Determine whether number is odd or even without using conditional code

Question

How to find whether a number is odd or even, without using if condition or ternary operators in Java?

This question is given by my teacher. He also give me a hint that it is possible by using a bitwise operator.

Are you sure your teacher meant `without using if condition` and not `without using modulo operator`? — Pshemo, Dec 20 '13 at 13:58
Disclaimer: the ternary conditional operator "?:" IS a if-then-else — Toni Toni Chopper, Dec 20 '13 at 14:02
Do you mean "odd or even", or "odd or even parity" ? They are two different things. Your title says "parity", but the question just says "odd or even". — Paul R, Dec 20 '13 at 17:56
possible duplicate of [How do I check if an integer is even or odd using bitwise operators](http://stackoverflow.com/questions/5700911/how-do-i-check-if-an-integer-is-even-or-odd-using-bitwise-operators) — Paul R, Dec 20 '13 at 17:57
Be careful with the answers recommending `% 2`. It will give -1 for negative odd numbers. If you use that to index an array, or if you compare it to 1, you're in trouble. `& 1` doesn't have that problem. — harold, Dec 22 '13 at 07:34

score 18 · Accepted Answer · edited May 23 '17 at 12:10

There are few ways to not use if and get behavior that will be same as if if was used, like ternary operator condition ? valueIfTrue : valueIfFalse or switch/case.

But to be tricky you can also use arrays and try to figure some transformation of our value to proper array index. In this case your code could look like

int number = 13;
String[] trick = { "even", "odd" };
System.out.println(number + " is " + trick[number % 2]);

output:

13 is odd

You can change number % 2 with number & 1 to use suggestion of your teacher. Explanation of how it works can be found here.

score 10 · Answer 2 · answered Dec 20 '13 at 13:58

10

Consider a number's representation in binary format (E.g., 5 would be 0b101). An odd number has a "1" as its singles digit, an even number had a zero there. So all you have to do is bitwise-and it with 1 to extract only that digit, and examine the result:

public static boolean isEven (int num) {
    return (num & 1) == 0;
}

answered Dec 20 '13 at 13:58

Mureinik

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Then `if(isEven(7)){}` ?? – Suresh Atta Dec 20 '13 at 14:01
this is ok. But the if construct is now just within the outer method which invokes this method. I had also suggested num & 1 so +1 – Diversity Dec 20 '13 at 14:02
@sᴜʀᴇsʜᴀᴛᴛᴀ 7 & 1 == 1 and 1 != 0 so it would correctly return false – msam Dec 20 '13 at 14:04
@msam I'm talking about using the method :) – Suresh Atta Dec 20 '13 at 14:04
4

@sᴜʀᴇsʜᴀᴛᴛᴀ `println(isEven(7));` :) – msam Dec 20 '13 at 14:06
@msam Ha ha. That's it. :) Home work done for today :D – Suresh Atta Dec 20 '13 at 14:17

score 6 · Answer 3 · edited Dec 20 '13 at 14:10

6

int isOdd = (number & 1);

isOdd will be 1 if number is odd, otherwise it will be 0.

edited Dec 20 '13 at 14:10

Suresh Atta

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answered Dec 20 '13 at 13:58

umanganiello

756
4
7

score 2 · Answer 4 · answered Dec 20 '13 at 13:59

2

Did you mean something like this?

boolean isEven(int value) {
  return value % 2 == 0;
}

boolean isOdd(int value) {
  return value % 2 == 1;
}

answered Dec 20 '13 at 13:59

Arne Burmeister

20,046
8
53
94

1

and how would you check that return value? – kai Dec 20 '13 at 14:00

score 0 · Answer 5 · edited Dec 20 '13 at 14:11

0

Just saw now 'Without using IF'

boolean isEven(double num) { return (num % 2 == 0) }

edited Dec 20 '13 at 14:11

Suresh Atta

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answered Dec 20 '13 at 13:59

gabriel

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score 0 · Answer 6 · answered Dec 20 '13 at 14:02

0

Every odd number have 1 at the end of its binary representation.

Sample :

public static boolean isEven(int num) {
    return (num & 1) == 0;
}

answered Dec 20 '13 at 14:02

SANN3

9,459
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score 0 · Answer 7 · answered Dec 20 '13 at 14:38

0

Just a quick wrapper over the already defined process...

public String OddEven(int n){
String oe[] = new String[]{"even","odd"};
        return oe[n & 1];
}

answered Dec 20 '13 at 14:38

boxed__l

1,334
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score 0 · Answer 8 · answered Dec 29 '13 at 16:25

0

I would use:

( (x%2)==0 ? return "x is even" : return "x is odd");

One line code.

answered Dec 29 '13 at 16:25

iPot

23
5

score 0 · Answer 9 · answered Dec 23 '15 at 11:41

Method 1:

System.out.println(new String[]{"even","odd"}[Math.abs(n%2)]);

Method 2:

System.out.println(new String[]{"odd","even"}[(n|1)-n]);

Method 1 differs from the accepted answer in the way that it accounts for negative numbers as well, which are also considered for even/odd.

score 0 · Answer 10 · edited Jun 13 '17 at 14:22

0

import java.util.Scanner;
public class EvenOddExample
{
    public static void main(String[] args) 
    {
        System.out.println("\nEnter any Number To check Even or Odd");
        Scanner sc=new Scanner(System.in);

        int no=sc.nextInt();
        int no1=no;
        while (no>1) 
        {   
            no=no-2;
        }       
        if(no==0)
        {
            System.out.println(no1 +" is evenNumber");
        }
        else
        {
            System.out.println(no1 +" is odd Number");
        }

    }
}

edited Jun 13 '17 at 14:22

kiner_shah

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answered Jun 13 '17 at 12:32

Keshav Gera

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check when number is greater then 1 number = number -2; – Keshav Gera Jun 13 '17 at 12:34

score 0 · Answer 11 · answered Jul 29 '17 at 15:49

0

you can also use bitwise shift operators (number >> 1)<<1 == number then even else odd

answered Jul 29 '17 at 15:49

Prashu

1
2

score -1 · Answer 12 · answered Aug 03 '14 at 15:15

# /* **this program find number is odd or even without using if-else,
## switch-case, neither any java library function...*/
##//find odd and even number without using any condition

  class OddEven {
       public static void main(String[] args) {
       int number = 14;
       String[] trick = { "even", "odd" };
       System.out.println(number + " is " + trick[number % 2]);
     }
 }

  /**************OUTPUT*************
  // 14 is even
  // ...................
  //13 is odd

Determine whether number is odd or even without using conditional code

12 Answers12

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