How does size( ) works in java's ArrayList class?

问题

So, in order to get the most efficient code, I really wanted to know how does the size() method in Java ArrayList work... Does it count every element, going through all the positions, like a simple list? or does it just gets the size by the last index registered?

Thanks in advance!

回答1:

Never hurts to look in the source code:

public int size() {
    return size;
}

It returns an instance variable - pretty damn fast.

回答2:

As of the latest Java7, it does a little more than read a member field value:

public int size() {
    checkForComodification();
    return this.size;
}

private void checkForComodification() {
    if (ArrayList.this.modCount != this.modCount)
        throw new ConcurrentModificationException();
}

回答3:

In ArrayList there's an int attribute for storing the current size (say, called size). Clearly, calculating the size of an array list should be an O(1) operation, for efficiency. Even in a data structure such as a LinkedList (a double-linked list) the size is kept updated in an attribute, to avoid having to calculate it each time it's needed. To see it more clearly, take a look at the source code in OpenJDK, there you'll find this:

 /**
  * The size of the ArrayList (the number of elements it contains).
  *
  * @serial
  */
  private int size;

 /**
  * Returns the number of elements in this list.
  *
  * @return the number of elements in this list
  */
  public int size() {
      return size;
  }

回答4:

According to the source code for ArrayList, the size() method returns a private variable named size, which is just a counter that is incremented on each add.

回答5:

It reads a field variable. Java 1.6's ArrayList.size():

/**
 * Returns the number of elements in this list.
 *
 * @return the number of elements in this list
 */
public int size() {
    return size;
}

来源：https://stackoverflow.com/questions/15512508/how-does-size-works-in-javas-arraylist-class

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java

arraylist

performance