CODEWITHSUNDEEP
javabit-manipulation
CKJOKER
CKJOKER

Reputation: 93

Why does -9 & 0xaaaa get an unsigned result?

The method convertToBinary is used to show the 8bit binary number,

public static String convertToBinary(int a){
    String binary="";
    for(int i=0;i<8;++i){
        Integer bit=new Integer(a&1);
        binary=bit.toString()+binary;
        a>>=1;
    }
    return binary;
}

public static void main(String[] args) {
    int a=-9;
    System.out.println(convertBinary(a));            //11110111
    System.out.println(convertBinary(0xaaaa));       //10101010
    System.out.println(convertBinary(a&0xaaaa));     //10100010
    System.out.println(convertBinary((a&0xaaaa)>>1));//01010001
}

The last result prove that the result of a&0xaaaa is unsigned,can anyone explain why?

Upvotes: 3

Views: 231

Answers (2)

Marko Topolnik
Marko Topolnik

Reputation: 200148

a is an int and so is the literal a&0xaaaa. The result of the bitwise-and operation is also int. It has 32 bits. With

a&0xaaaa

you have masked out all but the lowest few bits from a. The result is a positive, 32-bit int value. So by printing out the lowest 8 bits of

(a&0xaaaa)>>1)

all you reveal is the value of the 9th bit in a&0xaaaa.

Note that the bit mask 0xaaaa, even with the flawed assumption that it's no wider than the digits used, is still a 16-bit value and you are printing just the lower 8 bits.

Upvotes: 4

user555045
user555045

Reputation: 64904

The last result does not prove that >> did an unsigned shift (saying that the value was unsigned is meaningless, values are just a bunch of bits, it's operations that are signed or unsigned).

Since you're only printing 8 bits, and so not looking at the 24 high bits, you have not seen the only bit that could be affected by the difference between x >> 1 and x >>> 1, which is the highest bit. There wouldn't have been any difference between a signed shift and an unsigned shift, so based on that result you can not say which happened.

Upvotes: 0

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