Java - 异常处理并强制完全退出系统

Question

Please excuse me for this kind of questions here but I am sure to get good explanation with sample which will make to have better understanding about java.

when the System.exit(0); gets executed, the system will break the execution flow and come out the system. this is what is my understanding as of now but I have came across some thing like the below :

Example 1 : 

class FinallySystemExit
{
public static void main(String args[])
{
try
{
int a=2/0;
System.exit(0);
}
catch(Exception e)
{
System.out.println("i am in catch block");
}
finally
{
System.out.println("finally");
}
}
}

my understanding about the above code is it will not print anything and exit from the system but the output is :

i am in catch block
finally

Example 2

class FinallySystemExit
{
public static void main(String args[])
{
try
{
int a=2/1;
System.exit(0);
}
catch(Exception e)
{
System.out.println("i am in catch block");
}
finally
{
System.out.println("finally");
}
}
}

when i execute the above code it prints nothing

The difference between two programs are :

First Program : 

int a=2/0;

and the

 Second Program : 

int a=2/1;

I am totally confused and my basic understanding is broken here.

Could some one explain the reason please.

Thanks

Answer 1

In the first snippet, there is Divide-by-zero error and so the System.exit() is not even called. In the second snippet System.exit() is called and so the JVM exited

Answer 2

In Example 1 : You perform int a=2/0;

This will throw java.lang.ArithmeticException as you are dividing a number by zero . As your code is surrounded by try - catch the exception is caught and it printed the statement in catch block and went to finally block

In Example 2: You perform int a=2/1; So there is no problem at all.

After executing the above line, your program executed System.exit(0); . So No chance of executing the finally block. That is the reason you don't get any output in this case.