Operator precedence && or ||.. which is greater

Question

http://www.csc.calpoly.edu/~csturner/courses/101/Hints/operator_precedence.html
http://www.uni-bonn.de/~manfear/javaoperators.php
The above two links are where i read about operator precedence. They say && has more precedence than ||, if thats the case then the output for the following should be true, true , true, but the actual output is as below. Can anyone give me a chart that lists the operator precedence correctly. or have i missed something..pls let me know.. thaks..
class ABC{
  static boolean a, b, c;
  public static void main (String[] args) {
    boolean x = (a = true) || (b = true) && (c = true);
    System.out.print(a + "," + b + "," + c);
output: true false false

Answer 1

If && has higher precedence then shouldnt (b = true)
&& (c = true) get executed first before evaluating
for || operator.??No.
a || b && c
The precedence only means that && binds tighter than || -- that is, that, as somebody said earlier, the above is equivalent to
a || (b && c)
rather than
(a || b) && c
It's still evaluated left to right though. First a, then, iff a is false, (b && c). But for us, since a was true, there was no need to eval (b && c)