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Command Line Invocation

Chadd Franck
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Joined: Nov 05, 2008
Posts: 50
This is from Sun Certified Programmer for Java 6 Study Guide, Page 320,


And the command-line invocation:
java Fork live2

What is the result?

A. test case
B. production live2
C. test case live2
D. Compilation fails
E. An exception is thrown at runtime

Answer:
E is correct. Because the short circuit (||) is not used, both operands are evaluated. Since args[1] is past the args array bounds, an ArrayIndexOutOfBoundsException is thrown. A, B, C, and D are incorrect based on the above. (Objective 7.6)

Why isn't args[1] = live2? therefore not outofbounds?
Henry Wong
author
Sheriff

Joined: Sep 28, 2004
Posts: 18505
    
  40

Why isn't args[1] = live2? therefore not outofbounds?


Arrays are zero based -- so the first element is args[0]. Hence, args[0] is live2, and args[1] is out of bounds.

Henry


Books: Java Threads, 3rd Edition, Jini in a Nutshell, and Java Gems (contributor)
Chadd Franck
Ranch Hand

Joined: Nov 05, 2008
Posts: 50
Hehe yea I see it, I was thinking that Fork was an arg for some reason, I can see that it is the name of the class (G)...
 
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