Statement Exceution

[Lakshmi Saradha]

I found this question in one of the forums. The ans is 2 .
How is line 1 executed ? I understand that the method getarray() is called and then assignment takes place, but looks a bit vague to me
Is there a place in the internet where I can learn abt execution of such statememts.?

class Test
{
public static int[ ] getArray() { return null; }
public static void main(String[] args)
{
int index = 1;
try
{
getArray()[index=2]++; //line 1
}
catch (Exception e){ } //empty catch
System.out.println("index = " + index);
}
}

[Lakshmi Saradha]

Is there anyone who can help me with this?

[Vad Fogel]

We've got a sequence of 3 postfix operators of the equal priority here:
postfix operators [] (params) expr++
The evaluation flows left to right. First, a call to getArray() gets resolved returning null. Second, index is assigned 2 for the array, and third - the postfix increment is supposed to occur, but it never does because of NullPointerException being thrown.
Here's a modified code to demonstrate that evaluation indeed occurs left to right:

class NullTest { 
    public static int[ ] getArray() { 
        //return null; 
        throw new ArithmeticException();
    } 
    public static void main(String[] args) { 
        int index = 1; 
        try{ 
            getArray()[index=2]++; //line 1
        } 
        catch (Exception e){ 
            System.out.println(e); 
        } 
        System.out.println("index = " + index); 
    } 
}

The result is:
java.lang.ArithmeticException
index = 1
You can see that index was not assigned 2 since getArray() blew up with an exception. Refer to Operators Precedence and to the Dan's exams to clear the concept.