字节到BitSet的转换

Question

I'm using the following code to do the conversion:

public static BitSet fromByte(byte b){
    BitSet bs = new BitSet(8);
    for (int i=0; i<8; i++){  
        if ((b & (1 << i)) > 0){  
            bs.set(i);              
        }  
    } 
    int length = bs.length();

    return bs;
}  

The output is {0, 3, 4, 5, 6}(from debugger display of the bs) - the indexes where the bits are set. I think this should represent 1001111 with length = 7, but is wrong since 1001111 is 79, not 121. Also I want the length to be 8. Basically I want a bitSet with length 8 which represents correctly any byte number. My expectation would be 01111001 and the display of the debugger to show {1,2,3,4,5,7}

Answer 1

Bits in a byte are numbered right-to-left, not left-to-right. That's why setting bits {0, 3, 4, 5, 6} defines this pattern:

7 6 5 4 3 2 1 0
0 1 1 1 1 0 0 1

Answer 2

You are checking the bits in this order:

00000001 = index 0
00000010 = index 1
00000100 = index 2
00001000 = index 3
etc.

ie from right to left, and storing them from left-to-right in the bit set.

Answer 3

{0, 3, 4, 5, 6} is equal to : (2 ^ 0) | (2 ^ 3) | (2 ^ 4) | (2 ^ 5) | (2 ^ 6)

2^0 = 00000001
2^3 = 00001000
2^4 = 00010000
2^5 = 00100000
2^6 = 01000000
--------------
      01111001

Like you guessed, 0 is also the indice of a bit equals to 1. But the bits are ordered from right to left.

Answer 4

The length of the bitset will be as few as bits as necessary to represent the bits you have set. For instance if you set the first three bits, the length will be 2, as only 2 bits are necessary for representation.

The constructor of the BitSet sets its Size , not its length . I suspect they are 2 different concepts.

When I run your code, I get the result I expect. Are you sure of the value you are passing in? Perhaps you are confused about the endianess of your bits? Usually its read from right to left, not left to right (or big endian bitwise order )