1

I am trying to create a array with Java that can hold as many numbers as the index 'i' is big.

for (int i = 0; i <= 10; i++)
    {
        int[] zahlenListe = new int[i];
        zahlenListe[i] = i + 5;
        System.out.println(zahlenListe[i]);

    }

but I am always getting the error message: Exception in thread "main" java.lang.ArrayIndexOutOfBoundsException: 0 at Start.main(Start.java:27)

Java:27 is this line of code: zahlenListe[i] = i + 5;.

But everything is working fine when I change this line

int[] zahlenListe = new int[i];

to this:

int[] zahlenListe = new int[11];

Anybody cares to explain where the error is?

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6 Answers 6

Reset to default
8

Array indices are zero based. Hence the maximum index for i sized array is i-1.

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4 Comments

After seeing a typo and then an edit, I have to note that the plural of "index" if technical contexts is "indices"
@Arc676 Edited and followed by old school :)
@sᴜʀᴇsʜᴀᴛᴛᴀ indexes and indices are both valid English, but indexes is used more in technical context, all db documentations refer to them as indexes
@SleimanJneidi Both are valid and indices take less calories to pronounce. Just try ;)
5

An array of length i doesn't have an i'th index. The valid indices go from 0 to i-1.

If you initialize your array with int[] zahlenListe = new int[i+1];, you'll be able to assign a value to zahlenListe[i].

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Comments

1 
int[] zahlenListe = new int[i];
zahlenListe[i] = i + 5;

Arrays start at index 0.

So index i will never be in an i-dimensional array. It stops at i-1.

For i == 0 the array is empty (no entries at all).

You may want to start your loop at i=1.

From your code it is also not clear why you need an array at all (but you probably have more code in there that you are not showing).

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Comments

1

If you creating an array using:

int[] zahlenListe = new int[i];

The last element in array zahlenListe would be zahlenListe[i-1] instead of zahlenListe[i]. In addition, assuming i should start with 1 instead of 0 because an array of length is pointless.

Therefore, use

zahlenListe[i-1] = i + 5;
System.out.println(zahlenListe[i-1]);
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Comments

1

You need to start 0 to size-1 in an array as you see;

int[] zahlenListe = new int[i];

Your array size is always i which means you can allowed to access max i-1

Assuming that i always bigger than 0

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Comments

-1

Basically because in your condition you are telling the computer to go to the 11th element. Think of it this way ...

int i = 0; 

First iteration i = 0
if i <= 10; then i = i + 1

Second iteration i = 1
if i <= 10; then i = i + 1

Third iteration i = 2
if i <= 10; then i = i + 1

Fourth iteration i = 3
if i <= 10; then i = i + 1

Fifth iteration i = 4
if i <= 10; then i = i + 1

Sixth iteration i = 5
if i <= 10; then i = i + 1

Seventh iteration i = 6
if i <= 10; then i = i + 1

Eighth iteration i = 7
if i <= 10; then i = i + 1

Ninth iteration i = 8
if i <= 10; then i = i + 1

Tenth iteration i = 9
if i <= 10; then i = i + 1

Eleventh iteration i = 10
if i <= 10; then i = i + 1

Once you get here you are trying to access an element that does not exist. Because your condition says that as long as i is less that or equal that 10, it should repeat the task. If you change <= for < then you stop at the last element available.

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3 Comments

The eleventh iteration is not a problem (at least not more than the first one is).
Not sure what you mean, Explain!
The current code crashes in its first iteration, when i==0. If you fix that problem, it will also work for i==10.

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