Как исправить array index out of bounds

What causes ArrayIndexOutOfBoundsException?

If you think of a variable as a «box» where you can place a value, then an array is a series of boxes placed next to each other, where the number of boxes is a finite and explicit integer.

Creating an array like this:

final int[] myArray = new int[5]

creates a row of 5 boxes, each holding an int. Each of the boxes has an index, a position in the series of boxes. This index starts at 0 and ends at N-1, where N is the size of the array (the number of boxes).

To retrieve one of the values from this series of boxes, you can refer to it through its index, like this:

myArray[3]

Which will give you the value of the 4th box in the series (since the first box has an index of 0).

An ArrayIndexOutOfBoundsException is caused by trying to retrieve a «box» that does not exist, by passing an index that is higher than the index of the last «box», or negative.

With my running example, these code snippets would produce such an exception:

myArray[5] //tries to retrieve the 6th "box" when there is only 5
myArray[-1] //just makes no sense
myArray[1337] //way to high

How to avoid ArrayIndexOutOfBoundsException

In order to prevent ArrayIndexOutOfBoundsException, there are some key points to consider:

Looping

When looping through an array, always make sure that the index you are retrieving is strictly smaller than the length of the array (the number of boxes). For instance:

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

Notice the <, never mix a = in there..

You might want to be tempted to do something like this:

for (int i = 1; i <= myArray.length; i++) {
    final int someint = myArray[i - 1]

Just don’t. Stick to the one above (if you need to use the index) and it will save you a lot of pain.

Where possible, use foreach:

for (int value : myArray) {

This way you won’t have to think about indexes at all.

When looping, whatever you do, NEVER change the value of the loop iterator (here: i). The only place this should change value is to keep the loop going. Changing it otherwise is just risking an exception, and is in most cases not necessary.

Retrieval/update

When retrieving an arbitrary element of the array, always check that it is a valid index against the length of the array:

public Integer getArrayElement(final int index) {
    if (index < 0 || index >= myArray.length) {
        return null; //although I would much prefer an actual exception being thrown when this happens.
    }
    return myArray[index];
}

What causes ArrayIndexOutOfBoundsException?

If you think of a variable as a «box» where you can place a value, then an array is a series of boxes placed next to each other, where the number of boxes is a finite and explicit integer.

Creating an array like this:

final int[] myArray = new int[5]

creates a row of 5 boxes, each holding an int. Each of the boxes has an index, a position in the series of boxes. This index starts at 0 and ends at N-1, where N is the size of the array (the number of boxes).

To retrieve one of the values from this series of boxes, you can refer to it through its index, like this:

myArray[3]

Which will give you the value of the 4th box in the series (since the first box has an index of 0).

An ArrayIndexOutOfBoundsException is caused by trying to retrieve a «box» that does not exist, by passing an index that is higher than the index of the last «box», or negative.

With my running example, these code snippets would produce such an exception:

myArray[5] //tries to retrieve the 6th "box" when there is only 5
myArray[-1] //just makes no sense
myArray[1337] //way to high

How to avoid ArrayIndexOutOfBoundsException

In order to prevent ArrayIndexOutOfBoundsException, there are some key points to consider:

Looping

When looping through an array, always make sure that the index you are retrieving is strictly smaller than the length of the array (the number of boxes). For instance:

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

Notice the <, never mix a = in there..

You might want to be tempted to do something like this:

for (int i = 1; i <= myArray.length; i++) {
    final int someint = myArray[i - 1]

Just don’t. Stick to the one above (if you need to use the index) and it will save you a lot of pain.

Where possible, use foreach:

for (int value : myArray) {

This way you won’t have to think about indexes at all.

When looping, whatever you do, NEVER change the value of the loop iterator (here: i). The only place this should change value is to keep the loop going. Changing it otherwise is just risking an exception, and is in most cases not necessary.

Retrieval/update

When retrieving an arbitrary element of the array, always check that it is a valid index against the length of the array:

public Integer getArrayElement(final int index) {
    if (index < 0 || index >= myArray.length) {
        return null; //although I would much prefer an actual exception being thrown when this happens.
    }
    return myArray[index];
}

[18:00:20] [debug] Run time error 4: "Array index out of bounds"
[18:00:20] [debug]  Attempted to read/write array element at index 200 in array of size 200
[18:00:20] [debug] AMX backtrace:
[18:00:20] [debug] #0 006411b4 in public BadEngine () at C:UsersrotkiDesktopсерверPRIME (1)gamemodesprime.pwn:73049
[18:00:20] [debug] Run time error 4: "Array index out of bounds"
[18:00:20] [debug]  Attempted to read/write array element at index 200 in array of size 200
[18:00:20] [debug] AMX backtrace:
[18:00:20] [debug] #0 006411b4 in public BadEngine () at C:UsersrotkiDesktopсерверPRIME (1)gamemodesprime.pwn:73049
[18:00:21] [debug] Run time error 4: "Array index out of bounds"
[18:00:21] [debug]  Attempted to read/write array element at index 200 in array of size 200
[18:00:21] [debug] AMX backtrace:
[18:00:21] [debug] #0 006411b4 in public BadEngine () at C:UsersrotkiDesktopсерверPRIME (1)gamemodesprime.pwn:73049
[18:00:21] [debug] Run time error 4: "Array index out of bounds"
[18:00:21] [debug]  Attempted to read/write array element at index 200 in array of size 200
[18:00:21] [debug] AMX backtrace:
[18:00:21] [debug] #0 006411b4 in public BadEngine () at C:UsersrotkiDesktopсерверPRIME (1)gamemodesprime.pwn:73049
[18:00:22] [debug] Run time error 4: "Array index out of bounds"
[18:00:22] [debug]  Attempted to read/write array element at index 200 in array of size 200
[18:00:22] [debug] AMX backtrace:
[18:00:22] [debug] #0 006411b4 in public BadEngine () at C:UsersrotkiDesktopсерверPRIME (1)gamemodesprime.pwn:73049

forward BadEngine();
public BadEngine()
{
    new veh, Float:spd[3], Float:hls;
    for(new i; i != GetMaxPlayers(); i++)
    {
        if( !BE_Play_Check[i] ) { continue; }
        veh = GetPlayerVehicleID( i );
        if( !veh ) { continue; }
        GetVehicleHealth( veh, hls );
        if( hls > BE_MIN_HLS ) { continue; }
        GetVehicleVelocity( veh, spd[0], spd[1], spd[2] );

        if( floatabs(spd[0]) > floatabs(spd[1]) )
        {
            if( floatabs(spd[ 0 ]) > BE_MAX_SPD )
            {
                hls = BE_MAX_SPD / floatabs(spd[ 0 ]);
                SetVehicleVelocity( veh, spd[0]*hls, spd[1]*hls, spd[2] );
            }
        }
        else
        {
            if( floatabs(spd[ 1 ]) > BE_MAX_SPD )
            {
                hls = BE_MAX_SPD / floatabs(spd[ 1 ]);
                SetVehicleVelocity( veh, spd[0]*hls, spd[1]*hls, spd[2] );
            }
        }
    }
}

[18:00:20] [debug] Run time error 4: "Array index out of bounds"
[18:00:20] [debug]  Attempted to read/write array element at index 200 in array of size 200

new Array[200];
for(new i; i < sizeof(Array); i++) {

	printf("%i", Array[i]);
}

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Posted by Marta on March 4, 2022 Viewed 6066 times

In this article, you will learn the main reason why you will face the java.lang.ArrayIndexOutOfBoundsException Exception in Java along with a few different examples and how to fix it.

I think it is essential to understand why this error occurs, since the better you understand the error, the better your ability to avoid it.

This tutorial contains some code examples and possible ways to fix the error.

Watch the tutorial on Youtube:

What causes a java.lang.ArrayIndexOutOfBoundsException?

The java.lang.ArrayIndexOutOfBoundsException error occurs when you are attempting to access by index an item in indexed collection, like a List or an Array. You will this error, when the index you are using to access doesn’t exist, hence the “Out of Bounds” message.

To put it in simple words, if you have a collection with four elements, for example, and you try to access element number 7, you will get java.lang.ArrayIndexOutOfBoundsException error.

To help you visualise this problem, you can think of arrays and index as a set of boxes with label, where every label is a number. Note that you will start counting from 0.

For instance, we have a collection with three elements, the valid indexes are 0,1 and 2. This means three boxes, one label with 0, another one with 1, and the last box has the label 2. So if you try to access box number 3, you can’t, because it doesn’t exist. That’s when you get the java.lang.ArrayIndexOutOfBoundsException error.

The Simplest Case

Let’s see the simplest code snippet which will through this error:

public class IndexError {

    public static void main(String[] args){
        String[] fruits = {"Orange", "Pear", "Watermelon"};
        System.out.println(fruits[4]);
    }
}

Output:

Exception in thread "main" java.lang.ArrayIndexOutOfBoundsException: Index 4 out of bounds for length 3

Although the error refer to arrays, you could also encounter the error when working with a List, which is also an indexed collection, meaning a collection where item can be accessed by their index.

import java.util.Arrays;
import java.util.List;
public class IndexError {

    public static void main(String[] args){
        List<String> fruits = Arrays.asList("Orange", "Pear", "Watermelon");
        System.out.println(fruits.get(4));
    }
}

Output:

Exception in thread "main" java.lang.ArrayIndexOutOfBoundsException: Index 4 out of bounds for length 3

Example #1: Using invalid indexes in a loop

An instance where you might encounter the ArrayIndexOutOfBoundsException exception is when working with loops. You should make sure the index limits of your loop are valid in all the iterations. The loop shouldn’t try to access an item that doesn’t exist in the collection. Let’s see an example.

import java.util.Arrays;
import java.util.List;

public class IndexError {

    public static void main(String[] args){
        List<String> fruits = Arrays.asList("Orange", "Pear", "Watermelon");
        for(int index=0;index<=fruits.size();index++){
            System.out.println(fruits.get(index));
        }
    }
}

Output:

Exception in thread "main" java.lang.ArrayIndexOutOfBoundsException: Index 3 out of bounds for length 3
Orange
Pear
Watermelon
	at java.base/java.util.Arrays$ArrayList.get(Arrays.java:4351)
	at com.hellocodeclub.IndexError.main(IndexError.java:11)

The code above will loop through every item in the list and print its value, however there is an exception at the end. The problem in this case is the stopping condition in the loop: index<=fruits.size(). This means that it will while index is less or equal to 3, therefore the index will start in 0, then 1, then 2 and finally 3, but unfortunately 3 is invalid index.

How to fix it – Approach #1

You can fix this error by changing the stopping condition. Instead of looping while the index is less or equal than 3, you can replace by “while index is less than 3”. Doing so the index will go through 0, 1, and 2, which are all valid indexes. Therefore the for loop should look as below:

for(int index=0;index<fruits.size();index++){

That’s all, really straightforward change once you understand the root cause of the problem. Here is how the code looks after the fix:

import java.util.Arrays;
import java.util.List;

public class IndexError {
    public static void main(String[] args){
        List<String> fruits = Arrays.asList("Orange", "Pear", "Watermelon");
        for(int index=0;index<fruits.size();index++){
            System.out.println(fruits.get(index));
        }
    }
}

Output:

How to fix it – Approach #2

Another way to avoid this issue is using the for each syntax. Using this approach, you don’t have to manage the index, since Java will do it for you. Here is how the code will look:

import java.util.Arrays;
import java.util.List;

public class IndexError {

    public static void main(String[] args){
        List<String> fruits = Arrays.asList("Orange", "Pear", "Watermelon");
        for(String fruit: fruits){
            System.out.println(fruit);
        }
    }
}

Example #2: Loop through a string

Another example. Imagine that you want to count the appearances of character ‘a’ in a given sentence. You will write a piece of code similar to the one below:

public class CountingA {

    public static void main(String[] args){
        String sentence = "Live as if you were to die tomorrow. Learn as if you were to live forever";
        char [] characters = sentence.toCharArray();
        int countA = 0;
        for(int index=0;index<=characters.length;index++){
            if(characters[index]=='a'){
                countA++;
            }
        }
        System.out.println(countA);
    }
}

Output:

Exception in thread "main" java.lang.ArrayIndexOutOfBoundsException: Index 73 out of bounds for length 73
	at com.hellocodeclub.dev.CountingA.main(CountingA.java:8)

The issue in this case is similar to the one from the previous example. The stopping condition is the loop is right. You should make sure the index is within the boundaries.

How to fix it

As in the previous example, you can fix it by removing the equal from the stopping condition, or by using a foreach and therefore avoid managing the index. See below both fixes:

public class CountingA {
    public static void main(String[] args){
        String sentence = "Live as if you were to die tomorrow. Learn as if you were to live forever";
        int countA = 0;
        for(Character charater: sentence.toCharArray()){ // FOR EACH
            if(charater=='a'){
                countA++;
            }
        }
        System.out.println(countA);
    }
}

Output:

Conclusion

In summary, we have seen what the “java.lang.ArrayIndexOutOfBoundsException” error means . Additionally we covered how you can fix this error by making sure the index is within the collection boundaries, or using foreach syntax so you don’t need to manage the index.

I hope you enjoy this article, and understand this issue better to avoid it when you are programming.

Thank you so much for reading and supporting this blog!

Happy Coding!

ArrayIndexOutOfBoundsException – это исключение, появляющееся во время выполнения. Оно возникает тогда, когда мы пытаемся обратиться к элементу массива по отрицательному или превышающему размер массива индексу. Давайте посмотрим на примеры, когда получается ArrayIndexOutOfBoundsException в программе на Java.

Попробуйте выполнить такой код:

    static int number=11;
    public static String[][] transactions=new String[8][number];
    public static void deposit(double amount){
        transactions[4][number]="deposit";
        number++;
    }

    public static void main(String[] args) {
        deposit(11);
    }
}

Вы увидите ошибку:

Caused by: java.lang.ArrayIndexOutOfBoundsException: 0
	at sample.Main.deposit(Main.java:22)
	at sample.Main.main(Main.java:27)
Exception running application sample.Main

Process finished with exit code 1

Что здесь произошло? Ошибка в строке 27 – мы вызвали метод deposit(), а в нем уже, в строке 22 – попытались внести в поле массива значение «deposit». Почему выкинуло исключение? Дело в том, что мы инициализировали массив размера 11 (number = 11), н опопытались обратиться к 12-му элементу. Нумерация элементов массива начинается с нуля. Так что здесь надо сделать, например, так

public static String[][] transactions=new String[8][100];

Но вообще, это плохой код, так писать не надо. Давайте рассмотрим еще один пример возникновения ошибки ArrayIndexOutOfBoundsException:

public static void main(String[] args) {
    Random random = new Random();
    int [] arr = new int[10];
    for (int i = 0; i <= arr.length; i++) {
       arr[i] =  random.nextInt(100);
       System.out.println(arr[i]);
    }
}

Здесь массив заполняется случайными значениями. При выполнении IntelliJ IDEA выдаст ошибку

Caused by: java.lang.ArrayIndexOutOfBoundsException: 10
	at sample.Main.main(Main.java:37)

В строке 37 мы заносим значение в массив. Ошибка возникла помтому, что индекса 10 нет в массиве arr, поэтому условие цикла i <= arr.length надо поменять на i < arr.length

Конструкция try для ArrayIndexOutOfBoundsException

ArrayIndexOutOfBoundsException можно обработать с помощью конструкции try-catch. Для этого оберните try то место, где происходит обращение к элементу массива по индексу, например, заносится значение. Как-то так:

try {
    array[index] = "что-то";
}
catch (ArrayIndexOutOfBoundsException ae){
    System.out.println(ae);
}

Но я бы рекомендовал вам все же не допускать данной ошибки, писать код таким образом, чтобы не пришлось ловить исключение ArrayIndexOutOfBoundsException.

Автор этого материала — я — Пахолков Юрий. Я оказываю услуги по написанию программ на языках Java, C++, C# (а также консультирую по ним) и созданию сайтов. Работаю с сайтами на CMS OpenCart, WordPress, ModX и самописными. Кроме этого, работаю напрямую с JavaScript, PHP, CSS, HTML — то есть могу доработать ваш сайт или помочь с веб-программированием. Пишите сюда.

заметки, ArrayIndexOutOfBoundsException, java, ошибки, исключения

Как исправить Error
Автор: neka

Значение Error можно посмотреть здесь.

error 040: duplicate «case» label (value 28)

Это означает что case стаким значением повторяется. Решение этой проблемы простое — нам нужно цифру 28 изменит на другую (в той строчке на которую жалуется )

error 032: array index out of bounds (variable «JoinPed»)

Это означает что индекс массива превышен (но не всегда, смотрим дальше) Пример:

131 — массив поигравшись с ним я понял что дело не в нем, а в чём же спросите вы? Пример данной ошибки:

как видим — JoinPed[123] сначало с таким значением, а потом JoinPed[98]. Решение простое: JoinPed[123] число в данных скобках должно быть одинаковым. Пример:

error 037: invalid string (possibly non-terminated string)

Это означает что строка неправильная, а точнее где то допущена ошибка:

как видим нам после слова «модера» не хватает «. Правим:

error 001: expected token: «,», but found «;»

Это значит что мы пропустили знак или скобку (в данном примере скобку) Пример:

смотрим на:

и вим что мы ппропустили )

Правим:

И в итоге:

error 002: only a single statement (or expression) can follow each «case»

Это означает что у вас после «case» идет if(dialogid == ). Пример:

Решение простое: if(dialogid == 7504) это нам нужно заменить на case как и последующий диалог !

error 004: function «%s» is not implemented

Это означает что мы пропустили скобку. Мой совет:

• проверить весь код в ручную
• на форуме был урок как найти не по ставленую скобку
• Можно воспользоватся notepad++ там показы линии открытых скобок и тогда можно найти эту скобку

error 017: undefined symbol %s

Это означает что мы не поставили переменную new. Пример:

Решение — ко всем new добавим: