args[] problem?

[sonir shah]

What will happen when you compile and run the following program using the command line:
java TestClass 1 2

public class TestClass
{
public static void main(String[] args) 
{
int i = Integer.parseInt(args[1]);
System.out.println(args[i]);
}
}

Answer : It will throw ArrayIndexOutOfBoundsException
Since at the command line java TestClass 1 2 is passed.
and the args[1]; which means 2 elements i.e 0 and 1
When" java TestClass 1 2 "is passed , here "1 2" are like other objects or strings or...etc
so when in the main args[i] is called i.e args[1] is called ,i.e second index, "2" should be printed

Am I right?
Please explain.
Sonir

[Roy Ben Ami]

you are printing args[i]. not i !
first line you get the i as 2 like you said, no prioblem.
second line look at it carefully: it says:
system.out.prinltn(args[i]); !
not i but args[i]! it tries to do args[2] and thats out of bounds!
tricky one!

[Rob Ross]

public class TestClass
{
    public static void main(String[] args)
    {
        int i = Integer.parseInt(args[1]);
        System.out.println(args[i]);
    }
}

if you run this program with
java TestClass 1 2
then there are 2 arguments passed to the program, and args has 2 elements
args[0] = 1
args[1] = 2
Now, the first line creates an int value by parsing the string in args[1] (remember that each element in args is a String)
When it parses the string, it takes the String value "2" and returns an int value = 2.This 2 gets assigned to the variable named 'i'.
Then this int value is used as an index into the args array....
System.out.println(args[i]);
becomes
System.out.println(args[2]);
But args only has 2 elements, so args[2] is not in the array, it is outside the range of elements it contains, and this access throws an ArrayIndexOutOfBoundsException.
Rob
[ January 13, 2002: Message edited by: Rob Ross ]