I have some code and when it executes, it throws a IndexOutOfRangeException, saying,
Index was outside the bounds of the array.
What does this mean, and what can I do about it?
Depending on classes used it can also be ArgumentOutOfRangeException
An exception of type 'System.ArgumentOutOfRangeException' occurred in mscorlib.dll but was not handled in user code Additional information: Index was out of range. Must be non-negative and less than the size of the collection.
What Is It?
This exception means that you're trying to access a collection item by index, using an invalid index. An index is invalid when it's lower than the collection's lower bound or greater than or equal to the number of elements it contains.
When It Is Thrown
Given an array declared as:
byte[] array = new byte[4];
You can access this array from 0 to 3, values outside this range will cause IndexOutOfRangeException to be thrown. Remember this when you create and access an array.
Array Length
In C#, usually, arrays are 0-based. It means that first element has index 0 and last element has index Length - 1 (where Length is total number of items in the array) so this code doesn't work:
array[array.Length] = 0;
Moreover please note that if you have a multidimensional array then you can't use Array.Length for both dimension, you have to use Array.GetLength():
int[,] data = new int[10, 5];
for (int i=0; i < data.GetLength(0); ++i) {
for (int j=0; j < data.GetLength(1); ++j) {
data[i, j] = 1;
}
}
Upper Bound Is Not Inclusive
In the following example we create a raw bidimensional array of Color. Each item represents a pixel, indices are from (0, 0) to (imageWidth - 1, imageHeight - 1).
Color[,] pixels = new Color[imageWidth, imageHeight];
for (int x = 0; x <= imageWidth; ++x) {
for (int y = 0; y <= imageHeight; ++y) {
pixels[x, y] = backgroundColor;
}
}
This code will then fail because array is 0-based and last (bottom-right) pixel in the image is pixels[imageWidth - 1, imageHeight - 1]:
pixels[imageWidth, imageHeight] = Color.Black;
In another scenario you may get ArgumentOutOfRangeException for this code (for example if you're using GetPixel method on a Bitmap class).
Arrays Do Not Grow
An array is fast. Very fast in linear search compared to every other collection. It is because items are contiguous in memory so memory address can be calculated (and increment is just an addition). No need to follow a node list, simple math! You pay this with a limitation: they can't grow, if you need more elements you need to reallocate that array (this may take a relatively long time if old items must be copied to a new block). You resize them with Array.Resize<T>(), this example adds a new entry to an existing array:
Array.Resize(ref array, array.Length + 1);
Don't forget that valid indices are from 0 to Length - 1. If you simply try to assign an item at Length you'll get IndexOutOfRangeException (this behavior may confuse you if you think they may increase with a syntax similar to Insert method of other collections).
Special Arrays With Custom Lower Bound
First item in arrays has always index 0. This is not always true because you can create an array with a custom lower bound:
var array = Array.CreateInstance(typeof(byte), new int[] { 4 }, new int[] { 1 });
In that example, array indices are valid from 1 to 4. Of course, upper bound cannot be changed.
Wrong Arguments
If you access an array using unvalidated arguments (from user input or from function user) you may get this error:
private static string[] RomanNumbers =
new string[] { "I", "II", "III", "IV", "V" };
public static string Romanize(int number)
{
return RomanNumbers[number];
}
Unexpected Results
This exception may be thrown for another reason too: by convention, many search functions will return -1 (nullables has been introduced with .NET 2.0 and anyway it's also a well-known convention in use from many years) if they didn't find anything. Let's imagine you have an array of objects comparable with a string. You may think to write this code:
// Items comparable with a string
Console.WriteLine("First item equals to 'Debug' is '{0}'.",
myArray[Array.IndexOf(myArray, "Debug")]);
// Arbitrary objects
Console.WriteLine("First item equals to 'Debug' is '{0}'.",
myArray[Array.FindIndex(myArray, x => x.Type == "Debug")]);
This will fail if no items in myArray will satisfy search condition because Array.IndexOf() will return -1 and then array access will throw.
Next example is a naive example to calculate occurrences of a given set of numbers (knowing maximum number and returning an array where item at index 0 represents number 0, items at index 1 represents number 1 and so on):
static int[] CountOccurences(int maximum, IEnumerable<int> numbers) {
int[] result = new int[maximum + 1]; // Includes 0
foreach (int number in numbers)
++result[number];
return result;
}
Of course, it's a pretty terrible implementation but what I want to show is that it'll fail for negative numbers and numbers above maximum.
How it applies to List<T>?
Same cases as array - range of valid indexes - 0 (List's indexes always start with 0) to list.Count - accessing elements outside of this range will cause the exception.
Note that List<T> throws ArgumentOutOfRangeException for the same cases where arrays use IndexOutOfRangeException.
Unlike arrays, List<T> starts empty - so trying to access items of just created list lead to this exception.
var list = new List<int>();
Common case is to populate list with indexing (similar to Dictionary<int, T>) will cause exception:
list[0] = 42; // exception
list.Add(42); // correct
IDataReader and Columns
Imagine you're trying to read data from a database with this code:
using (var connection = CreateConnection()) {
using (var command = connection.CreateCommand()) {
command.CommandText = "SELECT MyColumn1, MyColumn2 FROM MyTable";
using (var reader = command.ExecuteReader()) {
while (reader.Read()) {
ProcessData(reader.GetString(2)); // Throws!
}
}
}
}
GetString() will throw IndexOutOfRangeException because you're dataset has only two columns but you're trying to get a value from 3rd one (indices are always 0-based).
Please note that this behavior is shared with most IDataReader implementations (SqlDataReader, OleDbDataReader and so on).
You can get the same exception also if you use the IDataReader overload of the indexer operator that takes a column name and pass an invalid column name.
Suppose for example that you have retrieved a column named Column1 but then you try to retrieve the value of that field with
var data = dr["Colum1"]; // Missing the n in Column1.
This happens because the indexer operator is implemented trying to retrieve the index of a Colum1 field that doesn't exist. The GetOrdinal method will throw this exception when its internal helper code returns a -1 as the index of "Colum1".
Others
There is another (documented) case when this exception is thrown: if, in DataView, data column name being supplied to the DataViewSort property is not valid.
How to Avoid
In this example, let me assume, for simplicity, that arrays are always monodimensional and 0-based. If you want to be strict (or you're developing a library), you may need to replace 0 with GetLowerBound(0) and .Length with GetUpperBound(0) (of course if you have parameters of type System.Array, it doesn't apply for T[]). Please note that in this case, upper bound is inclusive then this code:
for (int i=0; i < array.Length; ++i) { }
Should be rewritten like this:
for (int i=array.GetLowerBound(0); i <= array.GetUpperBound(0); ++i) { }
Please note that this is not allowed (it'll throw InvalidCastException), that's why if your parameters are T[] you're safe about custom lower bound arrays:
void foo<T>(T[] array) { }
void test() {
// This will throw InvalidCastException, cannot convert Int32[] to Int32[*]
foo((int)Array.CreateInstance(typeof(int), new int[] { 1 }, new int[] { 1 }));
}
Validate Parameters
If index comes from a parameter you should always validate them (throwing appropriate ArgumentException or ArgumentOutOfRangeException). In the next example, wrong parameters may cause IndexOutOfRangeException, users of this function may expect this because they're passing an array but it's not always so obvious. I'd suggest to always validate parameters for public functions:
static void SetRange<T>(T[] array, int from, int length, Func<i, T> function)
{
if (from < 0 || from>= array.Length)
throw new ArgumentOutOfRangeException("from");
if (length < 0)
throw new ArgumentOutOfRangeException("length");
if (from + length > array.Length)
throw new ArgumentException("...");
for (int i=from; i < from + length; ++i)
array[i] = function(i);
}
If function is private you may simply replace if logic with Debug.Assert():
Debug.Assert(from >= 0 && from < array.Length);
Check Object State
Array index may not come directly from a parameter. It may be part of object state. In general is always a good practice to validate object state (by itself and with function parameters, if needed). You can use Debug.Assert(), throw a proper exception (more descriptive about the problem) or handle that like in this example:
class Table {
public int SelectedIndex { get; set; }
public Row[] Rows { get; set; }
public Row SelectedRow {
get {
if (Rows == null)
throw new InvalidOperationException("...");
// No or wrong selection, here we just return null for
// this case (it may be the reason we use this property
// instead of direct access)
if (SelectedIndex < 0 || SelectedIndex >= Rows.Length)
return null;
return Rows[SelectedIndex];
}
}
Validate Return Values
In one of previous examples we directly used Array.IndexOf() return value. If we know it may fail then it's better to handle that case:
int index = myArray[Array.IndexOf(myArray, "Debug");
if (index != -1) { } else { }
How to Debug
In my opinion, most of the questions, here on SO, about this error can be simply avoided. The time you spend to write a proper question (with a small working example and a small explanation) could easily much more than the time you'll need to debug your code. First of all, read this Eric Lippert's blog post about debugging of small programs, I won't repeat his words here but it's absolutely a must read.
You have source code, you have exception message with a stack trace. Go there, pick right line number and you'll see:
array[index] = newValue;
You found your error, check how index increases. Is it right? Check how array is allocated, is coherent with how index increases? Is it right according to your specifications? If you answer yes to all these questions, then you'll find good help here on StackOverflow but please first check for that by yourself. You'll save your own time!
A good start point is to always use assertions and to validate inputs. You may even want to use code contracts. When something went wrong and you can't figure out what happens with a quick look at your code then you have to resort to an old friend: debugger. Just run your application in debug inside Visual Studio (or your favorite IDE), you'll see exactly which line throws this exception, which array is involved and which index you're trying to use. Really, 99% of the times you'll solve it by yourself in a few minutes.
If this happens in production then you'd better to add assertions in incriminated code, probably we won't see in your code what you can't see by yourself (but you can always bet).
The VB.NET side of the story
Everything that we have said in the C# answer is valid for VB.NET with the obvious syntax differences but there is an important point to consider when you deal with VB.NET arrays.
In VB.NET, arrays are declared setting the maximum valid index value for the array. It is not the count of the elements that we want to store in the array.
' declares an array with space for 5 integer
' 4 is the maximum valid index starting from 0 to 4
Dim myArray(4) as Integer
So this loop will fill the array with 5 integers without causing any IndexOutOfRangeException
For i As Integer = 0 To 4
myArray(i) = i
Next
The VB.NET rule
This exception means that you're trying to access a collection item by index, using an invalid index. An index is invalid when it's lower than the collection's lower bound or greater than equal to the number of elements it contains. the maximum allowed index defined in the array declaration
Simple explanation about what a Index out of bound exception is:
Just think one train is there its compartments are D1,D2,D3.
One passenger came to enter the train and he have the ticket for D4.
now what will happen. the passenger want to enter a compartment that does not exist so obviously problem will arise.
Same scenario: whenever we try to access an array list, etc. we can only access the existing indexes in the array. array[0] and array[1] are existing. If we try to access array[3], it's not there actually, so an index out of bound exception will arise.
To easily understand the problem, imagine we wrote this code:
static void Main(string[] args)
{
string[] test = new string[3];
test[0]= "hello1";
test[1]= "hello2";
test[2]= "hello3";
for (int i = 0; i <= 3; i++)
{
Console.WriteLine(test[i].ToString());
}
}
Result will be:
hello1
hello2
hello3
Unhandled Exception: System.IndexOutOfRangeException: Index was outside the bounds of the array.
Size of array is 3 (indices 0, 1 and 2), but the for-loop loops 4 times (0, 1, 2 and 3). So when it tries to access outside the bounds with (3) it throws the exception.
A side from the very long complete accepted answer there is an important point to make about IndexOutOfRangeException compared with many other exception types, and that is:
Often there is complex program state that maybe difficult to have control over at a particular point in code e.g a DB connection goes down so data for an input cannot be retrieved etc... This kind of issue often results in an Exception of some kind that has to bubble up to a higher level because where it occurs has no way of dealing with it at that point.
IndexOutOfRangeException is generally different in that it in most cases it is pretty trivial to check for at the point where the exception is being raised. Generally this kind of exception get thrown by some code that could very easily deal with the issue at the place it is occurring - just by checking the actual length of the array. You don't want to 'fix' this by handling this exception higher up - but instead by ensuring its not thrown in the first instance - which in most cases is easy to do by checking the array length.
Another way of putting this is that other exceptions can arise due to genuine lack of control over input or program state BUT IndexOutOfRangeException more often than not is simply just pilot (programmer) error.
These two exceptions are common in various programming languages and as others said it's when you access an element with an index greater than the size of the array. For example:
var array = [1,2,3];
/* var lastElement = array[3] this will throw an exception, because indices
start from zero, length of the array is 3, but its last index is 2. */
The main reason behind this is compilers usually don't check this stuff, hence they will only express themselves at runtime.
Similar to this:
Why don't modern compilers catch attempts to make out-of-bounds access to arrays?
I need help! am creating an app in flutter where I need 8 unique ranodm numbers. So how can I create a list of 8 random numbers which are all unique in Dart? If I go through a for loop using the random class and then append it to a list there is still the chance of repeating numbers. Anybody can help?
You can check if the random number is already existing in the list before adding it. I created an example, I hope it helps you.
In this example, I am trying to generate 8 completely unique random numbers between 0 and 8. You can change these numbers as you want it will not affect the code (But make sure the random range is equal or bigger than the list length you want, or the code will break).
import 'dart:math';
void main() {
List<int> randomNumbers = [];
//While the length of the list is smaller than 8 (the length I want),
//keep trying to add unique random numbers.
while(randomNumbers.length < 8){
//Get a random number between 0 and 8 (Edit this numbers as you want)
//from the method we created below at the bottom.
int randomNum = randomInt(0,8);
//Check if the random number is already existing in the list.
bool condition = randomNumbers.where((number) => number == randomNum).toList().isEmpty;
//If the number is not existing, add the number to the list.
//Else do nothing (repeat the while loop to find another number).
if(condition){
randomNumbers.add(randomNum);
}
}
print(randomNumbers);
}
// A method to get a random integer. Max and min are inclusive.
int randomInt(int min, int max) {
return Random().nextInt(max - min + 1) + min;
}
The output will be completely unique 8 numbers such as : [0, 1, 5, 2, 4, 6, 3, 8].
Given that a number can contain only digits from 1 to 8 (with no repetition), and is of length 8, how can we hash such numbers without using a hashSet?
We can't just directly use the value of the number of the hashing value, as the stack size of the program is limited. (By this, I mean that we can't directly make the index of an array, represent our number).
Therefore, this 8 digit number needs to be mapped to, at maximum, a 5 digit number.
I saw this answer. The hash function returns a 8-digit number, for a input that is an 8-digit number.
So, what can I do here?
There's a few things you can do. You could subtract 1 from each digit and parse it as an octal number, which will map one-to-one every number from your domain to the range [0,16777216) with no gaps. The resulting number can be used as an index into a very large array. An example of this could work as below:
function hash(num) {
return parseInt(num
.toString()
.split('')
.map(x => x - 1), 8);
}
const set = new Array(8**8);
set[hash(12345678)] = true;
// 12345678 is in the set
Or if you wanna conserve some space and grow the data structure as you add elements. You can use a tree structure with 8 branches at every node and a maximum depth of 8. I'll leave that up to you to figure out if you think it's worth the trouble.
Edit:
After seeing the updated question, I began thinking about how you could probably map the number to its position in a lexicographically sorted list of the permutations of the digits 1-8. That would be optimal because it gives you the theoretical 5-digit hash you want (under 40320). I had some trouble formulating the algorithm to do this on my own, so I did some digging. I found this example implementation that does just what you're looking for. I've taken inspiration from this to implement the algorithm in JavaScript for you.
function hash(num) {
const digits = num
.toString()
.split('')
.map(x => x - 1);
const len = digits.length;
const seen = new Array(len);
let rank = 0;
for(let i = 0; i < len; i++) {
seen[digits[i]] = true;
rank += numsBelowUnseen(digits[i], seen) * fact(len - i - 1);
}
return rank;
}
// count unseen digits less than n
function numsBelowUnseen(n, seen) {
let count = 0;
for(let i = 0; i < n; i++) {
if(!seen[i]) count++;
}
return count;
}
// factorial fuction
function fact(x) {
return x <= 0 ? 1 : x * fact(x - 1);
}
kamoroso94 gave me the idea of representing the number in octal. The number remains unique if we remove the first digit from it. So, we can make an array of length 8^7=2097152, and thus use the 7-digit octal version as index.
If this array size is bigger than the stack, then we can use only 6 digits of the input, convert them to their octal values. So, 8^6=262144, that is pretty small. We can make a 2D array of length 8^6. So, total space used will be in the order of 2*(8^6). The first index of the second dimension represents that the number starts from the smaller number, and the second index represents that the number starts from the bigger number.
class Distance
{
public:
int a;
};
int main()
{
Distance d1; //declaring object
char *p=(char *)&d1;
*p=1;
printf("\n %d ",d1.a);
return 0;
}
This is my code.
When I am passing the value of 'a' to be like 256,512 , I am getting 257,513 respectively but for values like 1000 i get 769 and for values like 16,128,100 I am getting 1.
First I thought it might be related to powers of 2 being incremented by 1 due to changes in their binary representation. But adding 1 to binary representation of 1000 won't give me 769.
Please help me to understand this code.
*p = 1 sets the last byte(char) to 000000001
As you're type casting int to char,
binary for (int)1000 is (binary)0000001111101000
you're assigning (int)1 for last 8 bits i,e (binary)0000001100000001 which is 769.
Using 256512 worked because last 8 bit that you change are all zeros i.e (int)256512 is (binary)111110101000000000 so making last bit as 1 gives you (binary)111110101000000001 which is (int)256513
And I think(not sure) you get 1 for 16,128,100 because this integer is well out of int range and thus not assigned and a is set to 0 as class object is created. and thus setting last bit to 1 makes a = 1
I want to count the number of set bits in a uint in Specman:
var x: uint;
gen x;
var x_set_bits: uint;
x_set_bits = ?;
What's the best way to do this?
One way I've seen is:
x_set_bits = pack(NULL, x).count(it == 1);
pack(NULL, x) converts x to a list of bits.
count acts on the list and counts all the elements for which the condition holds. In this case the condition is that the element equals 1, which comes out to the number of set bits.
I don't know Specman, but another way I've seen this done looks a bit cheesy, but tends to be efficient: Keep a 256-element array; each element of the array consists of the number of bits corresponding to that value. For example (pseudocode):
bit_count = [0, 1, 1, 2, 1, ...]
Thus, bit_count2 == 1, because the value 2, in binary, has a single "1" bit. Simiarly, bit_count[255] == 8.
Then, break the uint into bytes, use the byte values to index into the bit_count array, and add the results. Pseudocode:
total = 0
for byte in list_of_bytes
total = total + bit_count[byte]
EDIT
This issue shows up in the book Beautiful Code, in the chapter by Henry S. Warren. Also, Matt Howells shows a C-language implementation that efficiently calculates a bit count. See this answer.