147483642**));
   
   
    }
   

   }

question from:https://stackoverflow.com/questions/65912660/java-intwhy-returning-0-when-number-is-within-range

Answer 1

This is happens because the reversed number of 2147483642 is 2463847412, and this number is greater then Intrgre.MAX_VALUE which is 2147483647, so the number became less than 0. This is happens to 2147483623 too, because his reversed number is 3263847412, and this number is greater then Intrgre.MAX_VALUE. To fix that, I see two possible solutions:

1. Use long instead of int.
2. Rewrite the method to work with String, because you aren't really do any calculations (You can use string.charAt(int index) to get the digits one bt one).