Why is this Java operator precedence being ignored here?

Question

The following code prints out \"3\", not \"4\" as you might expect.

public class Foo2 {
    public static void main(String[] args) {
        int a=1, b=2;            


        

Answer 1

I had the same problem with this operator precedence definition (as defined here) and I think none of the above replies are exactly explaining and clarifying the paradox in this definition. This is what I think the higher precedence of postfix operator to other operators (in this example to binary plus operator) means.

Consider the following code fragments:

    int x = 1, y =4 , z;
    z = x+++y;  // evaluates as: x++ + y
    System.out.println("z : " + z); // z: 5
    System.out.println("y : " + y); // y: 4
    System.out.println("x : " + x); // x: 2

    x = 1; y =4 ; 
    z = x + ++y;
    System.out.println("z : " + z); // z: 6
    System.out.println("y : " + y); // y: 5
    System.out.println("x : " + x); // x: 1

As you can see, a single expression z = x+++y; which has two possible evaluations, will be evaluated as z = x++ + y; by java compiler. This means that from three plus sign which came together, compiler assumes the first two of them as a postfix operator and the third one as a binary plus operator. This is in fact a result of higher precedence of postfix operator over other operators.

The second code fragment shows how the outputs differ by writing the expression as z = x + ++y; which explicitly specifies which plus sign is a binary operator.