Check if a number is even or odd without using any conditional statement

Given a number, determine if it is even or odd without using any conditional statement (if–else) or ternary operator.

Method 1: Using string array

The idea is to use the modulo operator or LSB to determine if a number is even or odd. Instead of using a conditional statement to output odd/even, we can use a string array index as demonstrated below in C/C++:

#include <iostream>

using namespace std;

int main()

{

int n = 6;

string arr[] = {"even", "odd"};

// using modulo operator

cout << arr[n % 2];

return 0;

}

#include <iostream>

using namespace std;

int main()

{

int n = 6;

string arr[] = {"even", "odd"};

// checking LSB of number for even/odd

cout << arr[n & 1];

return 0;

}

#include <stdio.h>

int main()

{

int n = 8;

char* arr[] = {"odd", "even"};

// repeated subtraction

while (n-- && --n > 0) {}

// if `n` is 0, the number is even

// if `n` is -1, the number is odd

printf("%s", arr[n + 1]);

return 0;

}

Method 2: Using short-circuiting in boolean expressions

We can take advantage of short-circuiting in boolean expressions. We know that in boolean AND operation such as x && y, y is evaluated only if x is true. If x is false, then y is not evaluated because the whole expression would become false, which can be deduced without even evaluating y. Similarly, for boolean OR operation such as x || y, y is evaluated only if x is false. If x is true, then y is not evaluated. This is called short-circuiting in boolean expressions.

We can apply this logic to solve a given problem. Consider the following code snippet:

((n & 1) && printf("odd")) || printf("even");

Initially, check if n is odd or even. If n is odd, the first subexpression of AND becomes true, and its second subexpression printf("odd") gets evaluated. Otherwise, if n is even, the first subexpression of OR will become false, and its second subexpression printf("even") gets evaluated.

This approach is demonstrated below in C:

#include <stdio.h>

int main()

{

int n = 6;

((n & 1) && printf("odd")) || printf("even");

return 0;

}

#include <stdio.h>

int main()

{

int n = 6;

((n % 2) && printf("odd")) || printf("even");

return 0;

}

#include <stdio.h>

int main()

{

int n = 9;

while (n-- && --n > 0) {}

((n == 0) && printf("even")) || printf("odd");

return 0;

}

Don’t forget to share with us if you find any more interesting methods to solve this problem.