About Question enthuware.ocajp.i.v7.2.1029 :

[qmwuzapz]

from the exam explanation:

In an array access, the expression to the left of the brackets appears to be fully evaluated before any part of the expression within the brackets is evaluated. Note that if evaluation of the expression to the left of the brackets completes abruptly, no part of the expression within the brackets will appear to have been evaluated.

Is this the case ?

Code: Select all

class X {
    public static int [] apply(){
       if (Y.i==Y.i)throw new NullPointerException();
        return new int[]{10,8,4,6};
    }
}
public class Y {
    public static int i = 0;
    public static void main(String[] args) {
        try {
            System.out.println(X.apply()[i = 3]);
        } finally {
            System.out.println(i);
        }
    }
}

[EricLeesburg]

Code: Select all

     class Test  {

         public static int[ ] getArray() {  return null;  } 

         public static void main(String[] args)    {      
         
               int index = 1;     

               try{         
 
                     getArray()[index=2]++;     //1 
                   
                     } catch (Exception e){  }  //empty catch      
  
             System.out.println("index = " + index);   
             }
      }

For the statement 1, my understanding is that it equals the statements below:

Code: Select all

                 
                int temp= getArray()[index=2];
                
                 temp++;

and getArray()[index=2] actually is a element of the array. So by the explanation " In an array access, the expression to the left of the brackets appears to be fully evaluated before any part of the expression within the brackets is evaluated. Note that if evaluation of the expression to the left of the brackets completes abruptly, no part of the expression within the brackets will appear to have been evaluated.", the getArray () should be evaluated first, then the index=2 statement. In this case, getArray() throws a NullPointerException which leaves the index=2 statement unevaluated. So, what's wrong here?

[admin]

from the exam explanation:

In an array access, the expression to the left of the brackets appears to be fully evaluated before any part of the expression within the brackets is evaluated. Note that if evaluation of the expression to the left of the brackets completes abruptly, no part of the expression within the brackets will appear to have been evaluated.

Is this the case ?

Code: Select all

class X {
    public static int [] apply(){
       if (Y.i==Y.i)throw new NullPointerException();
        return new int[]{10,8,4,6};
    }
}
public class Y {
    public static int i = 0;
    public static void main(String[] args) {
        try {
            System.out.println(X.apply()[i = 3]);
        } finally {
            System.out.println(i);
        }
    }
}

Yes, i will remain 0. i = 3 will not be executed.

[admin]

Your getArray method doesn't throw a NullPointerException. It returns null. So the statement, "if evaluation of the expression to the left of the brackets completes abruptly," doesn't apply here.

[MariaRoxana]

Hello,

Can someone please give an example when it applies this rule "if evaluation of the expression to the left of the brackets completes abruptly"? Meaning in which cases the evaluation completes abruptly?

[admin]

For example, if the expression to the left of the brackets is a method call and that method throws an exception.

[Mikhail Volkov]

Guys, I think I understand... It's very easy in fact!
Look at two examples:
1) public static int[ ] getArray() { return null; } // [index=2] will work!!! This is case form question.
2) public static int[ ] getArray() { throw new RuntimeException(); } //[index=2] will not work. RTE - it is abnormal finish!

For 1:
If the array reference expression produces null(!!!!!!) instead of a reference to an array, then a NullPointerException is thrown at runtime, but only(!!!) after all parts of the array reference expression have been evaluated and only if these evaluations completed normally.
return null - it is normal finish!

For 2:
Note that if evaluation of the expression to the left of the brackets completes abruptly, no part of the expression within the brackets will appear to have been evaluated.
RunTimeException - it is abnormal finish!