Enthuware

Why does the following code prints 5? I thought that every exception would stop the thread.

int k = 0;
try{
int i = 5/k;
}
catch (ArithmeticException e){
System.out.println("1");
}
catch (RuntimeException e){
System.out.println("2");
return ;
}
catch (Exception e){
System.out.println("3");
}
finally{
System.out.println("4");
}
System.out.println("5");

If an exception is left uncaught, then that stops the thread. But here, we are catching the exception. You might want to check this out: http://www.javamex.com/tutorials/except ... dler.shtml

HTH,
Paul.

In the explanation to the question it sais: "Remember : finally is always executed even if try or catch return; (Except when there is System.exit() in try.)"

You can also have System.exit() in the catch clause. This will prevent finally from being executed as well.

Yes, that is a good point.

If I put a return statement in the catch (AritmeticException e) clause instead of the catch (RuntimeException e) clause, then 5 will not print:
The printout here is:
1
4

instead of:
1
4
5

Can you tell me what the logic is with regard to the return statement?
I'm guessing that the following things will happen, in order:
- of course the ArithmeticException is being caught,
- then 1 gets printed,
- then it encounters the return statement,
- but finally always runs (except in case of an System.exit()) so 4 gets printed,
- and now what? the code returns to where?

The control returns to whoever called the method. In this case, it is the JVM that invoked the main method. So the control goes back to the JVM.