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q on primitives

 
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class JMM101 {
public static void main(String[] args) {
int i = 0;
while (i++ < args.length) {
System.out.print(args[i]);
}}}
java JMM101 A B C D E F
What is the result of attempting to compile and run the program using the specified command line?

a. Prints: JMM101ABCDEF
b. Prints: ABCDEF
c. Compile-time error
d. Run-time error
e. None of the above


I answered b, but the correct ans is d.
In the while loop, we take the value of i, compare the same and then increment right?
wht wud be the effect if it was while(++i,args.length)?
 
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BCDEF
 
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The result will be BCDE.

Thanks.
 
Mike Gershman
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Edoil wrote:

The result will be BCDE.

Thanks.



I executed the code myself.

With i++ you get an ArrayIndexOutOfBoundsException.

With ++i you get BCDEF.
 
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Why is the result an ArrayIndexOutOfBound Exception? Please explain.
Thanks
 
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You get the ArrayOutOfBoundsException because:
arg.length = 6
When you get to the situation where i=5, you move into the while loop and i gets incremented to 6. However, arrays are indexed beginning at 0, so the six elements indexes range from 0-5. Hence when you try to access element 6, you get an ArrayOutOfBoundsException.

Hopefully I explained this well!
 
Erin Goettl
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Originally posted by Erin Goettl:
You get the ArrayOutOfBoundsException because:
arg.length = 6
When you get to the situation where i=5, you move into the while loop and i gets incremented to 6. However, arrays are indexed beginning at 0, so the six elements indexes range from 0-5. Hence when you try to access element 6, you get an ArrayOutOfBoundsException.

Hopefully I explained this well!



Whoops I meant args.length :roll:
 
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Sumitabh,

The above code can be manipulated as follows

I've modified the code but maintained the functionlity to give you better understanding how post increment works in this case.
Try walking through this code and you will figure out why we get exception at runtime.
 
Don't get me started about those stupid light bulbs.
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