What does ArrayIndexOutOfBoundsException mean and how do I get rid of it?
Here is a code sample that triggers the exception:
String[] names = { "tom", "bob", "harry" };
for (int i = 0; i <= names.length; i++) {
System.out.println(names[i]);
}
Your first port of call should be the documentation which explains it reasonably clearly:
Thrown to indicate that an array has been accessed with an illegal index. The index is either negative or greater than or equal to the size of the array.
So for example:
int[] array = new int[5];
int boom = array[10]; // Throws the exception
As for how to avoid it... um, don't do that. Be careful with your array indexes.
One problem people sometimes run into is thinking that arrays are 1-indexed, e.g.
int[] array = new int[5];
// ... populate the array here ...
for (int index = 1; index <= array.length; index++)
{
System.out.println(array[index]);
}
That will miss out the first element (index 0) and throw an exception when index is 5. The valid indexes here are 0-4 inclusive. The correct, idiomatic for statement here would be:
for (int index = 0; index < array.length; index++)
(That's assuming you need the index, of course. If you can use the enhanced for loop instead, do so.)
if (index < 0 || index >= array.length) {
// Don't use this index. This is out of bounds (borders, limits, whatever).
} else {
// Yes, you can safely use this index. The index is present in the array.
Object element = array[index];
}
See also:
The Java Tutorials - Language Basics - Arrays
Update: as per your code snippet,
for (int i = 0; i<=name.length; i++) {
The index is inclusive the array's length. This is out of bounds. You need to replace <= by <.
for (int i = 0; i < name.length; i++) {
From this excellent article: ArrayIndexOutOfBoundsException in for loop
To put it briefly:
In the last iteration of
for (int i = 0; i <= name.length; i++) {
i will equal name.length which is an illegal index, since array indices are zero-based.
Your code should read
for (int i = 0; i < name.length; i++)
^
It means that you are trying to access an index of an array which is not valid as it is not in between the bounds.
For example this would initialize a primitive integer array with the upper bound 4.
int intArray[] = new int[5];
Programmers count from zero. So this for example would throw an ArrayIndexOutOfBoundsException as the upper bound is 4 and not 5.
intArray[5];
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];
}
To avoid an array index out-of-bounds exception, one should use the enhanced-for statement where and when they can.
The primary motivation (and use case) is when you are iterating and you do not require any complicated iteration steps. You would not be able to use an enhanced-for to move backwards in an array or only iterate on every other element.
You're guaranteed not to run out of elements to iterate over when doing this, and your [corrected] example is easily converted over.
The code below:
String[] name = {"tom", "dick", "harry"};
for(int i = 0; i< name.length; i++) {
System.out.print(name[i] + "\n");
}
...is equivalent to this:
String[] name = {"tom", "dick", "harry"};
for(String firstName : name) {
System.out.println(firstName + "\n");
}
In your code you have accessed the elements from index 0 to the length of the string array. name.length gives the number of string objects in your array of string objects i.e. 3, but you can access only up to index 2 name[2],
because the array can be accessed from index 0 to name.length - 1 where you get name.length number of objects.
Even while using a for loop you have started with index zero and you should end with name.length - 1. In an array a[n] you can access form a[0] to a[n-1].
For example:
String[] a={"str1", "str2", "str3" ..., "strn"};
for(int i=0; i<a.length(); i++)
System.out.println(a[i]);
In your case:
String[] name = {"tom", "dick", "harry"};
for(int i = 0; i<=name.length; i++) {
System.out.print(name[i] +'\n');
}
For your given array the length of the array is 3(i.e. name.length = 3). But as it stores element starting from index 0, it has max index 2.
So, instead of 'i**<=name.length' you should write 'i<**name.length' to avoid 'ArrayIndexOutOfBoundsException'.
So much for this simple question, but I just wanted to highlight a new feature in Java which will avoid all confusions around indexing in arrays even for beginners. Java-8 has abstracted the task of iterating for you.
int[] array = new int[5];
//If you need just the items
Arrays.stream(array).forEach(item -> { println(item); });
//If you need the index as well
IntStream.range(0, array.length).forEach(index -> { println(array[index]); })
What's the benefit? Well, one thing is the readability like English. Second, you need not worry about the ArrayIndexOutOfBoundsException
The most common case I've seen for seemingly mysterious ArrayIndexOutOfBoundsExceptions, i.e. apparently not caused by your own array handling code, is the concurrent use of SimpleDateFormat. Particularly in a servlet or controller:
public class MyController {
SimpleDateFormat dateFormat = new SimpleDateFormat("MM/dd/yyyy");
public void handleRequest(ServletRequest req, ServletResponse res) {
Date date = dateFormat.parse(req.getParameter("date"));
}
}
If two threads enter the SimplateDateFormat.parse() method together you will likely see an ArrayIndexOutOfBoundsException. Note the synchronization section of the class javadoc for SimpleDateFormat.
Make sure there is no place in your code that are accessing thread unsafe classes like SimpleDateFormat in a concurrent manner like in a servlet or controller. Check all instance variables of your servlets and controllers for likely suspects.
You are getting ArrayIndexOutOfBoundsException due to i<=name.length part. name.length return the length of the string name, which is 3. Hence when you try to access name[3], it's illegal and throws an exception.
Resolved code:
String[] name = {"tom", "dick", "harry"};
for(int i = 0; i < name.length; i++) { //use < insteadof <=
System.out.print(name[i] +'\n');
}
It's defined in the Java language specification:
The public final field length, which contains the number of components
of the array. length may be positive or zero.
That's how this type of exception looks when thrown in Eclipse. The number in red signifies the index you tried to access. So the code would look like this:
myArray[5]
The error is thrown when you try to access an index which doesn't exist in that array. If an array has a length of 3,
int[] intArray = new int[3];
then the only valid indexes are:
intArray[0]
intArray[1]
intArray[2]
If an array has a length of 1,
int[] intArray = new int[1];
then the only valid index is:
intArray[0]
Any integer equal to the length of the array, or bigger than it: is out of bounds.
Any integer less than 0: is out of bounds;
P.S.: If you look to have a better understanding of arrays and do some practical exercises, there's a video here: tutorial on arrays in Java
For multidimensional arrays, it can be tricky to make sure you access the length property of the right dimension. Take the following code for example:
int [][][] a = new int [2][3][4];
for(int i = 0; i < a.length; i++){
for(int j = 0; j < a[i].length; j++){
for(int k = 0; k < a[j].length; k++){
System.out.print(a[i][j][k]);
}
System.out.println();
}
System.out.println();
}
Each dimension has a different length, so the subtle bug is that the middle and inner loops use the length property of the same dimension (because a[i].length is the same as a[j].length).
Instead, the inner loop should use a[i][j].length (or a[0][0].length, for simplicity).
For any array of length n, elements of the array will have an index from 0 to n-1.
If your program is trying to access any element (or memory) having array index greater than n-1, then Java will throw ArrayIndexOutOfBoundsException
So here are two solutions that we can use in a program
Maintaining count:
for(int count = 0; count < array.length; count++) {
System.out.println(array[count]);
}
Or some other looping statement like
int count = 0;
while(count < array.length) {
System.out.println(array[count]);
count++;
}
A better way go with a for each loop, in this method a programmer has no need to bother about the number of elements in the array.
for(String str : array) {
System.out.println(str);
}
ArrayIndexOutOfBoundsException whenever this exception is coming it mean you are trying to use an index of array which is out of its bounds or in lay man terms you are requesting more than than you have initialised.
To prevent this always make sure that you are not requesting a index which is not present in array i.e. if array length is 10 then your index must range between 0 to 9
ArrayIndexOutOfBounds means you are trying to index a position within an array that is not allocated.
In this case:
String[] name = { "tom", "dick", "harry" };
for (int i = 0; i <= name.length; i++) {
System.out.println(name[i]);
}
name.length is 3 since the array has been defined with 3 String objects.
When accessing the contents of an array, position starts from 0. Since there are 3 items, it would mean name[0]="tom", name[1]="dick" and name[2]="harry
When you loop, since i can be less than or equal to name.length, you are trying to access name[3] which is not available.
To get around this...
In your for loop, you can do i < name.length. This would prevent looping to name[3] and would instead stop at name[2]
for(int i = 0; i<name.length; i++)
Use a for each loop
String[] name = { "tom", "dick", "harry" };
for(String n : name) {
System.out.println(n);
}
Use list.forEach(Consumer action) (requires Java8)
String[] name = { "tom", "dick", "harry" };
Arrays.asList(name).forEach(System.out::println);
Convert array to stream - this is a good option if you want to perform additional 'operations' to your array e.g. filter, transform the text, convert to a map etc (requires Java8)
String[] name = { "tom", "dick", "harry" };
--- Arrays.asList(name).stream().forEach(System.out::println);
--- Stream.of(name).forEach(System.out::println);
ArrayIndexOutOfBoundsException means that you are trying to access an index of the array that does not exist or out of the bound of this array. Array indexes start from 0 and end at length - 1.
In your case
for(int i = 0; i<=name.length; i++) {
System.out.print(name[i] +'\n'); // i goes from 0 to length, Not correct
}
ArrayIndexOutOfBoundsException happens when you are trying to access
the name.length indexed element which does not exist (array index ends at length -1). just replacing <= with < would solve this problem.
for(int i = 0; i < name.length; i++) {
System.out.print(name[i] +'\n'); // i goes from 0 to length - 1, Correct
}
According to your Code :
String[] name = {"tom", "dick", "harry"};
for(int i = 0; i<=name.length; i++) {
System.out.print(name[i] +'\n');
}
If You check
System.out.print(name.length);
you will get 3;
that mean your name length is 3
your loop is running from 0 to 3
which should be running either "0 to 2" or "1 to 3"
Answer
String[] name = {"tom", "dick", "harry"};
for(int i = 0; i<name.length; i++) {
System.out.print(name[i] +'\n');
}
Each item in an array is called an element, and each element is accessed by its numerical index. As shown in the preceding illustration, numbering begins with 0. The 9th element, for example, would therefore be accessed at index 8.
IndexOutOfBoundsException is thrown to indicate that an index of some sort (such as to an array, to a string, or to a vector) is out of range.
Any array X, can be accessed from [0 to (X.length - 1)]
I see all the answers here explaining how to work with arrays and how to avoid the index out of bounds exceptions. I personally avoid arrays at all costs. I use the Collections classes, which avoids all the silliness of having to deal with array indices entirely. The looping constructs work beautifully with collections supporting code that is both easier to write, understand and maintain.
If you use an array's length to control iteration of a for loop, always remember that the index of the first item in an array is 0. So the index of the last element in an array is one less than the array's length.
ArrayIndexOutOfBoundsException name itself explains that If you trying to access the value at the index which is out of the scope of Array size then such kind of exception occur.
In your case, You can just remove equal sign from your for loop.
for(int i = 0; i<name.length; i++)
The better option is to iterate an array:
for(String i : name )
System.out.println(i);
This error is occurs at runs loop overlimit times.Let's consider simple example like this,
class demo{
public static void main(String a[]){
int[] numberArray={4,8,2,3,89,5};
int i;
for(i=0;i<numberArray.length;i++){
System.out.print(numberArray[i+1]+" ");
}
}
At first, I have initialized an array as 'numberArray'. then , some array elements are printed using for loop. When loop is running 'i' time , print the (numberArray[i+1] element..(when i value is 1, numberArray[i+1] element is printed.)..Suppose that, when i=(numberArray.length-2), last element of array is printed..When 'i' value goes to (numberArray.length-1) , no value for printing..In that point , 'ArrayIndexOutOfBoundsException' is occur.I hope to you could get idea.thank you !
You can use Optional in functional style to avoid NullPointerException and ArrayIndexOutOfBoundsException :
String[] array = new String[]{"aaa", null, "ccc"};
for (int i = 0; i < 4; i++) {
String result = Optional.ofNullable(array.length > i ? array[i] : null)
.map(x -> x.toUpperCase()) //some operation here
.orElse("NO_DATA");
System.out.println(result);
}
Output:
AAA
NO_DATA
CCC
NO_DATA
In most of the programming language indexes is start from 0.So you must have to write i<names.length or i<=names.length-1 instead of i<=names.length.
You could not iterate or store more data than the length of your array. In this case you could do like this:
for (int i = 0; i <= name.length - 1; i++) {
// ....
}
Or this:
for (int i = 0; i < name.length; i++) {
// ...
}
Basically, my program will try to generate the list of all possible lowercase 5-letter words. Including all combinations that clearly are not real words like jshcc or mmdzq.
I do that by stacking up a massive amount of calls for a function, which does the word work.
But that's simply too much, and I get a stack overflow error.
How would someone control that?
Basically, convert from recursion to iteration. Typically that involves creating a Stack<T> as a "logical" stack, or something similar.
However, I'd have expected a method generating a list of all possible 5-letter words to only have a stack about 5 deep - one for each letter. Each stack level would be responsible for one level of letter - so the "top" of the stack would iterate through each possible last letter; the next stack frame down would iterate through every possible fourth letter, calling the method recursively to iterate through all possible last letters etc. Something like this (C# code, but hopefully you can understand it and apply it to VB):
const string Letters = "abcdefghijklmnopqrstuvwxyz";
public static List<string> GenerateValidWords(int length)
{
List<string> words = new List<string>();
GenerateValidWords(0, new char[length], words);
return words;
}
private static void GenerateValidWords(int depth, char[] current,
List<string> words)
{
foreach (char letter in letters)
{
current[depth] = letter;
if (depth == current.Length - 1)
{
string word = new string(current);
if (IsValid(word))
{
words.Add(word);
}
}
else
{
GenerateValidWords(depth + 1, current, words);
}
}
}
Now if you don't have any sort of filtering, that's going to generate 11,881,376 words - which at 24 bytes each (on x86) is about 285MB - plus all the space for the list etc. That shouldn't kill a suitably big machine, but it is quite a lot of memory. Are you sure you need all of these?
As a simple solution, I would use an iterative method with multiple loops in order to generate these words:
Dim words As New List(Of String)
Dim first As Integer = Asc("a")
Dim last As Integer = Asc("z")
For one As Integer = first To last
For two As Integer = first To last
For three As Integer = first To last
For four As Integer = first To last
For five As Integer = first To last
words.Add(Chr(one) & Chr(two) & Chr(three) & Chr(four) & Chr(five))
Next
Next
Next
Next
Next
MsgBox(words.Count)
I am trying to compare two long bytearrays in VB.NET and have run into a snag. Comparing two 50 megabyte files takes almost two minutes, so I'm clearly doing something wrong. I'm on an x64 machine with tons of memory so there are no issues there. Here is the code that I'm using at the moment and would like to change.
_Bytes and item.Bytes are the two different arrays to compare and are already the same length.
For Each B In item.Bytes
If B <> _Bytes(I) Then
Mismatch = True
Exit For
End If
I += 1
Next
I need to be able to compare as fast as possible files that are potentially hundreds of megabytes and even possibly a gigabyte or two. Any suggests or algorithms that would be able to do this faster?
Item.bytes is an object taken from the database/filesystem that is returned to compare, because its byte length matches the item that the user wants to add. By comparing the two arrays I can then determine if the user has added something new to the DB and if not then I can just map them to the other file and not waste hard disk drive space.
[Update]
I converted the arrays to local variables of Byte() and then did the same comparison, same code and it ran in like one second (I have to benchmark it still and compare it to others), but if you do the same thing with local variables and use a generic array it becomes massively slower. I’m not sure why, but it raises a lot more questions for me about the use of arrays.
What is the _Bytes(I) call doing? It's not loading the file each time, is it? Even with buffering, that would be bad news!
There will be plenty of ways to micro-optimise this in terms of looking at longs at a time, potentially using unsafe code etc - but I'd just concentrate on getting reasonable performance first. Clearly there's something very odd going on.
I suggest you extract the comparison code into a separate function which takes two byte arrays. That way you know you won't be doing anything odd. I'd also use a simple For loop rather than For Each in this case - it'll be simpler. Oh, and check whether the lengths are correct first :)
EDIT: Here's the code (untested, but simple enough) that I'd use. It's in C# for the minute - I'll convert it in a sec:
public static bool Equals(byte[] first, byte[] second)
{
if (first == second)
{
return true;
}
if (first == null || second == null)
{
return false;
}
if (first.Length != second.Length)
{
return false;
}
for (int i=0; i < first.Length; i++)
{
if (first[i] != second[i])
{
return false;
}
}
return true;
}
EDIT: And here's the VB:
Public Shared Function ArraysEqual(ByVal first As Byte(), _
ByVal second As Byte()) As Boolean
If (first Is second) Then
Return True
End If
If (first Is Nothing OrElse second Is Nothing) Then
Return False
End If
If (first.Length <> second.Length) Then
Return False
End If
For i as Integer = 0 To first.Length - 1
If (first(i) <> second(i)) Then
Return False
End If
Next i
Return True
End Function
The fastest way to compare two byte arrays of equal size is to use interop. Run the following code on a console application:
using System;
using System.Runtime.InteropServices;
using System.Security;
namespace CompareByteArray
{
class Program
{
static void Main(string[] args)
{
const int SIZE = 100000;
const int TEST_COUNT = 100;
byte[] arrayA = new byte[SIZE];
byte[] arrayB = new byte[SIZE];
for (int i = 0; i < SIZE; i++)
{
arrayA[i] = 0x22;
arrayB[i] = 0x22;
}
{
DateTime before = DateTime.Now;
for (int i = 0; i < TEST_COUNT; i++)
{
int result = MemCmp_Safe(arrayA, arrayB, (UIntPtr)SIZE);
if (result != 0) throw new Exception();
}
DateTime after = DateTime.Now;
Console.WriteLine("MemCmp_Safe: {0}", after - before);
}
{
DateTime before = DateTime.Now;
for (int i = 0; i < TEST_COUNT; i++)
{
int result = MemCmp_Unsafe(arrayA, arrayB, (UIntPtr)SIZE);
if (result != 0) throw new Exception();
}
DateTime after = DateTime.Now;
Console.WriteLine("MemCmp_Unsafe: {0}", after - before);
}
{
DateTime before = DateTime.Now;
for (int i = 0; i < TEST_COUNT; i++)
{
int result = MemCmp_Pure(arrayA, arrayB, SIZE);
if (result != 0) throw new Exception();
}
DateTime after = DateTime.Now;
Console.WriteLine("MemCmp_Pure: {0}", after - before);
}
return;
}
[DllImport("msvcrt.dll", CallingConvention = CallingConvention.Cdecl, EntryPoint="memcmp", ExactSpelling=true)]
[SuppressUnmanagedCodeSecurity]
static extern int memcmp_1(byte[] b1, byte[] b2, UIntPtr count);
[DllImport("msvcrt.dll", CallingConvention = CallingConvention.Cdecl, EntryPoint = "memcmp", ExactSpelling = true)]
[SuppressUnmanagedCodeSecurity]
static extern unsafe int memcmp_2(byte* b1, byte* b2, UIntPtr count);
public static int MemCmp_Safe(byte[] a, byte[] b, UIntPtr count)
{
return memcmp_1(a, b, count);
}
public unsafe static int MemCmp_Unsafe(byte[] a, byte[] b, UIntPtr count)
{
fixed(byte* p_a = a)
{
fixed (byte* p_b = b)
{
return memcmp_2(p_a, p_b, count);
}
}
}
public static int MemCmp_Pure(byte[] a, byte[] b, int count)
{
int result = 0;
for (int i = 0; i < count && result == 0; i += 1)
{
result = a[0] - b[0];
}
return result;
}
}
}
If you don't need to know the byte, use 64-bit ints that gives you 8 at once. Actually, you can figure out the wrong byte, once you've isolated it to a set of 8.
Use BinaryReader:
saveTime = binReader.ReadInt32()
Or for arrays of ints:
Dim count As Integer = binReader.Read(testArray, 0, 3)
Better approach... If you are just trying to see if the two are different then save some time by not having to go through the entire byte array and generate a hash of each byte array as strings and compare the strings. MD5 should work fine and is pretty efficient.
I see two things that might help:
First, rather than always accessing the second array as item.Bytes, use a local variable to point directly at the array. That is, before starting the loop, do something like this:
array2 = item.Bytes
That will save the overhead of dereferencing from the object each time you want a byte. That could be expensive in Visual Basic, especially if there's a Getter method on that property.
Also, use a "definite loop" instead of "for each". You already know the length of the arrays, so just code the loop using that value. This will avoid the overhead of treating the array as a collection. The loop would look something like this:
For i = 1 to max Step 1
If (array1(i) <> array2(i))
Exit For
EndIf
Next
Not strictly related to the comparison algorithm:
Are you sure your bottleneck is not related to the memory available and the time used to load the byte arrays? Loading two 2 GB byte arrays just to compare them could bring most machines to their knees. If the program design allows, try using streams to read smaller chunks instead.
I am coding VB.NET in VS2008.
I have a comma delimited string of numbers, i.e. 16,7,99,1456,1,3
I do this in VB:
Dim MyArr() As String = MyString.Split(",")
Will MyArr keep the items in the order they were in the string?
If I do this:
For Each S as String in MyString.Split(",")
'Do something with S
'Will my items be in the same order they were
'in the string?
Next
I tested it and it appears to keep the sort order but will it ~always~ keep the order?
If it does not maintain the order then what is a good way to split a string and keep order?
I'm asking because MSDN Array documentation says: "The Array is not guaranteed to be sorted." So I'm a bit unsure.
Yes, in your example the items will stay in the original order.
The MSDN documentation indicates that an Array is not necessarily sorted just because it's an Array, but once the items are in the Array, they won't be rearranged. The Split() operation will break it down based on the given token, preserving the order.
Yes, order will be maintained for these operations.
Yes, String.Split walks down the string, everything will stay in order. From .NET Reflector:
private string[] InternalSplitKeepEmptyEntries(int[] sepList, int[] lengthList, int numReplaces, int count)
{
int startIndex = 0;
int index = 0;
count--;
int num3 = (numReplaces < count) ? numReplaces : count;
string[] strArray = new string[num3 + 1];
for (int i = 0; (i < num3) && (startIndex < this.Length); i++)
{
strArray[index++] = this.Substring(startIndex, sepList[i] - startIndex);
startIndex = sepList[i] + ((lengthList == null) ? 1 : lengthList[i]);
}
if ((startIndex < this.Length) && (num3 >= 0))
{
strArray[index] = this.Substring(startIndex);
return strArray;
}
if (index == num3)
{
strArray[index] = Empty;
}
return strArray;
}
In .NET strings are immutable objects. Long story short, the string S and those returned by Split(",") live in different memory.