Using Array and not ArrayList. User has to write the exact number of data that is provided by the array. - Artificial Intelligence

AlvinAravinnthShankar
August 21, 2015
226 views
3 votes
3 Answers

Using Array and not ArrayList. User has to write the exact number of data that is provided by the array. The error that I get is NullPointerException, For example, if my array Dinfo and array Linfo has 1, i only can write 1 data for each class. But if i put 100 for my array Dinfo and Linfo, i have to write 100 data so that my information will be shown. If not i will get a nullpointerexception.

    import java.util.*;
public class testMain
{
public static void main(String args[])
{
    String User;
    Scanner scanner = new Scanner(System.in);
    Desktop[] Dinfo = new Desktop[100];
    Laptop[] Linfo = new Laptop[100];
    int D = 0;
    int L = 0;

    int A;
    int B;

    do
    {
        System.out.println("****************** Artificial Intelligence Co. **********");
        System.out.println("Computer Menu");  
        System.out.println("1. Add a new Desktop Information");
        System.out.println("2. Add a new Laptop Information");
        System.out.println("3. Display all Computer Information");
        System.out.println("4. Quit");
        System.out.println("*********************************************************");
        System.out.print("Please enter either 1 to 4: "); 
        User =(scanner.nextLine());

        if (User.equals("1"))
        {
            Dinfo[D] = new Desktop();
            Dinfo[D].setDisplayDesktopInfo();
            D++;


        }

        else if (User.equals("2"))
        {

            Linfo[L] = new Laptop();
            Linfo[L].setDisplayLaptopInfo();
            L++;


         }

         else if(User.equals("3"))
         {
             for(A= 0; A < Dinfo.length; A++)
             {

                     System.out.println("=======================Desktop===========================");
                     Dinfo[A].getDisplayDesktopInfo();

                     System.out.println("");



             }

             for(B = 0; B < Linfo.length; B++)
             {

                     System.out.println("=======================Laptop============================");
                     Linfo[B].getDisplayLaptopInfo();

                     System.out.println("");


             }

         }


         else if(User.equals("4"))
         {
             System.out.println("Thank You!");
            }
    }
    while(!User.equals("4"));


}

    }

Answers

- RobbyCornelissen
- August 21, 2015 at 3:21 am
- 0 votes
0
Just put a null check in place:
```
if (Dinfo[A] != null) {
    Dinfo[A].getDisplayDesktopInfo();

    System.out.println("");
}
```
And similar for Linfo[B].

Obviously, using an ArrayList is far easier.
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- Swapnil
- August 21, 2015 at 3:45 am
- 0 votes
0
NullPointerException is
Thrown when an application attempts to use null in a case where an object is required. These include:
- Calling the instance method of a null object.
- Accessing or modifying the field of a null object.
- Taking the length of null as if it were an array.
- Accessing or modifying the slots of null as if it were an array.
- Throwing null as if it were a Throwable value.
So the easiest way to avoid is to check for null and then access the value of that reference / variable if it isn’t null.
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- BriceDjilo
- August 21, 2015 at 4:38 am
- 0 votes
0
If you don’t want to worry about the predefined size of the area, you may declare the arrays like this: create arrays with dummy size of 1
```
Desktop[] Dinfo = new Desktop[1];
Laptop[] Linfo = new Laptop[1];
```
and then for selection 1, 2, … you may do something like this:
```
 if (User.equals("1")) {
       Desktop[] temp = new Desktop[Dinfo.length+1];
       System.arraycopy( Dinfo, 0, temp, 0, Dinfo.length );
       Dinfo = temp
       Dinfo[D] = new Desktop();
       Dinfo[D].setDisplayDesktopInfo();
       D++;
    } 
```
Finally, in order to print, do what Robby Cornelissen specified above.
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