Even though this is not a direct answer to the question, it is an addition to the .length
vs .size()
argument. I was researching something related to this question so when I came across it I noticed that the definition(s) provided here
The public final field length, which contains the number of components of the array.
is not "exactly" correct.
The field length contains the number of available places to put a component, not the number of components present in the array. So it represents the total available memory allocated to that array, not how much of that memory is filled.

Example:
static class StuffClass {
int stuff;
StuffClass(int stuff) {
this.stuff = stuff;
}
}
public static void main(String[] args) {
int[] test = new int[5];
test[0] = 2;
test[1] = 33;
System.out.println("Length of int[]:\t" + test.length);
String[] test2 = new String[5];
test2[0] = "2";
test2[1] = "33";
System.out.println("Length of String[]:\t" + test2.length);
StuffClass[] test3 = new StuffClass[5];
test3[0] = new StuffClass(2);
test3[1] = new StuffClass(33);
System.out.println("Length of StuffClass[]:\t" + test3.length);
}
Output:
Length of int[]: 5
Length of String[]: 5
Length of StuffClass[]: 5
However, the .size()
property of the ArrayList
does give the number of elements in the list:
ArrayList<Integer> intsList = new ArrayList<Integer>();
System.out.println("List size:\t" + intsList.size());
intsList.add(2);
System.out.println("List size:\t" + intsList.size());
intsList.add(33);
System.out.println("List size:\t" + intsList.size());
Output:
List size: 0
List size: 1
List size: 2