Execution from right to left : a=b=c;

Question

Help!
I am using a tutorial to study for the certification exam.
Unfortunately, I do not understand the answer of a very simple question.
Can you give me a hand on this one?
************* This is the Question of the tutorial ************************************
After the execution of the code-fragment below, what will be the values of i and a?
1.     public class CalcMe {
2.     public static void main(String[] args) {
3.     int [] b = {3,7};
4.     int i = 2, a = 4;
5.     b[i] = i = 0;
6.     a += b;
7.     }
8.     }
Answer 1 i: 0 a: 11
Answer 2 i: 0 a: 4
Answer 3 i: 3 a: 4
Answer 4 An exception is thrown
Answer 5 None of the above
************* This is the Answer to the Question of the tutorial ************************************
After the execution of the code-fragment below, what will be the values of i and a?
1. i: 0 a: 11
2. i: 0 a: 4
3. i: 3 a: 4
4. An exception is thrown
5. None of the above
Answer 4 is correct, an ArrayIndexOutOfBoundsException is thrown at line 5. The value of i
evaluated from the right hand side to the left, it is element b[2] that should get a value assigned,
arrays in Java starts from 0, element 2 is out of scope.
*************************** This is my answer *********************************************************
Despite I read the answer (and created and run it) I still do not understand how line 5 is executed because it seems
that the expression is evaluated from left to right instead of right to left.
I believe that correct answer should be:
in line 5. b[i]=i=0 if expression is executed from right to left then
first i=0 and after that b[0]=0
which leads to believe that the values for i & a will be : i=0 a=4
[Copied from Complete Java 2 Certification Study Guide ISBN 0-7821-4077-7 Phillip Heller & Simon Roberts Chapter #2 Operators and Assignments Page 66]
Given three int varibles a,b,c, the statement a=b=0; is entirely legal. It is executed from right to left,
so that first 0 is assigned into variable c. After it has been executed, hte expression c=0 takes the value
that was assigned to the left side - that is 0. Next, the assignment of b takes place, using the value of
the expression to the right to the equal sign-again, 0. SImilarly, that expression takes the value that was assigned ,
so finally the variable a is also set to 0.
I'd appreciate an explanation

Answer 1

Does this mean that the assignment of indexes in arrays have higher priority than assignment operations?
in other words,
b[2]= ...whatever
will happen before anything else is evaluated in the right side of the expression.
Is that right?
Bingo. Section [url
http://java.sun.com/docs/books/jls/second_edition/html/
xpressions.doc.html#5295]15.26.1:"If the left-hand operand expression [of a simple
assignment operator] is an array access expression
(15.13), then many steps are required:
# First, the array reference subexpression of the
left-hand operand array access expression is
evaluated. If this evaluation completes abruptly, then
the assignment expression completes abruptly for the
same reason; the index subexpression (of the left-hand
operand array access expression) and the right-hand
operand are not evaluated and no assignment occurs.
# Otherwise, the index subexpression of the left-hand
operand array access expression is evaluated. If this
evaluation completes abruptly, then the assignment
expression completes abruptly for the same reason and
the right-hand operand is not evaluated and no
assignment occurs.
# Otherwise, the right-hand operand is evaluated. If
this evaluation completes abruptly, then the
assignment expression completes abruptly for the same
reason and no assignment occurs."
It then goes on to detail null checks on the array
reference, ArrayIndexOutOfBoundsException checks, etc.