Question on Array

[Bhaswati Karmakar]

Hi All,
What gets printed on the standard output when the class below is compiled and executed by entering "java test lets see what happens". Select the one correct answer.
public class test {
public static void main(String args[]) {
System.out.println(args[0]+" "+args[args.length-1]);
}
}

A.The program will throw an ArrayIndexOutOfBounds exception.
B.The program will print "java test"
C.The program will print "java happens";
D.The program will print "test happens"
E.The program will print "lets happens"
The correct answer is E. I don't understand why ? I think it's C. Please can anybody explain, why it is E.
Thanks.

[Syed Hussain]

Hi,
The correct answer is E because in
java test lets see what happens
"java" is the command name for running the application
"test" is the class name u want to run
"lets see what happens" are the arguments for the application
args.length = 4
therfore
args[0] = lets
args[4-1] = happens

[Ravindra Mohan]

Hi Bhaswati,
Well to get the right answer you need to get the fundamentals
of C and C++ out in java .

public class test {
public static void main(String args[]) {
System.out.println(args[0]+" "+args[args.length-1]); // args[0]  in java holds the first array element to the main 
// unlike C and C++ where it is the name of the program
// so we have args[0] = "lets"
// args.length = 4 
// so args[4-1] = args[3] = "happens"
// please note that args[4] will throw an ArrayIndexOutOfBounds exception.
}
}

The correct answer is E. I don't understand why ? I think it's C. Please can anybody explain, why it is E.

Dont worry its is not your fault many C and C++ still get stumpted because of "these" and many other differences that
exists between Java vis-a-vis C and C++.
Hope your querry is answered.
Ravindra Mohan

[Bhaswati Karmakar]

Thanks to all for the explanation