The code is outputing the String if username != "user", and exits without returning anything when username = "user". The code still will not pass?

Question about the task

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see code below:

This is the signIn() method that greets website users. Currently, it greets all users, but it should only greet registered users. All unregistered users have the name "user".
Add username validation at the beginning of the signIn() method.
If the name is "user", use the return keyword to abort execution of the method.

Requirements:

If the username is "user", then the signIn() method should exit without printing anything on the console.
If the username is not "user", then the signIn() method should continue and print greetings on the console.
The return keyword must be used to interrupt the signIn() method.

package en.codegym.task.pro.task06.task0607; /* Welcome! But not everyone. */ public class Solution { public static void main(String[] args) { String username = "user"; signIn(username); } public static void signIn(String username) { //write your code here if(username.equals("user")){ return; }else{ System.out.println("Welcome, " + username); } } }

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Jill L

Level 10 , China, China

3 March 2023, 15:27

Do I need to enter the username by BufferedReader or scanner here?

Thomas Yates

Level 16 , USA

12 March 2023, 05:12

Scanner

Thomas

Level 41 , Bayreuth, Germany

24 February 2023, 17:40

if you use return to drop out of the method, then no other conditional statements are necessary (or else if or else). That's probably what the codegym people try to tell you. Then you have all conditions at the top of the method and do not need to look through a bunch of nested conditions. Some pseudo code to give you a rough idea...

public void enterAdminMode(User user) {
    if (user == null)
        return;

    if (user.getGroup() != UserGroup.ADMIN)
        return;

    if (!user.getIp().equals("localhost"))
        return;

    AdminConsole console = AdminConsole.getInstance();
    console.log("Access by: " + user + ", " + LocalDateTime.now());
    console.open();
}