You cannot make an empty array and then let it grow dynamically whenever the user enters a number in the command line. You should read the numbers and put them in an ArrayList instead. An ArrayList does not require a initial size and does grow dynamically. Something like this:

public void main(String[] args) {
    ArrayList<Integer> numbers = new ArrayList<Integer>();
    BufferedReader console = 
            new BufferedReader(new InputStreamReader(System.in));
    while(true) {
        String word = console.readLine();
        if (word.equalsIgnoreCase("end") {
            break;
        } else {
            numbers.add(Integer.parseInt(word);
        }
    }

Ofcourse you won't use while(true)and you won't put this in main, but it's just for the sake of the example

If you are going to use a collection that you don't know the size of in advance, there are better options than arrays.

Use a List<string> instead - it will allow you to add as many items as you need and if you need to return an array, call ToArray() on the variable.

var listOfStrings = new List<string>();

// do stuff...

string[] arrayOfStrings = listOfStrings.ToArray();

If you must create an empty array you can do this:

string[] emptyStringArray = new string[0]; 

As of .NET 4.6, the following would be recommended for an empty array:

String[] a = Array.Empty<string>();

With C# 12/.NET8, there's an even more succinct way (using collection expressions):

String[] a = [];

1) To create new array which, you cannot iterate over, you can use array constructor:

Array(100) or new Array(100)

2) You can create new array, which can be iterated over like below:

a) All JavaScript versions

• Array.apply: Array.apply(null, Array(100))

b) From ES6 JavaScript version

• Destructuring operator: [...Array(100)]
• Array.prototype.fill Array(100).fill(undefined)
• Array.from Array.from({ length: 100 })

You can map over these arrays like below.

• Array(4).fill(null).map((u, i) => i) [0, 1, 2, 3]

• [...Array(4)].map((u, i) => i) [0, 1, 2, 3]

• Array.apply(null, Array(4)).map((u, i) => i) [0, 1, 2, 3]

• Array.from({ length: 4 }).map((u, i) => i) [0, 1, 2, 3]