Have a look at the lines of code below and determine if they can be used to precisely identify if a given number is Odd?

public boolean oddOrNot(intnum) {
 return num %2==1;
}

These lines seem correct but they will return incorrect results one of every four times (Statistically speaking). Consider a negative Odd number, the remainder of division with 2 will not be 1. So, the returned result will be false which is incorrect!

This can be fixed as follows:

public boolean oddOrNot(intnum) {
 return (num &1) !=0;
}

Using this code, not only is the problem of negative odd numbers solved, but this code is also highly optimized.

Since, Arithmetic and Logical operations are much faster compared to division and multiplication, the results are achieved faster so in second snippet.


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If you use a modulus or absolute value to convert negetive No to positive then it will work fine.

What about:

public boolean oddOrNot(intnum) {
return (num |1) ==num;
}

It won't have to load the literal 0;

GetAvenue:

 public static void checkNum(int Num){
 if(num%2 == 0){
 System.out.println("The number" +Num+ "is even";
 }else{
 System.out.println("The number" +Num+ " is odd";
 }
 }
 

That's how I would have done it personally.