call a method that prints an array error java

Question

I need to call a method that returns an array of strings but I keep getting an error. I did the Arrays.toString but it still is not working.

public class MyStore {

public static void main(String[] args) {
    SalesAssociate salesAssoc = new SalesAssociate("Bob", "Jones", "001");

    System.out.println(Arrays.toString(salesAssoc.getCashPosition()));
}//main
}//class

This is my class and method.

public class SalesAssociate extends FloorAssociate {
// Constructor
public SalesAssociate(String firstName, String lastName, String employeeId) {
        super(firstName, lastName, employeeId);
}
public String[] getCashPosition(){
        String cp[] = new String[3];
        cp[0]= super.getStoreLocation();
        cp[1]= super.getEmployeeId();
        cp[2]= "$3500";
        cp[3]= timeStamp();
        return cp;
    }
}

And this is my error:

Exception in thread "main" java.lang.ArrayIndexOutOfBoundsException: 3
at indassn3.SalesAssociate.getCashPosition(SalesAssociate.java:38)
at indassn3.MyStore.main(MyStore.java:21)
Java Result: 1

By the way, the super.getStoreLocation, super.getEmployeeId, and timeStamp methods all return strings.

Your array has 3 indices. cp[3] is a 4th index. Just change new String[3]; to new String[4]; — Drew Kennedy
– Drew Kennedy, Commented Apr 13, 2016 at 15:13

André Stannek · Accepted Answer · 2016-04-13 15:14:49Z

4

    String cp[] = new String[3];
    cp[0]= super.getStoreLocation();
    cp[1]= super.getEmployeeId();
    cp[2]= "$3500";
    cp[3]= timeStamp();

You are creating an array of length 3 and then try to add 4 elements. An array of length 3 has the indicies 0 to 2

answered Apr 13, 2016 at 15:14

André Stannek

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1 Comment

araviv1 Over a year ago

thanks, sorry about the stupid mistake, just missed it i guess

Sh4d0wsPlyr · Accepted Answer · 2016-04-13 15:15:12Z

1

You declared an array of size new String[3].

This means there is only 3 elements in that string, but you are attempting to set 4 (0, 1, 2, 3). Increse it to new String[4] and it should work just fine.

answered Apr 13, 2016 at 15:15

Sh4d0wsPlyr

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Comments

user1176726 · Accepted Answer · 2016-04-13 15:21:24Z

1

Use ArrayList without ArrayIndexOutOfBoundsException :)

public static void main(String[] args)
{
    ArrayList<String> cp = new ArrayList();
    cp.add("1"); 
    cp.add("2"); 
    cp.add("3");
    cp.add("4");
    cp.add("5");
}

answered Apr 13, 2016 at 15:21

user1176726

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Comments

Adrien Brunelat · Accepted Answer · 2016-04-13 15:30:39Z

1

Create your String array with:

String cp[] = new String[4];

Explanations

The following line:

String cp[] = new String[3];

Creates an array with 3 possible elements:

cp[0]
cp[1]
cp[2]

But a little bit further into the code, you wrote :

cp[3]= timeStamp();

This tries to assign a value to the 4rth element, this is out of bound and thus throws an ArrayIndexOutOfBoundsException. Remember that the first element is at position 0 in your array.

edited Apr 13, 2016 at 15:30

answered Apr 13, 2016 at 15:15

Adrien Brunelat

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Comments

Vikrant Kashyap · Accepted Answer · 2016-04-13 15:30:57Z

In your given code

   String cp[] = new String[3];
    cp[0]= super.getStoreLocation();
    cp[1]= super.getEmployeeId();
    cp[2]= "$3500";
    cp[3]= timeStamp(); //problem with this code actually.

This Statement new String[3] Creates a array of String of length 3. So, accessible index would be 0, 1 and 2.

but cp[3]= timeStamp(); in this code you try to access index 3 which is illegal to java.

So, cp[3]= timeStamp(); this code will raise an Exception ArrayIndexBoundException.

Better to use ArrayList<String> to avoid these kind of exception generally. because ArrayList is auto grow-able. you will never face these kind of issue in future.

ArrayList<String > cp= new ArrayList<>();
cp.add(super.getStoreLocation());
cp.add(super.getEmployeeId());
cp.add("$3500");
cp.add(timeStamp()); // and so on. No need to define the size at early.

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