Setting a scanner as a global variable

Question

Method at the top, is it declared as a global variable as its not actually taking in anything.

 public class java_1 {

   static Scanner stdin = new Scanner(System.in);

import java.util.Scanner;

Thats the code used to declare it, if you can declare it another way could i get a link to the documentation.

public class java_1 {

   static Scanner stdin = new Scanner(System.in);

   static String getLastName (String Name){
       String lastName;
       int spacePos,length;

       spacePos = fullName.indexOf("");
       length = fullName.length();
       lastName = fullName.substring(spacePos + 1);
       return lastName;
   }


   static String getInitial (String fullName){
       String initial;
       initial = fullName.substring(0,1); 
       return initial;

   }
    static String =Name (){

       String fullName;
       String userName;
       String initial;
       String lastName;

          System.out.println("name");
          fullName = stdin.nextLine();

          initial = getInitial(fullName);
          lastName = getLastName(fullName);

          userName = initial + lastName;

          System.out.println(userName);
          return userName;
      }

static String printuserName (){

       String fullName,userName,initial,lastName;

          System.out.println("enter name");
          fullName = stdin.nextLine();

          initial = getInitial(fullName);

          userName = initial + lastName;

          System.out.println("username: " + userName);
          return userName;
      }


   static int menu(){

          int choice;
            System.out.println("Input a number from the table, corresponding to the task required");
            System.out.println("1 = User name");
            System.out.println("2 = Factor");
            System.out.println("3 = Quit");

            choice = stdin.nextInt();

            while (choice != 1){
                System.out.println("Re enter");
                choice = stdin.nextInt();
        }
            return choice;
   }




      }
     }
    }

I answered your other question and you didn't vote for or accept the answer. — TypeIA, Dec 23 '13 at 21:25
Also you should probably post the text of the error you say you're getting? — TypeIA, Dec 23 '13 at 21:26
I think you've changed this completely. What is the actual question here? You don't actually seem to be asking anything. — Dawood ibn Kareem, Jan 13 '14 at 21:52

score 1 · Answer 1 · answered Dec 23 '13 at 21:36

The error is generated by this line of code, inside getInitial():

initial = fullName.substring(0,1);

The StringIndexOutOfBoundsException means that 1 is beyond the end of the string (meaning the input string is empty). If you're not already familiar with line-by-line debugging in whatever IDE you're using, this would be a good time to try. You can step through your code to determine why an empty string is being passed into getInitial().

score 1 · Accepted Answer · answered Dec 23 '13 at 21:43

1

The problem is that when you read the menu option with stdin.nextInt(), there is still a newline character waiting in the scanner. When you call stdin.nextLine() some time later, you get the rest of the line - which is just an empty string. The name itself won't be read until the FOLLOWING call to nextLine().

To remedy this, you should call nextLine() immediately after your call to nextInt() in the Menu() method, just to clear out that extra newline character.

answered Dec 23 '13 at 21:43

Dawood ibn Kareem

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Be aware that there are a handful of other bugs in your code. I recommend learning to use a debugger before you go too much further with this program. – Dawood ibn Kareem Dec 23 '13 at 21:52

Setting a scanner as a global variable

2 Answers2

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