compilation error on boolean expression

Question

This is a snippet of Java code:

static boolean a; // gets false
static boolean b; 
static boolean c;

    public void printA(){

         boolean bool = (a = tru         


        

Answer 1

If I might....

It looks like some people aren't quite getting what the question is.

x = (a=true) || (b=true) && (c==true);

Since && has a higher precedence than ||, it seems like (b=true) && (c==true) should be evaulated first, thus setting b and c to true. If && has higher precedence, why is one of the operands for the || evaluated first?

And Rohit Jain has explained it the best so far. All I might add is that precedence of operators doesn't dictate the order in which the operands are evaluated -- merely what operations must be completed as operands for for other operators if not rendered unnecessary by short-circuit operators. The precedence determines the tree for the expression (with, ironically, higher-precedent operators going lower in the tree), but then the tree is evaluated depth-first and left-to-right, regardless of operators precedence.

     ||
   /    \
 =      &&
/ \    /   \
a t    =   =
      / \ / \
      b t c t

First the a=true is evaluated, with the "side effect" of doing the assignment, to a value of true. Since || short circuits, the other side isn't even looked at.

If you really want the && (and its operands) to be evaluated first, you'd have to rewrite the expression:

x = (b=true) && (c=true) || (a=true);

Of course, then b and c are set to true and a remains false because || short circuits.

I don't think I've explained anything new, just the same info in smaller steps.