CODEWITHSUNDEEP
javabinarybit-manipulationshift
Furkan Aktaş
Furkan Aktaş

Reputation: 91

Bitwise Operator

public class UnsignedShift {

public static void main(String[] args) {

    char hex[] = new char[] {'0','1','2','3','4','5','6','7','8','9','a','b','c','d','e','f'};


    byte b = (byte) 0xf1;
    byte d = (byte)(b>>>4); 


    System.out.println("b>>>4=0x" + hex[(d>>4)&0x0f] + hex[d&0x0f]);

    }

}

The result = 0xff

can anyone explain how it is possible in Java?

I think,it is 0x0f

Upvotes: 1

Views: 528

Answers (2)

kamoor
kamoor

Reputation: 2949

byte b = (byte) 0xf1 will be 1111 0001

byte d = (byte)(b>>>4) will be 1111 1111

d>>4 will be 11111111

0x0f will be 00001111

(d>>4)&0x0f will be 00001111 == 15

d will be 11111111

0f will be 00001111

hex[d&0x0f] will be 00001111 == 15

So Final Answer : 0xff

I think you are expecting (byte)(b>>>4) to shift 0 from left to right 4 times. But b is an integer with 32 bits, it will move 4 bytes from left but get ignored by (byte) conversion. byte conversion take 8 least significant bits of an integer.

Upvotes: 0

ulix
ulix

Reputation: 999

There are no binary operators in java that can operate directly with bytes (8 bits). Variables of type byte, short or char automatically suffer a "numeric promotion" to a 32 bit integer before operations like these are performed as detailed in here. So here is what happens in your code: 

public static void main(String[] args) {

    char hex[] = new char[] {'0','1','2','3','4','5','6','7','8','9','a','b','c','d','e','f'};

    byte b = (byte) 0xf1;    // b is a byte with 0xf1
    byte d = (byte)(b>>>4);  // b is converted to int, becoming 0xfffffff1 then shifted
                             // to the right by 4 bits, resulting in 0x0fffffff

    System.out.println("b>>>4=0x" + hex[(d>>4)&0x0f] + hex[d&0x0f]);

}

If you want to get that right its just easier to use 32 bit variables for all binary operations, like in the example below: 

public static void main(String[] args) {
       char hex[] = new char[] {'0','1','2','3','4','5','6','7','8','9','a','b','c','d','e','f'};

        byte b = (byte) 0xf1;
        int ib = b & 0xff;
        byte d = (byte)(ib>>>4); 

        System.out.println("b>>>4=0x" + hex[(d>>4)&0x0f] + hex[d&0x0f]);
}

Note: Just in case you don't know, you can easily print an integer number in the hexadecimal format by calling Integer.toHexString(n).

Upvotes: 1

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