storing a string into an array causes an ArrayIndexOutOfBoundsException error

Question

int x = 0;
String[] QEquivalent = {};

String s = sc.nextLine();
String[] question2 = s.split(" ");

for (int i = 0; i < question2.length; i++) {
    System.out.println(question2[i]);
    x++;
}                                                                   //debug
System.out.println(x);

String s2 = sc2.nextLine();
String[] Answer = s2.split(" ");

for (int c = 0; c < Answer.length; c++) {
    System.out.println(Answer[c]);
}                                                                   //debug
int y;

String u = sn.nextLine();
String[] t = u.split(" ");
for (y = 0; y < question2.length; y++) {
    for (int w = 0; w < t.length; w++) {
        if (t[w].equals(question2[y])) {
            QEquivalent[y] = "ADJ";
            System.out.println(QEquivalent[y]);
            break;
        }
    }
}

this is the line of codes that I have as of now. when a string in question2 is found in String[] t, it should store the string "ADJ" in String[] QEquivalent. I can't seem to fix the error. can someone please help me?

The variables in Java should start with a lowercase. Like Answer and QEquivalent. — araknoid
– araknoid, Commented Jul 31, 2013 at 15:18
The String[] QEquivalent is an empty array - obviously any attempt to access its members by index will result in an java.lang.ArrayIndexOutOfBoundsException. — Nick Holt
– Nick Holt, Commented Jul 31, 2013 at 15:19

Rohit Jain · Accepted Answer · 2013-07-31 15:29:41Z

3

You are creating an empty array here:

String[] QEquivalent = {};

So, any index you try to access will be out of bounds. You should creating an array using a fixed size.

Or, you can better use an ArrayList instead, which can dynamically grow in size:

List<String> qEquivalent = new ArrayList<String>();

and then add elements using:

qEquivalent.add("ADJ");

And please follow Java Naming conventions. Variable names should start with lowercase letters.

edited Jul 31, 2013 at 15:29

answered Jul 31, 2013 at 15:17

Rohit Jain

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4 Comments

user2610661 Over a year ago

should I remove ={} ?

Rohit Jain Over a year ago

@user2610661. Either you create an array of some size, which is still unknown - String[] qEuivalent = new String[50];. Or, use an ArrayList as I suggest.

user2610661 Over a year ago

It works perfectly! Thanks alot! I'm just starting to learn java so thanks again for helping me!

Rohit Jain Over a year ago

@user2610661. You're welcome :) And keep consulting Java API, time to time during your learning. You might get something useful for the problem you are solving.

Jack · Accepted Answer · 2013-07-31 15:19:25Z

1

You create an empty array:

String[] QEquivalent = {};

and then set some elements at index y > 0:

QEquivalent[y] = "ADJ";

You can either:

compute the final dimension of the array and be sure to instantiate it: String[] QEquivalent = new String[SIZE];
use a dynamic structure like an ArrayList

eg:

ArrayList<String> QEquivalent = new ArrayList<QEquivalent>(); 
QEquivalent.add("ADJ");

answered Jul 31, 2013 at 15:19

Jack

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Comments

AllTooSir · Accepted Answer · 2013-07-31 15:28:29Z

Your array QEquivalent is an empty array . It is of length 0 , hence even QEquivalent[0] will throw ArrayIndexOutOfBoundsException.

One fix I can see is assign it a length :

String[] question2 = s.split(" ");
// Just assign the dimension till which you will iterate finally
// from your code `y < question2.length` it seems it should be question2.length
// Note you are always indexing the array using the outer loop counter  y 
// So even if there are n number of nested loops , assigning the question2.length
// as dimension will work fine , unless there is something subtle you missed 
// in your code
String[] QEquivalent = new String[question2.length];

Better use any implementation of List , like an ArrayList.

List<String> qEquivalent = new ArrayList<String>();
......
if (t[w].equals(question2[y])) {
       qEquivalent.add("ADJ");
       System.out.println(qEquivalent.get(y));
       break;
}

Prasad Kharkar · Accepted Answer · 2013-07-31 15:18:50Z

0

Give some size to the array String[] QEquivalent = new String[100];

You statement String[] QEquivalent = {}; creates an array with zero size.

answered Jul 31, 2013 at 15:18

Prasad Kharkar

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Comments

Mena · Accepted Answer · 2013-07-31 15:21:06Z

0

You are declaring your QEquivalent array as an empty String array.

When you access the index QEquivalent[y], that index doesn't exist, hence the ArrayIndexOutOfBoundsException.

I strongly suggest you use a List<String> instead.

Such as:

List<String> qEquivalent = new ArrayList<String>(); // replaces the array declaration and uses Java conventional naming

...

qEquivalent.add("ADJ"); // replaces the indexing of the array and adds the item

answered Jul 31, 2013 at 15:21

Mena

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Comments

ihsan kocak · Accepted Answer · 2013-07-31 15:21:39Z

0

Possibly QEquivalent variable makes the error.Because when you declare that variable, its length is 0.So declare the variable as with new and a size.

answered Jul 31, 2013 at 15:21

ihsan kocak

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Comments

Bumptious Q Bangwhistle · Accepted Answer · 2013-07-31 15:21:45Z

0

Or move it after you split the string into question2 and use:

String[] QEquivalent = new String[question2.length];

answered Jul 31, 2013 at 15:21

Bumptious Q Bangwhistle

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4 Comments

Rohit Jain Over a year ago

question2.length won't be enough. There are more elements going to be added. Take a look at the loop which populates array. It is nested, with outer one itself running till question2.length.

AllTooSir Over a year ago

@RohitJain But he is indexing it with j that is the outer loop counter !

Bumptious Q Bangwhistle Over a year ago

@RohitJain I disagree. He uses the loop for (y = 0; y < question2.length; y++) and only assigns QEquivalent[y] = "ADJ";. So regardless of the inner loop, QEquivalent[y] will never be greater than question2.length.

Rohit Jain Over a year ago

@TheNewIdiot. OOPs, didn't notice that. :(

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