Simple Array Cause Exception

simple array cause exception

Your array is simply too large to fit on the stack. You don't have enough stack space for 1000 * 1000 elements.

You'll need to allocate your array on the heap. You can do this using the new keyword, but an easier way is to just use std::vector.

std::vector<std::vector<float> > floats(1000);
for (unsigned i = 0; i != floats.size(); ++i) floats[i].resize(1000);

This will give you a two-dimensional vector of floats, with 1000 elements per vector.

Also see: Segmentation fault on large array sizes

Array out of bounds exception for basic array

public DataSet(int maximumNumberOfValues) {
  double[] list = {maximumNumberOfValues}; 
}

Here, list s a local variable declaration (and only has one element), which has nothing to do with the member variable of the same name.

Also this expression {maximumNumberOfValues} actually means "create an array with a single element with value maximumNumberOfValues"

Assign a value to the member variable instead, where that value is a new array with the desired number of elements:

list = new double[maximumNumberOfValues];

What causes a java.lang.ArrayIndexOutOfBoundsException and how do I prevent it?

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.)

Accessing an array out of bounds gives no error, why?

Welcome to every C/C++ programmer's bestest friend: Undefined Behavior.

There is a lot that is not specified by the language standard, for a variety of reasons. This is one of them.

In general, whenever you encounter undefined behavior, anything might happen. The application may crash, it may freeze, it may eject your CD-ROM drive or make demons come out of your nose. It may format your harddrive or email all your porn to your grandmother.

It may even, if you are really unlucky, appear to work correctly.

The language simply says what should happen if you access the elements within the bounds of an array. It is left undefined what happens if you go out of bounds. It might seem to work today, on your compiler, but it is not legal C or C++, and there is no guarantee that it'll still work the next time you run the program. Or that it hasn't overwritten essential data even now, and you just haven't encountered the problems, that it is going to cause — yet.

As for why there is no bounds checking, there are a couple aspects to the answer:

• An array is a leftover from C. C arrays are about as primitive as you can get. Just a sequence of elements with contiguous addresses. There is no bounds checking because it is simply exposing raw memory. Implementing a robust bounds-checking mechanism would have been almost impossible in C.
• In C++, bounds-checking is possible on class types. But an array is still the plain old C-compatible one. It is not a class. Further, C++ is also built on another rule which makes bounds-checking non-ideal. The C++ guiding principle is "you don't pay for what you don't use". If your code is correct, you don't need bounds-checking, and you shouldn't be forced to pay for the overhead of runtime bounds-checking.
• So C++ offers the std::vector class template, which allows both. operator[] is designed to be efficient. The language standard does not require that it performs bounds checking (although it does not forbid it either). A vector also has the at() member function which is guaranteed to perform bounds-checking. So in C++, you get the best of both worlds if you use a vector. You get array-like performance without bounds-checking, and you get the ability to use bounds-checked access when you want it.

array out of bound exception in java

Accessing an array with an index that is greater than or equal to the
length of the array causes an 'ArrayIndexOutOfBoundsException'.

In your code, you're trying to access the 11th (and 12th, etc...) index a 10 length arrays. (A is 11, B is 10).

Considering you're apparently trying to output numerical value for a given "in letters" number, the best solution seems to be looping 21 times on a single array containing the 21 elements.

Last thing, you should always use array.length in a loop, using a litteral value like "20" means you'll have to edit it as soon as you edit your array.

To sum up, it would look like this :

  public static void main(String[] args){
        String n;
        String a[]={"zero","one","two","three","four","five","six","seven","eight","nine","ten", "eleven","twelve","thirteen","fourteen","fifteen","sixteen","seventeen","eighteen","nineteen","twenty"}; 
        System.out.println("Enter number in words");
        Scanner sc=new Scanner(System.in);
        n=sc.next();
        for(int i=0;i<a.length;i++){
            if(n.equals(a[i])){
                System.out.println(i);
                break;
            }
        }
    }

Java- Array Index Out of Bounds Exception

I'd be willing to bet you're not passing the name of your data file to your program.

if args.length == 0 then args[0] throws ArrayIndexOutOfBoundsException

That's about the only way you can get that error on that line.

The Arrays.fill() method causes an exception

Arrays.fill expects a single-dimensional array, you're passing in a jagged array.

Instead, do this:

for(int x=0;x<board.length;x++)
    for(int y=0;y<board[x].length;y++)
        board[x][y] = ' ';

or this:

for(int x=0;x<board.length;x++)
    Arrays.fill( board[x], ' ' );

casting Arrays.asList causing exception: java.util.Arrays$ArrayList cannot be cast to java.util.ArrayList

For me (using Java 1.6.0_26), the first snippet gives the same exception as the second one. The reason is that the Arrays.asList(..) method does only return a List, not necessarily an ArrayList. Because you don't really know what kind (or implementation of) of List that method returns, your cast to ArrayList<String> is not safe. The result is that it may or may not work as expected. From a coding style perspective, a good fix for this would be to change your stuff declaration to:

List<List<String>> stuff = new ArrayList<List<String>>();

which will allow to add whatever comes out of the Arrays.asList(..) method.

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