Sun Certified Programmer for the Java 2 Platform, Standard Edition 6 question

Question

I am studying for my first Java Cert! Hurah! Sorry, don't mean to sound so cheezy, but I can't help it :) Anywho, trying to understand the below question and answer. I don't get it. . . I don't really use the command line that much (mainly NetBeans and I am getting acquainted with Eclipse). Any pointers are greatly appreciated.

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OBJECTIVE: 7.2: Given an example of a class and a command-line, determine the expected runtime behavior. 10)

Given:

1. class x {

2.   public static void main(String [] args) {
3.     String p = System.getProperty("x");
4.     if(p.equals(args[1]))
5.       System.out.println("found");
6.   }
7. }

Which command-line invocation will produce the output found?

a) java -Dx=y x y z
b) java -Px=y x y z
c) java -Dx=y x x y z (*)
d) java -Px=y x x y z
e) java x x y z -Dx=y
f) java x x y z -Px=y

//So the answer is C

REFERENCE:
API for java command
Option C is correct. -D sets a property and args[1] is the second argument (whose value is y)

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Uhm. . . so how come args[1] is the second argument? i thought "p" would be equal to the String {y x x y z}?

Answer 1

Uhm. . . so how come args[1] is the second argument? i thought "p" would be equal to the String {y x x y z}?

The question is tricky. In c), we are starting the class x with arguments x y z. And the -D option will cause the x property to be set to y.

So when the class x starts:

• args will be `String[]{"x", "y", "z"}, and
• p will be "y".

We then compare p with args[1] ... and the two strings are equal.

(Of course, in the real world, no person who named their class x would survive long enough to receive their first pay check :-) )