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confused about the output of the program on exceptions?  RSS feed

 
Ranch Hand
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public class test {
public static void main(String args[]) {
int i=1, j=1;
try {
i++;
j--;
if(i/j > 1)
i++;
}
catch(ArithmeticException e) {
System.out.println(0);
}
catch(ArrayIndexOutOfBoundsException e) {
System.out.println(1);
}
catch(Exception e) {
System.out.println(2);
}
finally {
System.out.println(3);
}
System.out.println(4);
}
}

The above program produces the output as shown below:

0
3
4

I was expecting the output as:
0
3

After the execution of the "finally" block the program should have terminated but its not doing so....WHY..?Please help.......
 
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Please use code tags.
 
Bartender
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Naveen Megharaj wrote:
After the execution of the "finally" block the program should have terminated but its not doing so....WHY..?Please help.......



When the program reaches the point where it throws an exception execution jumps out of the try block which the exception occurs and either you catch it (as happens in this case) or throw it to the caller. Here as you are catching it inside the method(main in this case) so, the execution continues from end of the finally block.
 
Greenhorn
Posts: 9
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If your program throws any exceptions then it will handle by try catch block and it wont be affect the entire program.
It will be continue the process until it reaches the end.

finally block does not consider about the exception . It will be excuting even there is no exception in program
 
Don't get me started about those stupid light bulbs.
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