Exception in thread “main” java.lang.ArrayIndexOutOfBoundsException: 0

Question

public class TestSample {
    public static void main(String[] args) {
        System.out.print(\"Hi, \");
        System.out.print(args[0]);
        System.out.prin         


        

Answer 1

1. ArrayIndexOutOfBoundsException: 0

It was thrown because args.length == 0 therefore args[0] is outside the arrays range of valid indices (learn more about arrays).

Add a check for args.length>0 to fix it.

public class TestSample {
    public static void main(String[] args) {
        System.out.print("Hi, ");
        System.out.print(args.length>0 ? args[0] : " I don't know who you are");
        System.out.println(". How are you?");
   }
}

2. Command line args as int

You will have to parse the arguments to int[] yourself as the command line arguments are passed only as a String[]. To do this, use Integer.parseInt() but you will need exception handling to make sure the parsing went OK (learn more about exceptions). Ashkan's answer shows you how to do this.

Answer 2

with regards to second part of your question:

from http://download.oracle.com/javase/tutorial/essential/environment/cmdLineArgs.html :

Parsing Numeric Command-Line Arguments

If an application needs to support a numeric command-line argument, it must convert a String argument that represents a number, such as "34", to a numeric value. Here is a code snippet that converts a command-line argument to an int:

int firstArg;
if (args.length > 0) {
    try {
        firstArg = Integer.parseInt(args[0]);
    } catch (NumberFormatException e) {
        System.err.println("Argument must be an integer");
        System.exit(1);
    }
}

Answer 3

1. The error is because no arguments were added when the program started.
2. Since the signature of the called main method (by JVM) is public static void main(String[] args) and not public static void main(int[] args) if you want ints, you'll need to parse them from the arguments.