CODEWITHSUNDEEP
javaarrays
Angry Red Panda
Angry Red Panda

Reputation: 439

Java - create a array with the length of an index

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?

Upvotes: 1

Views: 1254

Answers (6)

Gacci
Gacci

Reputation: 1398

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.

Upvotes: -1

Fatih Donmez
Fatih Donmez

Reputation: 4347

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

Upvotes: 1

Bon
Bon

Reputation: 3103

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]);

Upvotes: 1

Suresh Atta
Suresh Atta

Reputation: 121998

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

Upvotes: 8

Thilo
Thilo

Reputation: 262534

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).

Upvotes: 1

Eran
Eran

Reputation: 393846

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].

Upvotes: 5

Related Questions