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could some one explain how the control flows on the following code..

 
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What is the result of attempting to compile and run the program?

a. Prints: 0,0,0,1,0,0
b. Prints: 0,0,1,1,0,1
c. Prints: 0,1,1,1,0,1
d. Prints: 1,0,1,1,0,1
e. Prints: 1,1,1,1,0,1
f. Compile-time error
g. Run-time error
h. None of the above


Answer is b.

thanks in advance,
reena


EDIT by mw: Added Code Tags to original poster's indentation. (Please use Code Tags.)
[ August 14, 2006: Message edited by: marc weber ]
 
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What's your interpretation of how this would execute? What exception is thrown? Where is it caught? Etc...
 
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This is what happening in the above code:

At the inner try block, the value of x is 1 therefore 'Level1Exception' will be thrown in case 1.
The catch is only for Level2Exception therefore it won't enter there.
The 'finally(c++)' will be executed and the value of c = 1
The outer catch(Level1Exception) will catch the above exception and thereby increments the values of d. d =1.
Outer finally(g++) makes g=1.

Therefore the result will be 0,0,1,1,0,1
 
Greenhorn
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The important thing to remember here Reena is that Level1Exception IS-NOT-A Level2Exception, so the catch block for Level2Exception will not be run. BUT, the finally blocks will always run, no matter what.

If the catch block for Level1Exception was not there, then the catch block for Exception would have been run, as Level1Exception IS-A Exception. And, as before, the finally block always runs.
 
R .sourav nayak
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thanks guys for the reply. Krishnan your explaination was really helpful.
 
arch rival
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You may find that a useful tool to answer questions like this is to run the code through a debugger such as the ones built into NetBeans or Eclipse. The debuggers are possibly the most userful features of these tools, and for most learning purposes I recommend the naked JDK.
 
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