Determine whether number is odd or even without us

问题:

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.

回答1:

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.

回答2:

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;
}

回答3:

int isOdd = (number & 1);      

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

回答4:

Did you mean something like this?

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

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

回答5:

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

Sample :

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

回答6:

Just saw now 'Without using IF'

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

回答7:

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

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

回答8:

I would use:

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

One line code.

回答9:

# /* **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

回答10:

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.

回答11:

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");
        }

    }
}

回答12:

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

回答13:

Well, yes you can make use of bitwise operator. Here's an example,

public class EvenOddBitwise
{
   public static void main(String[] args)
   {
      int number = 64;
      if((number & 1) == 1)
      {
         System.out.println(number + " is Odd.");
      }
      else
      {
         System.out.println(number + " is Even.");
      }
   }
}