Super new at Java

**Marauder** · May 14th, 2019, 12:45 PM

Hey, I'm just learning java I wrote this super simple program. I encountered a bug that I fixed but don't know why it fixed it and am trying to understand why, on top of the countless other mistakes I've probably made. If you remove () from
if((name.equalsIgnoreCase(username) ||name.equalsIgnoreCase(username2)) && pass.equalsIgnoreCase(password))
to look like
if(name.equalsIgnoreCase(username) || name.equalsIgnoreCase(username2) && pass.equalsIgnoreCase(password))
for the first username "brennan" you can input any password but for the username2 "Chip" it works, and putting () fixes it.

public class New {
    static Scanner sc = new Scanner(System.in);
    public static void main(String []args){
        boolean click = true;
        String username = "Brennan";
        String username2 = "Chip";
        String name;
        String password = "Let me in";
        String pass;
        do{
            System.out.println("Username: ");
            name = sc.nextLine();
            System.out.println("Password: ");
            pass = sc.nextLine();
            if((name.equalsIgnoreCase(username) || name.equalsIgnoreCase(username2)) && pass.equalsIgnoreCase(password)){
                System.out.println("Correct!");
                sc.close();
                break;
            }else if(!name.equals(username) && !pass.equals(password)){
                System.out.println("Incorrect username or password!");
            }
        }while(click);
    }
}

**jim829** · May 14th, 2019, 02:11 PM

Without the parentheses, the evaluation order is strictly left to right. Once you get a valid first user name, the rest is not evaluated because if the first part is true, the rest is irrelevant so the if statement short circuits. When you type in a valid second user name, the && comes into play so a valid password is required.

When you group the first two statements, one or the other must be true at which point the && becomes important because if it is false, the if statement false, if true, the if statement is true.

In the last case, if either username is invalid, the the && short circuits since (false || false) is false and thus the last expression is not necessary to evaluate.

	  System.out.println(true || false && false); // true
	  System.out.println(true || false && true);  // true
	  System.out.println(false || true && true);  // true
	  System.out.println(false || true && false); // false
	  System.out.println((true || false) && false); // false
	  System.out.println((true || false) && true); // true
	  System.out.println((false || true) && true); // true
	  System.out.println((false || true) && false);// false

Regards,
Jim

**Marauder** · May 14th, 2019, 03:55 PM

Ah ok I see, thank you.

Thread: Super new at Java

LinkBack

Thread Tools

Display

Super new at Java

Re: Super new at Java

Re: Super new at Java

Similar Threads

keyword super in java

Super Simple?

please help super new to Java

Super Friendly & Nice Java Forum!

Java student, super newb, totally stumped. Compile, no errors. Run shows nothing.