Converting 32-bit unsigned integer (big endian) to long and back

Question

I have a byte[4] which contains a 32-bit unsigned integer (in big endian order) and I need to convert it to long (as int can't hold an unsigned number).

Also, how do I do it vice-versa (i.e. from long that contains a 32-bit unsigned integer to byte[4])?

where does the byte array come from? – Raffaele Mar 24 '12 at 20:10 — Raffaele, Mar 24 '12 at 20:10

Edwin Dalorzo · Accepted Answer · 2012-03-24T20:17:45.217

10

Sounds like a work for the ByteBuffer.

Somewhat like

public static void main(String[] args) {
    byte[] payload = toArray(-1991249);
    int number = fromArray(payload);
    System.out.println(number);
}

public static  int fromArray(byte[] payload){
    ByteBuffer buffer = ByteBuffer.wrap(payload);
    buffer.order(ByteOrder.BIG_ENDIAN);
    return buffer.getInt();
}

public static byte[] toArray(int value){
    ByteBuffer buffer = ByteBuffer.allocate(4);
    buffer.order(ByteOrder.BIG_ENDIAN);
    buffer.putInt(value);
    buffer.flip();
    return buffer.array();
}

edited Mar 24 '12 at 20:17

answered Mar 24 '12 at 20:04

Edwin Dalorzo

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Correct me if I'm wrong, but if I do `int value = buffer.getInt();` then int might not be able to contain the whole number (if it is unsigned and not signed). – Aviram Mar 24 '12 at 20:10
@Aviram An integer in Java is 32-bits (4 bytes), as long as your ByteBuffer is 4 bytes long, I do not see why there should be a problem. I have improved my answer and I tested it with positives and negatives and it works just fine so far. May I be missing something? If you intend to use unsigned integers then use longs and not integers, because integers in Java are signed. – Edwin Dalorzo Mar 24 '12 at 20:19
3

You can use `return buffer.getInt() & 0xFFFFFFFFL;` as you will always get the unsigned value. ByteBuffer's are BIG_ENDIAN by default. You don't need to call `flip()` to use `array()` – Peter Lawrey Mar 25 '12 at 08:22

score 10 · Answer 2 · answered Mar 24 '12 at 20:09

10

You can use ByteBuffer, or you can do it the old-fashioned way:

long result = 0x00FF & byteData[0];
result <<= 8;
result += 0x00FF & byteData[1];
result <<= 8;
result += 0x00FF & byteData[2];
result <<= 8;
result += 0x00FF & byteData[3];

answered Mar 24 '12 at 20:09

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score 1 · Answer 3 · answered Mar 24 '12 at 20:10

1

Guava has useful classes for dealing with unsigned numeric values.

http://docs.guava-libraries.googlecode.com/git/javadoc/com/google/common/primitives/UnsignedInts.html#toLong(int)

answered Mar 24 '12 at 20:10

Sam Barnum

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Good point, that is overwise just using the & 0xFFFFFFFFL method described by Peter Lawrey higher in one comment of this question. – Emmanuel Touzery Jan 08 '13 at 12:57
Why toLong(Integer.MIN_VALUE) doesn't work? and return negative long http://ideone.com/cERLo1 – gavioto Jul 17 '14 at 15:25

Converting 32-bit unsigned integer (big endian) to long and back

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