CODEWITHSUNDEEP
javabit-manipulationbitwise-operatorsoperator-precedence
LearnFromGenius
LearnFromGenius

Reputation: 31

What is the precedence for bitwise operators (& | ^ ~)?

I would assume in Java all the bit-wise operators have the same precedence. However, in fact the bit-wise operator AND (&) has higher precedence than the bit-wise operator OR (|). See the program below:

public class HelloWorld {

    public static void main(String[] args) {

        int a = 1 | 2 ^ 3 & 5;
        int b = ((1 | 2) ^ 3) & 5;
        int c = 1 | (2 ^ (3 & 5));

        System.out.print(a + "," + b + "," + c);
    }
}

The result of the above program is 3,0,3. So it also proves that XOR (^) has higher precedence.

Why does XOR (^) have higher precedence than OR (|) according to the result of the above result? How do they define the precedence?

Upvotes: 3

Views: 1330

Answers (2)

sandymatt
sandymatt

Reputation: 5612

Because & is defined to have higher precedence than ^, and ^ is defined to have higher precedence than |.

Look at Oracle's Java tutorial.

Upvotes: 1

kviiri
kviiri

Reputation: 3302

In Java, bitwise operators have different precedences as defined by the Java specification:

These [bitwise] operators have different precedence, with & having the highest precedence and | the lowest precedence.

So & comes before ^ and ^ comes before |.

Upvotes: 1

Related Questions