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Print “Even” or “Odd” without using conditional statement
  • Difficulty Level : Easy
  • Last Updated : 08 Feb, 2021

Write a C/C++ program that accepts a number from the user and prints “Even” if the entered number is even and prints “Odd” if the number is odd. You are not allowed to use any comparison (==, <,>,…etc) or conditional (if, else, switch, ternary operator,..etc) statement.
Method 1 
Below is a tricky code can be used to print “Even” or “Odd” accordingly. 


using namespace std;
int main()
  char arr[2][5] = {"Even", "Odd"};
  int no;
  cout << "Enter a number: ";
  cin >> no;
  cout << arr[no%2];
  return 0;


import java.util.Scanner;
class GFG
    public static void main(String[] args)
        String[] arr = {"Even", "Odd"};
        Scanner s = new Scanner(;
        System.out.print("Enter the number: ");
        int no = s.nextInt();
// This code is contributed by divyeshrabadiya07.


arr = ["Even", "Odd"]
print ("Enter the number")
no = input()
print (arr[int(no) % 2])


using System;
class GFG {
  static void Main() {
    string[] arr = {"Even", "Odd"};
    Console.Write("Enter the number: ");
    string val;
    val = Console.ReadLine();
    int no = Convert.ToInt32(val);
// This code is contributed by divyesh072019.


$arr = ["Even", "Odd"];
$input = 5;
echo ($arr[$input % 2]);
// This code is contributed
// by Aman ojha

Method 2 
Below is another tricky code can be used to print “Even” or “Odd” accordingly. Thanks to student for suggesting this method.


int main()
    int no;
    printf("Enter a no: ");
    scanf("%d", &no);
    (no & 1 && printf("odd"))|| printf("even");
    return 0;

Please write comments if you find the above code incorrect, or find better ways to solve the same problem

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