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Thread: Need help with fixing this error: ArrayIndexOutOfBoundsException: 1

  1. #1
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    Question Need help with fixing this error: ArrayIndexOutOfBoundsException: 1

    Here is my code for this method:
    (for passed-in desc, it has format as "brand: model"
    For example, if the brand is Whirlpool and the model is LX1279, the description will be Whirlpool:LX1279. If only the brand is set, the description should contain the brand followed by a colon. If only the model is set, the description should contain a colon followed by the model. If both the brand and the model are null, that is equivalent to a null description.)

    public void setDescription(String desc)
    {
    if (desc == null)
    {
    throw new IllegalArgumentException
    ("description is null.");
    }
    else if (!desc.contains(":"))
    {
    throw new IllegalArgumentException
    ("description does not contain a colon.");
    }
    else if (desc.contains(":"))
    {
    String[] tok = new String[2];
    tok = desc.split(":");

    if (tok[0] != null && tok[1] != null)
    {
    appliance.setBrand(tok[0]);
    appliance.setModel(tok[1]);
    }


    }
    }
    The error suggested the line that contains such problem is at line 53, which I highlighted with bold red color.

    symptom: java.lang.ArrayIndexOutOfBoundsException: 1
    at ApplianceAdapter.setDescription(ApplianceAdapter.j ava:53)

    I am not really familiar with such error, but any hint/explanation/examples will be helpful.


  2. #2
    Super Moderator Norm's Avatar
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    Default Re: Need help with fixing this error: ArrayIndexOutOfBoundsException: 1

    symptom: java.lang.ArrayIndexOutOfBoundsException: 1
    at ApplianceAdapter.setDescription(ApplianceAdapter.j ava:53)
    At line 53 the code used an index whose value was past the end of the array.
    The array has less than 2 elements and index 1 is invalid
    Remember array indexes range from 0 to the array's length-1;

    Make sure that the array is long enough by using its .length attribute before trying to index into it.

    The split() method creates an array depending on the String it is working on. The value of the array variable will be replaced by what is returned by the split() method.

    Use this for debugging to see the contents of an array:
    System.out.println("an ID "+ java.util.Arrays.toString(theArrayName));
    If you don't understand my answer, don't ignore it, ask a question.

  3. The Following User Says Thank You to Norm For This Useful Post:

    mikuya707 (April 13th, 2013)

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