operator precedence and arrays

[preeti khane]

 
Q55.What will be the result of compiling and running the given program?
    Select one correct answer.
  1. class Q55 
  2. {
  3.    public static void main(String[] args) 
  4.    {
  5.       int[] a={11,12,13,14};
  6.       int[] b={0,1,2,3};
  7.       System.out.println(a[(a=b)[3]]); 
    8.    }
  9. }
 Compile time error at line number 7. 
 Runtime error at line number 7. 
 Program will compile correctly and print 3 when executed. 
 Program will compile correctly and print 14 when executed. 

 

Can someone explain what's happening at line 7

[Praveen Kumar Mathaley]

System.out.println(a[(a=b)[3]]);
let's consider how it's interpreted.
processing of the statement start from left to right
(a=b) get's evaluated which
assigns a to the reference held by b
this means now 'a' is pointing to array {0,1,2,3}
now the computation would be a[a[3] but here lies the trick, the outer 'a' reference was held before evaluating the argument. so the outer reference prints the 4th element of the array {11,12,13,14)
try inserting the line System.out.println(a[3]) at the end, it will print 3!