Converting from BitSet to Byte array

Question

I have picked up this example which converts BitSet to Byte array.

public static byte[] toByteArray(BitSet bits) {
    byte[] bytes = new byte[bits.length()/8+1];
    for (int i=0; i<bits.length(); i++) {
        if (bits.get(i)) {
            bytes[bytes.length-i/8-1] |= 1<<(i%8);
        }
    }
    return bytes;
}

But in the discussion forums I have seen that by this method we wont get all the bits as we will be loosing one bit per calculation. Is this true? Do we need to modify the above method?

the link exampledepot.com/egs/java.util/Bits2Array.html is expired, do you remember what was the example? — Vishrant
– Vishrant, Commented Jul 18, 2018 at 1:02

Jon Skeet · Accepted Answer · 2011-06-01 07:02:40Z

14

No, that's fine. The comment on the post was relating to the other piece of code in the post, converting from a byte array to a BitSet. I'd use rather more whitespace, admittedly.

Also this can end up with an array which is longer than it needs to be. The array creation expression could be:

byte[] bytes = new byte[(bits.length() + 7) / 8];

This gives room for as many bits are required, but no more. Basically it's equivalent to "Divide by 8, but always round up."

answered Jun 1, 2011 at 7:02

Jon Skeet

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Lars Lynch · Accepted Answer · 2012-12-14 11:24:40Z

9

If you need the BitSet in reverse order due to endian issues, change:

bytes[bytes.length-i/8-1] |= 1<<(i%8);

to:

bytes[i/8] |= 1<<(7-i%8);

answered Dec 14, 2012 at 11:24

Lars Lynch

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Joachim Sauer · Accepted Answer · 2011-06-01 07:45:54Z

3

This works fine for me. if you are using Java 1.7 then it has the method toByteArray().

edited Jun 1, 2011 at 7:45

Joachim Sauer

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answered Jun 1, 2011 at 7:13

Kamahire

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3 Comments

Joachim Sauer Over a year ago

By the way: the official name is "Java 7" (just as it was since Java 5, but Java 5 was still often called Java 1.5. Java 6 was rarely called Java 1.6).

Kamahire Over a year ago

@Joachim Sauer, Yeah official name Java 7. I just mention the version. Any way thanks to correct me.

Ryan Over a year ago

Be careful with BitSet.toByteArray() because it may not serialize the bytes in the order you expect. BitSet notEqual = BitSet.valueOf(bitset.toByteArray()); // This doesn't work.

Community · Accepted Answer · 2017-05-23 12:09:20Z

2

FYI, using

bits.length()

to fetch the size of the bitset may return incorrect results; I had to modify the original example to utilize the size() method to get the defined size of the bitset (whereas length() returns the number of bits set). See the thread below for more info.

java.util.BitSet -- set() doesn't work as expected

edited May 23, 2017 at 12:09

CommunityBot

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answered Jan 20, 2015 at 19:57

ryanbwork

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1 Comment

alwi Over a year ago

length also does something else: BitSet bits = new BitSet(16); System.out.println("BITSET LEN=" + bits.size()); gives: BITSET LEN=64 (bit-size of memory used I guess).

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