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javabit-manipulationbitwise-operators
NorvayHais
NorvayHais

Reputation: 611

Java "int i = byte1 | 0x0200" vs "int i = byte1"?

In the page Wikipedia - Shifts in Java:

In bit and shift operations, the type byte is implicitly converted to int. If the byte value is negative, the highest bit is one, then ones are used to fill up the extra bytes in the int. So 

byte b1=-5; int i = b1 | 0x0200;

will give i == -5 as result.

I understand that 0x0200 is equal to 0b0000 0010 0000 0000. But what is the significance of 0x0200 in the passage shown above?

I mean—b1 | 0x0200 will always be equal to i (see "My Test" below), then in the passage above, why not simply write byte b1=-5; int i = b1?

My Test:

public static void main(final String args[]) {
    final byte min_byte = Byte.MIN_VALUE; // -128
    final byte limit = 0; // according to the bolded words in the passage
    for (byte b = min_byte; b < limit; ++b) {
        final int i1 = b;
        final int i2 = b | 0x0200;
        if (i1 != i2) { // this never happens!
            System.out.println(b);
        }
    }
}

Upvotes: 2

Views: 721

Answers (2)

Elliott Frisch
Elliott Frisch

Reputation: 201447

I understand that 0x0200 is equal to 0b1111 1110 0000 0000

No, it isn't. The correct value is given by,

int i = 0x0200; // <-- decimal 512
System.out.println(Integer.toBinaryString(i));

Which outputs

1000000000

If we examine your second value,

byte b1 = -5;
System.out.println(Integer.toBinaryString(b1));

We get

11111111111111111111111111111011

Lining up both numbers

11111111111111111111111111111011
00000000000000000000001000000000

It seems clear that the result will be the bit value of -5 (since the only 0 in -5 is also 0 in 0x0200). To determine the significance we can examine

int i = 0x0200; // <-- Decimal 512
System.out.println("Dec: " + Integer.toBinaryString(i).length());

Output

Dec: 10

So, the given bitwise OR will force the tenth bit to be true. It was true in your input byte, but if you used - Decimal 1535 (0b 101 1111 1111) then you would get,

System.out.println(1535 | 0x0200);

Output is

2047

Because if you perform a bitwise-or on the two numbers

01000000000
10111111111

you get

11111111111

Upvotes: 2

Sergey Kalinichenko
Sergey Kalinichenko

Reputation: 726639

But what is the significance of 0x0200 in the passage shown above?

This is done for illustration purposes only: the value 0x200 ORs in a one in a position that is equal to 1 already. The idea is to show that the result is not 0x000002FB, but actually -5, i.e. 0xFFFFFFFB.

Upvotes: 4

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