# Counting the amount of zeroes, odd and even numbers in an integer

• December 4th, 2011, 04:40 PM
54stickers
Counting the amount of zeroes, odd and even numbers in an integer
Code :

```public class NumSearch {   public static void main(String[] args) { Scanner scan = new Scanner(System.in); int input; int i=0,g=0,f=0; int even = 0; int odd = 0; int zero = 0; String temp = "";   System.out.println("Enter a number: "); input = scan.nextInt();   temp = Integer.toString(input);   while(i <= (temp.length()-1)){ if(temp.charAt(i) == 0){ zero++; i++; }else{ i++; } }   while(f <=(temp.length()-1)){ if((temp.charAt(f) % 2) != 0){ odd++; f++; }else{ f++; } } while(g <= (temp.length()-1)){ if((temp.charAt(g) %2) == 0 ){ even++; g++; }else{ g++; }   } System.out.println("The number of 0's: " + zero + "\nThe number of odd's: " + odd + "\nThe number of even's: " + even); } }```

This code is not counting the amount of zeroes in the integer. Can anyone see why?
Thanks!
• December 4th, 2011, 05:08 PM
Norm
Re: Counting the amount of zeroes, odd and even numbers in an integer
Quote:

This code is not counting the amount of zeroes in the integer.
Can you post the console that shows the input and output from your program?

One thing I see is that you do not understand the difference between a character and an int.
The char '0' does not have the integer value of 0. Look at the ASCII character integer values table.
When testing for the character zero compare against '0' not 0
• December 4th, 2011, 06:17 PM
54stickers
Re: Counting the amount of zeroes, odd and even numbers in an integer
Comparing it to '0' worked this time... I tried that earlier, but the code didn't compile.

Thanks!