asked on

java arrays

Hi,

I am trying below example

http://www.learnjavaonline.org/en/Arrays

public class Main {
public static void main(String[] args) {
int[] numbers = {1, 2, 3};
int length = numbers[3];
char[] chars = new char[length];
chars[numbers.length] = 'y';
System.out.println("Done!");
}
}

I get error as below even after changiing to 3 from 2 as below
int length=number[3]
from
int length=number[2]

Exception in thread "main" java.lang.ArrayIndexOutOfBoundsException: 3
at Main.main(Main.java:4)

Please advice what is meaning of above exception and why i am getting and how to fix it.

krakatoa

Array indexes start at 0.

krakatoa

int[] numbers = {1, 2, 3};
        int length = numbers[2];
        char[] chars = new char[length];
        chars[numbers.length-1] = 'y';
        System.out.println("Done!");

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'length' (somewhat confusingly) IS the real length, but when you use it as an array index, it will always be one position too long.

gudii9

ASKER

int length = numbers[3];

when we pass this length to

does it take 3 or 4

krakatoa

. . . but there is *no* numbers[3] - only numbers[0], numbers[1], and numbers[2]. (these ARE your three positions).

That is how arrays work - the FIRST element is ALWAYS 0 - zero.

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krakatoa

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gudii9

ASKER

I think variable name(length) is confusing which is same as method name.

I changed as below

public class Main {
public static void main(String[] args) {
int[] numbers = {1, 2, 3};
int lengthVar = numbers[3];
char[] chars = new char[lengthVar];
chars[numbers.length-1] = 'y';
System.out.println("Done!");
}
}

I still get same index out of bound exception. How to fix to so that Done! is printed to console. please advice

krakatoa

I still get same index out of bound exception. How to fix to so that Done! is printed to console. please advice

But we just did that in the first comment I made. You are going 'round in circles here!

You CAN'T do this :

int lengthVar = numbers[3];

because there is NO "numbers[3]" only numbers[0], numbers[1], and numbers[2].

Read this comment again : krakatoa2014-05-29 at 17:37:34ID: 40098766

and let me know why you are having trouble with it.

krakatoa

Tell us what you think the value of lengthVar is in the code here :

int[] numbers = {47, 49, 51};
int lengthVar = numbers[2];

gudii9

ASKER

i see your point
It is 51

I modified code like below

public class Main {
    public static void main(String[] args) {
        int[] numbers = {1, 2, 3};
        int lengthVar = numbers[2];
        System.out.println("length!"+lengthVar);
        char[] chars = new char[lengthVar];
        chars[numbers.length-1] = 'y';
        System.out.println("char[2] is-->"+chars[2]+"char[1] is-->"+chars[1]+"char[] is-->"+chars[0]);
        System.out.println("Done!");
    }
}

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I got below output as belowlength!3
char[2] is-->ychar[1] is-->char[] is-->
Done!

gudii9

ASKER

i modified still further

public class Main {
    public static void main(String[] args) {
        int[] numbers = {1, 2, 3};
        int lengthVar = numbers[2];
        System.out.println("length!"+lengthVar);
        char[] chars = new char[lengthVar];
        System.out.println("numbers.length-->"+numbers.length);
        chars[numbers.length-1] = 'y';
        System.out.println("char[2] is-->"+chars[2]+"char[1] is-->"+chars[1]+"char[] is-->"+chars[0]);
        System.out.println("Done!");
    }

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i got below output which makes more sense now
length!3
numbers.length-->3
char[2] is-->ychar[1] is-->char[] is-->
Done!

krakatoa

i see your point

Great.

Slightly odd concept, something starting at zero, but that is how it works.