CODEWITHSUNDEEP
javaoperatorsbitwise-operators
Java Beginer
Java Beginer

Reputation: 321

Detailed explanation for Bit wise or Operator works in different method

Here the code.

public class Test {

    public static void main(String[] args) {
        int x=2; int y = 3;
        if((y == x++) | (x < ++y)){
            System.out.println("x = " + x + "Y = " + y);
        }
    }
}

This outputs x=3 y=4

Another Criteria

public class Test {

    public static void main(String[] args) {
        System.out.println(4|3);
    }
}

This outputs 7

Here the second criteria | works as bit wise operator. But in first criteria | is not. It manipulates and giving the output. I need explanation how the first criteria works.

The thing i know is, | will compare both sides and || will compare only left side and it true it proceeds to next.

Hope my question is clear. Thankyou in advance..

Upvotes: 2

Views: 120

Answers (3)

Konstantin Yovkov
Konstantin Yovkov

Reputation: 62864

For the first example you will have:

if ((3 == 2) | (3 < 4))

which is equal to:

if ( false | true )

and evaluates to true (since the both sides of the expression are compared).

For the second example, since you already know that | will compare both sides, here's what it really happens. The numbers are converted into binary ones and then the operation is applied:

4 | 3

will be converted to:

100 | 011 = 111

which is equal to 7.

Upvotes: 2

AllTooSir
AllTooSir

Reputation: 49372

Follow the JLS 15.22.2:

When both operands of a &, ^, or | operator are of type boolean or Boolean, then the type of the bitwise operator expression is boolean. In all cases, the operands are subject to unboxing conversion (§5.1.8) as necessary.

For &, the result value is true if both operand values are true; otherwise, the result is false.

For ^, the result value is true if the operand values are different; otherwise, the result is false.

For |, the result value is false if both operand values are false; otherwise, the result is true.

Take the first case:

int x=2; int y = 3;
if((y == x++) | (x < ++y)){
       System.out.println("x = " + x + "Y = " + y);
}

(y == x++) is false because x++ is post-increment , (x < ++y) is true , hence false|true is true and you get the console o/p with the values of x=3 and y=4.

Second case , is natural bitwise OR :

System.out.println(4|3);

That in 3-bits is (011|100) = (111) = 7(base 10).

Upvotes: 1

darijan
darijan

Reputation: 9775

In the first example, first the left side of the expression is evaluated:

y == x++ // y == x; x=x + 1

y is 3, x is 2, the equality check is false, and the x is incremented to 3.

and then the right side of the expression is evaluated:

x < ++y; // y = y + 1; x < y

y is incremented and is 4, and then is checked if x is smaller than y. And yes, 3 is smaller than 4, and the result of the right part is true.

Since there is bitwise | operator between false on the left side and true on the right side, the result is 

false | true = true

And that's why the expression is true and syso is executed.

In the second example, a bitwise OR is executed, hence

100 | 011 = 111 //is equal to 7

Upvotes: 1

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