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.

Answer 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

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

Answer 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;
}