Why is wrong bit representation got by converting byte array to BitSet in Java

Question

I tried to convert a byte array in order to use its bit representation.

Input:

byte[] values = new byte[]{21, 117, 41, -1};

I would like to create one BitSet object from the byte array but I splitted up it to investigate the issue and tried to create multiple BitSet objects from each element of the array.

BitSet objects are created on the following way:

BitSet bitSetTwentyOne = BitSet.valueOf(new byte[]{21});
BitSet bitSetOneHundredSevenTeen = BitSet.valueOf(new byte[{117});
BitSet bitSetFourtyOne = BitSet.valueOf(new byte[]{41});
BitSet bitSetMinusOne = BitSet.valueOf(new byte[]{-1});

The bits are printed out by using the following method:

private String getBits(BitSet bitSet) {
    StringBuilder bitsBuilder = new StringBuilder();

    for (int i = 0; i < bitSet.length(); i++) {
        bitsBuilder.append(bitSet.get(i) == true ? 1 : 0);
    }

    return bitsBuilder.toString();
}

Output:

bitSetTwentyOne: 10101 but expected -> 00010101 (BitSet::length = 5)
bitSetOneHundredSevenTeen: 1010111 but expected -> 01110101 (BitSet::length = 7)
bitSetFourtyOne: 100101 but expected -> 00101001 (BitSet::length = 6)
bitSetMinusOne: 11111111 but it is as expected at least (BitSet::length = 8)

I want all values 8bit width even if it is needed to fill with zeros. I do not understand why it gives wrong binary values in case of converting 117 and 41.

Stored as little-endian but printed big-endian - that's why. — yegodm
– yegodm, Commented Feb 12, 2019 at 21:19

Joop Eggen · Accepted Answer · 2019-02-12 21:35:50Z

3

length() delivers the logical length using the highest bit 1.
size() would give 8.
cardinality() would give the number of bit 1s.

So you should have used size().

And then the bits are in little endian format, and you reversed it by outputting bit at 0 first.

private String getBits(BitSet bitSet) {
    StringBuilder bitsBuilder = new StringBuilder();
    for (int i = 0; i < bitSet.size(); ++i) {
        bitsBuilder.insert(0, bitSet.get(i) ? '1' : '0');
    }
    return bitsBuilder.toString();
}

answered Feb 12, 2019 at 21:35

Joop Eggen

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2 Comments

Clashsoft Over a year ago

insert(0, ...) has poor performance because all the elements have to be shifted every time. It is better to allocate bitSet.size() and then use StringBuilder.setCharAt.

Joop Eggen Over a year ago

@Clashsoft thanks for mentioning that; implementation matters. In fact originally I tended to change the original problem to use a new char[8] i.o. a StringBuilder.

rgettman · Accepted Answer · 2019-02-12 21:23:57Z

It appears that you are printing the bits backwards. Notice how your output (except for the desired leading zeros) is a mirror image of your expected input.

Using your code, passing in 4, which should be 100, I get 001. It appears that bit 0 is the least significant bit, not the most significant bit. To correct this, loop over the indices backwards. The reason that -1 and 41 are correct apart from the leading zeros is that their bitset representations are palindromes.

To prepend leading zeros, subtract the length() from 8 and print that many zeros. The length() method only returns the number of bits necessary to represent the number. You need to code this yourself; there is no BitSet functionality for leading zeros.

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