Java ArrayIndexOutOfBounds Exception

Question

Perhaps I have been looking at this for too long as I cannot find the problem, yet it should be something simple. I am receiving an ArrayIndexOutOfBounds exception on the line:

nextWord = MyArray[i + 1].toLowerCase();

Can anyone see why?

  String currentWord = "";
  String nextWord = "";

  for (int i = 0; i <= MyArray.length; i++) {

   // If not at the end of the array
   if (MyArray.length > 0 && i < MyArray.length) {

    currentWord = MyArray[i].toLowerCase();
    nextWord = MyArray[i + 1].toLowerCase(); /* EXCEPTION */

    System.out.println("CURRENT WORD: " + currentWord);
    System.out.println("NEXT WORD: " + nextWord);
   } 
  }

Thanks!

Petar Minchev · Accepted Answer · 2010-10-28 15:50:46Z

4

MyArray.length - 1 is the last element of the array. The biggest value of i which will go down in the if is MyArray.length - 1. And you increase it by one in i + 1, so you get MyArray.length. Of course you will receive an exception:)

answered Oct 28, 2010 at 15:50

Petar Minchev

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Comments

Carl · Accepted Answer · 2010-10-28 16:54:16Z

3

Array indices run from 0 to array.length - 1.

The typical loop construct for arrays is thus:

for (int i=0; i<array.length; i++) // do stuff

in your case, you've a got a single position look ahead, so to avoid out-of-bounds you need to restrict that loop by one position:

for (int i=0; i<array.length-1; i++) // do stuff

if you scope the index outside of the loop, after the loop it will have the right value to assign the last currentWord:

int i=0;
for (; i<array.length-1; i++) // do stuff
// here i == array.length - 1, provided you don't mess with i in the "do stuff" part

edited Oct 28, 2010 at 16:54

answered Oct 28, 2010 at 15:50

Carl

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Comments

Dr. Snoopy · Accepted Answer · 2010-10-28 15:50:27Z

0

Because if i < MyArray.Length, then i+1 CAN be out of bounds. For example, if i = MyArray.Length - 1 (Last valid index), then i + 1 = MyArray.Length, which is out of bounds.

answered Oct 28, 2010 at 15:50

Dr. Snoopy

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Comments

codaddict · Accepted Answer · 2010-10-28 15:53:21Z

0

For array MyArray, the valid index are [0,MyArray.length-1]. Since for a given i you are accessing element at index i+1, valid value for i are [0,MyArray.length-2].

So you can do:

for (int i = 0; i <= MyArray.length-2; i++) {

    // no need of the if check anymore.
    currentWord = MyArray[i].toLowerCase();
    nextWord = MyArray[i + 1].toLowerCase();

answered Oct 28, 2010 at 15:53

codaddict

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Comments

robev · Accepted Answer · 2010-10-28 17:17:12Z

0

Simply fix your check that you are not at the last member of the array. If you are at the last member of the array, adding one to it will go beyond the array and thus you will get that exception. Also you are skipping the first element, and looping past the end of the array (since you start at zero, going to the length is one extra loop)

for (int i = 0; i < MyArray.length; i++) {  
    currentWord = MyArray[i].toLowerCase();
    System.out.println("CURRENT WORD: " + currentWord);

    // If not at the end of the array  
    if (i != MyArray.length - 1) {  
       nextWord = MyArray[i + 1].toLowerCase();
       System.out.println("NEXT WORD: " + nextWord);
    }
}

edited Oct 28, 2010 at 17:17

answered Oct 28, 2010 at 15:57

robev

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4 Comments

Petar Minchev Over a year ago

Remove the =. It should be i < myArray.length

littleK Over a year ago

One thing that I still don't understand here is that on iteration #1, when i=0, currentWord="" and nextWord="". On iteration #2, when i=1, then currentWord and nextWord will be set correctly. How can I fix that? Also, what if I also wanted to include and set a "previousWord"?

robev Over a year ago

That shouldn't happen unless MyArray[0] is an empty string. Also, you can add another check like the one I put but before current word, and check if i > 0, and prevWord = MyArray[i-1].toLowerCase()

littleK Over a year ago

Ah it turns out that I just had my print statements before all of the assignments. Thanks!

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