• 0

[JAVA]Deleting Array Elements


Question

10 answers to this question

Recommended Posts

  • 0

If you're using an array with a set size like String[5], then you cannot delete an element and expect the array to have a length of 4. Setting an array element to null in this case just assigns that array element to have a null value. This could cause problems if you were iterating through it and didn't consider null values. If you have to use an array, then you would have to copy the entire array to some other array, minus the element you don't want.

If you're using a dynamic list like ArrayList or LinkedList, then you can call a method to delete an element that will cause the length of the array to decrease by one.

It sounds like you should use a dynamic list. The one that resembles an array the most is ArrayList.

  • 0

Ok let's say I don't know how to use an ArrayList and will figure out how to later, so I'm going to copy it into a temporary Array and then copy it back. If I set a larger array equal to a smaller array of the same type will it also set the size of the array?

  • 0

Setting a larger array to a smaller array only 'replaces' it:

int[] a = new int[10];
int[] b = new int[5];
b = a; // results in b and a referring to the same int[10].  int[5] is gone.

What you probably want is something like this:

// Comments omitted so you can figure out how it works.  =)
int[] a = new int[10];
int[] b = new int[9];

int i = 0;
int j = 0;
int elementToDelete = 5;
while (i < a.length && j < b.length) {
  if (i == elementToDelete) {
    i++;
  }
  else {
    b[j] = a[i];
    i++;
    j++;
  }
}
// At this point, b is equal to a without the 5th element

  • 0

To delete an item in an array, first you have to find the item. Here's a simple method.

	public static void main(String[] a)
	{
  int[] arr = new int[4];
  int i, size = 0, searchItem;
  
  arr[0] = 23;
  arr[1] = 12;
  arr[2] = 5767;
  size = 3;
  searchItem = 23;

  for(int d = 0; d < size; d++)
  	System.out.print(arr[d]+" ");
  System.out.println();
    
  for(i = 0; i < size; i++)
  	if(arr[i] == searchItem)
    break;
  for(int k = i; k < size; k++)
  	arr[k] = arr[k+1];
  size--;
  
  for(int d = 0; d < size; d++)
  	System.out.print(arr[d]+" ");
  System.out.println();
	}

  • 0

just reiterating, yes you need to create a new array when you delete an item.

basically you copy all the elements for the old array into the new one except the item that is being deleted.

sound like a lot of work to delete an element, but that is what you have to do. Or else you use a Vector or ArrayList, which do this for you.

  • 0
just reiterating, yes you need to create a new array when you delete an item.

basically you copy all the elements for the old array into the new one except the item that is being deleted.

sound like a lot of work to delete an element, but that is what you have to do. Or else you use a Vector or ArrayList, which do this for you.

586712019[/snapback]

Actually Vector and ArrayList will just move all the elements that have indices greater that the deleted element up one. If you actually want to clear the memory and not have "empty" elements, you would call ArrayList.trimToSize().

The reason is for performance: copying an array can take a bit of time depending on the amount of elements and where the deletion took place. By having a trimToSize() function the developer is given more control over memory/performance.

just my two cents.

  • 0

^

I believe "System.arraycopy" does something similar to what I was saying. maybe my wording was really incorrect "copy", and "new" are a bit misleading?

taken from Sun's ArrayList implementation

    /**
     * Removes the element at the specified position in this list.
     * Shifts any subsequent elements to the left (subtracts one from their
     * indices).
     *
     * @param index the index of the element to removed.
     * @return the element that was removed from the list.
     * @throws    IndexOutOfBoundsException if index out of range <tt>(index
     *     < 0 || index >= size())</tt>.
     */
    public Object remove(int index) {
	RangeCheck(index);

	modCount++;
	Object oldValue = elementData[index];

	int numMoved = size - index - 1;
	if (numMoved > 0)
     System.arraycopy(elementData, index+1, elementData, index, numMoved);
	elementData[--size] = null; // Let gc do its work

	return oldValue;
    }

    /**
     * Copies an array from the specified source array, beginning at the
     * specified position, to the specified position of the destination array.
     * A subsequence of array components are copied from the source
     * array referenced by <code>src</code> to the destination array
     * referenced by <code>dest</code>. The number of components copied is
     * equal to the <code>length</code> argument. The components at
     * positions <code>srcPos</code> through
     * <code>srcPos+length-1</code> in the source array are copied into
     * positions <code>destPos</code> through
     * <code>destPos+length-1</code>, respectively, of the destination
     * array.
     * <p>
     * If the <code>src</code> and <code>dest</code> arguments refer to the
     * same array object, then the copying is performed as if the
     * components at positions <code>srcPos</code> through
     * <code>srcPos+length-1</code> were first copied to a temporary
     * array with <code>length</code> components and then the contents of
     * the temporary array were copied into positions
     * <code>destPos</code> through <code>destPos+length-1</code> of the
     * destination array.
     * <p>
     * If <code>dest</code> is <code>null</code>, then a
     * <code>NullPointerException</code> is thrown.
     * <p>
     * If <code>src</code> is <code>null</code>, then a
     * <code>NullPointerException</code> is thrown and the destination
     * array is not modified.
     * <p>
     * Otherwise, if any of the following is true, an
     * <code>ArrayStoreException</code> is thrown and the destination is
     * not modified:
     * <ul>
     * <li>The <code>src</code> argument refers to an object that is not an
     *     array.
     * <li>The <code>dest</code> argument refers to an object that is not an
     *     array.
     * <li>The <code>src</code> argument and <code>dest</code> argument refer
     *     to arrays whose component types are different primitive types.
     * <li>The <code>src</code> argument refers to an array with a primitive
     *    component type and the <code>dest</code> argument refers to an array
     *     with a reference component type.
     * <li>The <code>src</code> argument refers to an array with a reference
     *    component type and the <code>dest</code> argument refers to an array
     *     with a primitive component type.
     * </ul>
     * <p>
     * Otherwise, if any of the following is true, an
     * <code>IndexOutOfBoundsException</code> is
     * thrown and the destination is not modified:
     * <ul>
     * <li>The <code>srcPos</code> argument is negative.
     * <li>The <code>destPos</code> argument is negative.
     * <li>The <code>length</code> argument is negative.
     * <li><code>srcPos+length</code> is greater than
     *     <code>src.length</code>, the length of the source array.
     * <li><code>destPos+length</code> is greater than
     *     <code>dest.length</code>, the length of the destination array.
     * </ul>
     * <p>
     * Otherwise, if any actual component of the source array from
     * position <code>srcPos</code> through
     * <code>srcPos+length-1</code> cannot be converted to the component
     * type of the destination array by assignment conversion, an
     * <code>ArrayStoreException</code> is thrown. In this case, let
     * <b><i>k</i></b> be the smallest nonnegative integer less than
     * length such that <code>src[srcPos+</code><i>k</i><code>]</code>
     * cannot be converted to the component type of the destination
     * array; when the exception is thrown, source array components from
     * positions <code>srcPos</code> through
     * <code>srcPos+</code><i>k</i><code>-1</code>
     * will already have been copied to destination array positions
     * <code>destPos</code> through
     * <code>destPos+</code><i>k</I><code>-1</code> and no other
     * positions of the destination array will have been modified.
     * (Because of the restrictions already itemized, this
     * paragraph effectively applies only to the situation where both
     * arrays have component types that are reference types.)
     *
     * @param      src      the source array.
     * @param      srcPos   starting position in the source array.
     * @param      dest     the destination array.
     * @param      destPos  starting position in the destination data.
     * @param      length   the number of array elements to be copied.
     * @exception  IndexOutOfBoundsException  if copying would cause
     *               access of data outside array bounds.
     * @exception  ArrayStoreException  if an element in the <code>src</code>
     *               array could not be stored into the <code>dest</code> array
     *               because of a type mismatch.
     * @exception  NullPointerException if either <code>src</code> or
     *               <code>dest</code> is <code>null</code>.
     */
    public static native void arraycopy(Object src,  int  srcPos,
                                        Object dest, int destPos,
                                        int length);

This topic is now closed to further replies.
  • Recently Browsing   0 members

    • No registered users viewing this page.
  • Posts

    • Yeah, when I saw that, I wanted to find the nearest nose. You can't find a good nose these days when you need one.
    • Anthropic launches Claude Fable 5, a state-of-the-art AI model that beats OpenAI's GPT-5.5 by Pradeep Viswanathan Back in April, Anthropic announced Claude Mythos Preview, a frontier model with state-of-the-art coding capabilities. Due to the cybersecurity implications that would occur due to the availability of such a powerful model, Anthropic made it available to only a select set of companies around the world. The company's plan was to prepare appropriate guardrails before releasing such a powerful model to everyone. Now, after nearly two months, Anthropic announced Claude Fable 5, its most capable AI model yet for general users. The company also announced Claude Mythos 5, the same underlying model as Fable 5, but with safeguards lifted, making it more suitable for selected cybersecurity and biology use cases. Claude Fable 5 sits a tier above its Opus models and it beats most other generally available models across areas including software engineering, knowledge work, vision, scientific research, and long-running autonomous tasks. To prevent model misuse, when Claude Fable 5 detects certain requests related to cybersecurity, biology, chemistry, or model distillation, the request will be routed to the Claude Opus 4.8 model. Anthropic claims that these safeguards trigger in less than 5% of sessions on average. However, for large organizations working on critical software, Claude Mythos 5 can be availed through Project Glasswing. Later, Anthropic has plans to expand access through a broader trusted access program. As you can notice in the benchmarks above, Fable 5 and Mythos 5 are state-of-the-art on most key AI benchmarks and they are well ahead of OpenAI's frontier model, GPT-5.5. For example, Fable 5 is the new state-of-the-art model for vision tasks. Also, Mythos 5 has the strongest cybersecurity capabilities of any model in the world. Claude Fable 5 and Claude Mythos 5 are priced at $10 per million input tokens and $50 per million output tokens, which is less than half the price of Claude Mythos Preview. Another big change is that Anthropic is making a change to the way they handle business customer data for both Fable 5 and Mythos 5 models. The company will now require 30-day retention for all traffic on both first- and third-party surfaces. Anthropic promises that it won't use the data to train Claude models, instead it will use it against complex and novel attacks. Claude Fable 5 is available today on the Claude API and consumption-based Enterprise plans. It is also included at no extra cost for Pro, Max, Team, and seat-based Enterprise customers from today through June 22. After that, users on those plans will need usage credits to continue using Fable 5, unless Anthropic extends the included access window based on capacity. Developers can access Fable 5 through the Claude API using the claude-fable-5 model name.
    • Dragon's Dogma 2: Dark Arisen expansion to bring snowy region, new updates also coming by Pulasthi Ariyasinghe Capcom had a surprise waiting for Dragon's Dogma fans today in the Nintendo Direct presentation. The company revealed an expansion for the second installment with a name that should be familiar to series veterans. Coming later this year, Dragon's Dogma 2: Dark Arisen is promising a massive new region to explore, new monsters, fresh skills to learn, and more. The studio says players will be heading to the Northern region of the world, named Norgan, to find new secrets about an undying "Fallen Dragon." There will be forgotten relics that the protagonist can find to unlock fresh weapons and skills the expansion is introducing. Players will also be able to find mysterious equipment from a previous Arisen as a part of the expansion, all part of 12 Lost Rites Dungeon Challenges they must complete to gain access. In Neowin's own review, I found Dragon's Dogma 2 to be an impressive RPG when it launched back in 2024, giving the title an 8.5/10 for its class variants, companion system, and immersive exploration. "Once a prosperous region of the kingdom of Vermund, it was abandoned many years ago for reasons unknown," says Capcom about the new region. "Long has it been since any soul traveled its paths. Blanketed in heavy snow, these frigid lands are home to savage hordes and creatures of unbelievable power. Those who are capable of vanquishing such fearsome foes, or those who possess a keen eye for exploration, will find themselves rewarded with powerful relics." Dragon’s Dogma 2: Dark Arisen expansion launches on October 9, 2026, with a $29.99 price tag. Ahead of the expansion release, Capcom is also planning to release two free updates to the base game. The first will land tomorrow, June 10, bringing more accessible fast travel with an Eternal Ferrystone and other quality-of-life adjustments. The second update will land sometime in August, aiming to improve frame rates, add more save slots, and bring even more community-requested adjustments. This expanded Dark Arisen edition is also launching on the Nintendo Switch 2 on the same day the content comes to PC, Xbox Series X|S, and PlayStation 5.
  • Recent Achievements

    • Week One Done
      rubentuben8 earned a badge
      Week One Done
    • Week One Done
      ARaclen earned a badge
      Week One Done
    • One Year In
      jojodbn earned a badge
      One Year In
    • One Month Later
      jojodbn earned a badge
      One Month Later
    • Week One Done
      jojodbn earned a badge
      Week One Done
  • Popular Contributors

    1. 1
      +primortal
      525
    2. 2
      PsYcHoKiLLa
      231
    3. 3
      +Edouard
      124
    4. 4
      ATLien_0
      87
    5. 5
      Steven P.
      83
  • Tell a friend

    Love Neowin? Tell a friend!