Exceptions - Beginning Java - page 351

The try and catch blocks are all part of the loop because the catch is inextricably bound to the try .
You can see this from the output. On the fourth iteration, you get an exception thrown because j is 0.
However, after the catch block is executed, you still get one more iteration with j having the value −1.
Even though the try and catch blocks are both within the for loop, they have separate scopes. Variables
declared within the try block cease to exist when an exception is thrown. You can demonstrate that this
is so by declaring an arbitrary variable — k, say — in the try block, and then adding a statement to
output k in the catch block. Your code does not compile in this case.
Suppose you wanted the loop to end when an exception was thrown. You can easily arrange for this. Just
put the whole loop in a try block, thus:

public static void main(String[] args) {
int i = 12;
try {
System.out.println("Try block entered.");
for(int j=3 ; j > =-1 ; --j) {
System.out.println("Try block entered i = " + i + " j = " +

j);

System.out.println(i/j); // Divide by 0 - exception

thrown

}
System.out.println("Ending try block");

} catch(ArithmeticException e) { // Catch the exception
System.out.println("Arithmetic exception caught: " + e);

}

System.out.println("After try block");

}

TestLoopTryCatch2.java

With this version of main() , the program produces the following output:
Try block entered.
Try block entered i = 12 j = 3
4
Try block entered i = 12 j = 2
6
Try block entered i = 12 j = 1
12
Try block entered i = 12 j = 0
Arithmetic exception caught: java.lang.ArithmeticException: / by zero
After try block

Now, you no longer get the output for the last iteration because control passes to the catch block when
the exception is thrown, and that is now outside the loop.