Java arrays - Why is the output '1' ?

Question

Why is the output in this example 1?

public static void main(String[] args){
 int[] a = { 1, 2, 3, 4 };   
 int[] b = { 2, 3, 1, 0 };   
 System.out.println( a [ (a = b)[3] ] );   
}

I thought it would be 2. i.e., the expression is evaluated as:

a[(a=b)[3]]
a[b[3]]    //because a is now pointing to b
a[0]   

Shouldn't a[0] be 2 because a is pointing to b?

Thanks in advance.

Answer 1

That weirded me out as well... however, check section 15.7.1 over here

Essentially, operands are evaluated from left to right. But also note this:

It is recommended that code not rely crucially on this specification. Code is usually clearer when each expression contains at most one side effect, as its outermost operation, and when code does not depend on exactly which exception arises as a consequence of the left-to-right evaluation of expressions.

Answer 2

a [ (a = b)[3] ] )

is interpreted as follows:

a = b => a = {2, 3, 1, 0};
( a = b )[3] => 0;

Here is the trick here: a is evaluated as the value before b is assigned to it.

a[(a = b)[3]) => a[0] = 1;

Think about the operator precedence in Java. It should be a bit more obvious.

Answer 3

The arguments to each operator are evaluated left-to-right. I.e., the a in front of the [...] is evaluated before its contents, at which point it still refers to the first array.

Answer 4

As Mr Marcelo Cantos Pointed out, Arguments to each operator are evaluated from left to right. Therefore here is what I think the execution is

a[(a=b)[3]]

Here the outer 'a' will fetch "1,2,3,4" and then its argument (a=b)[3] is evaluated. Thus now a=b and the element at index 3 in b array is returned which is also pointed by a.

Hence we get a '0' from the argument evaluation. As said previously, outer a still refers to old contents Thus gives a[0] in 1,2,3,4 array.

Therefore we get a '1'.

This is my understanding. Please let me know if its wrong.

Thanks,