Using input value for user validation

[Joseph Williamson]

I am trying to take an input from the user and validate it to make sure it is not a null value. I will also have to convert a string to an integer. I have been trying to use scanner to get this result but with limited success. Any suggestions?

import java.util.Scanner;

public class Test_2 {

	public static void main(String[] args) {
			
		Scanner input = new Scanner(System.in);
			
		String Num = args[0];
		int i = Integer.parseInt(Num);
			
		if (input == null){
				printUsage();
				return;
			   }
			
				else if (i % 2 == 0){ 
					System.out.println("even");
				}
			
					else if (i % 2 != 0){ 
						System.out.println("odd");
					}
			}	
		
			private static void printUsage() {
					System.out.println("You must select a number.");
				}
	  }

[Zachary Griggs]

Well, in the code you've posted, you check whether your Scanner named "input" is null. This can never possibly happen, as you initiate it three lines above:

Scanner input = new Scanner(System.in);

I assume you wanted to check whether the String was null, but this isn't the String variable. As a reminder, integers cannot be null as they are primitives, not objects.

String Num = args[0];

Do you know what this line of code does, and what the args are? Because, based on the line above (where you create a Scanner), it seems like you want to use the Scanner to get input, but the args is something very different. Also, if you're using a Scanner, the input can never really be null. It can only be an empty string (if the user presses Enter when it's empty). For args, it can be null, but I wouldn't use those anyways.

int i = Integer.parseInt(Num);

This line occurs before you check if it's null or not. So, if the user entered null, you would get a NullPointerException here despite your checks (because they take place afterwards)

The string being null isn't the only bad thing that can happen when casting a String to an int, though. What happens if the user typed "abc" in? This isn't null, but obviously not something we can make into a number either. The answer here is to use a "try" statement, which will attempt to execute commands in its block. If they fail, then the exception will be "caught" by the catch block further down, and then can be handled as desired.

Assuming everything worked as intended right now, what would happen if the user entered a wrong number? It would display the message, then exit. However based on the message it looks like you want to give the user another chance to enter data, in which case you would need to use a loop, so you'll need to change the logic of this program a bit.

                else if (i % 2 == 0){ 
                    System.out.println("even");
                }
             
                    else if (i % 2 != 0){ 
                        System.out.println("odd");
                    }

This code looks good. Only one change I would suggest: remove the second else if, and just put it as "else", since the second condition is always true if the first is not true.

            private static void printUsage() {
                    System.out.println("You must select a number.");
                }

You don't really need another function for a single line of code - it doesn't shorten the program at all, and might actually take away readability.

-------

I would recommend rewriting the program, keeping the above problems in mind. I've made a method stub to hopefully help you:

    public static void main(String[] args) {
        //1: Create a new Scanner
        //2: Create a boolean, set to false. The purpose of this is as an exit
        //   condition for a loop. When we're successful, set it to true and exit
        //   the loop. While unsuccessful, ask for the user's input.
        //3: Start loop
        //4: Get user's input as a String
        //5: Start a Try statement and convert the user's input to an integer
        //6: Check even/odd
        //7: Set success boolean to true
        //8: If NumberFormatException if thrown, tell the user to enter only numbers
        //9: End try, end loop
    }

[Campbell Ritchie]

Find out how this line of code works:-
System.out.println("The number is " + (i % 2 == 0 ? "even." : "odd."));
It can save you lots of writing.

Did you know you can use the & operator to find whether a number is odd or even? You simply find out whether the rightmost bit is 1 or 0:-
(i & 1) == 0
You must use the () in that expression because == has a higher precedence than &

The input from a Scanner pointing to System.in can never be null. You can get the empty String "" but never null. If there is no input available, it will still not return null. You can find out what happens with this sort of code.

System.out.print("Write something on screen: ");
while (myScanner.hasNextLine())
{
   System.out.printf("You entered \u201c%s\u201d\nWrite something else: ",
       myScanner.nextLine());
}

The only way to escape that loop is to feed ctrl‑D (*nix) or ctrl‑Z (Windows®); try it and see what happens.
I suspect a Scanner will not return null from other input sources, either.