Get unsigned integer from byte array in java

Question

I need to get unsigned integer from byte array. I understand that java doesn't support unsigned primitives, and I have to use higher primitive (long) to get unsigned int. Many people usually suggest solution like:

public static long getUnsignedInt(byte[] data)
{
    ByteBuffer bb = ByteBuffer.wrap(data);
    bb.order(ByteOrder.LITTLE_ENDIAN);
    return bb.getInt() & 0xffffffffl;
}

But this is not smart since we have to get signed integer then convert it to unsigned which of course may result in overflow exception. I saw other solutions using BigInteger or new java 8 unsigned feature, but I couldn't get it to do what I want.

I couldn't get it to do what I want Can you clarify what you want? — Sotirios Delimanolis
– Sotirios Delimanolis, Commented Jul 31, 2015 at 15:48
@SotiriosDelimanolis I believe OP is looking for an alternate way (using BigInteger) to generate an unsigned int from a byte[] — ControlAltDel
– ControlAltDel, Commented Jul 31, 2015 at 15:50
Take a look at this answer: stackoverflow.com/a/1576404/1011791 — Chthonic Project
– Chthonic Project, Commented Jul 31, 2015 at 15:50
You have 4 bytes. You can't skip to a long directly. You need to get the int first. — Sotirios Delimanolis
– Sotirios Delimanolis, Commented Jul 31, 2015 at 15:51
Are you sure your input bytes are in little endian byte order? Apart from that, I think the code in the question is better than the one you accepted... — Harald K
– Harald K, Commented Jul 31, 2015 at 18:05

Louis Wasserman · Accepted Answer · 2015-07-31 17:46:18Z

3

But this is not smart since we have to get signed integer then convert it to unsigned which of course may result in overflow exception.

There is no such thing as an "overflow exception." Your solution will always work exactly correctly and efficiently. Stop worrying.

answered Jul 31, 2015 at 17:46

Louis Wasserman

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1 Comment

Harald K Over a year ago

This is obviously the correct answer. However, I'm thinking maybe it should be a comment and the question closed, as it seems entirely based on a misconception.

hnefatl · Accepted Answer · 2017-11-20 21:24:33Z

1

You could do something like this:

public static long getUnsignedInt(byte[] data) {
    long result = 0;

    for (int i = 0; i < data.length; i++) {
        result += data[i] << 8 * (data.length - 1 - i);
    }
    return result;
}

You basically create an empty long and shift the bytes into it. You can see this in action in the java.io.DataInputStream.readInt() method.

edited Nov 20, 2017 at 21:24

hnefatl

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answered Jul 31, 2015 at 15:58

redxef

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4 Comments

Eki Over a year ago

this will may get a negative result or wrong result if the last byte is larger than 128.

redxef Over a year ago

@Eki Considering that an int in Java is 32 bits long and we return a long (64 bits) this will not happen unless you try to build an int with 8 bytes, which is nonesense.

Eki Over a year ago

no, i use a 4 bytes array as input and only last byte has a value larger than 128 and than get a negative output. I think the reason is when you do "result += somebyte", java will convert somebyte into long type implicitly so actually result will plus a negative long value converted by somebyte.

redxef Over a year ago

Hm, you are right, with unsigned bytes it wouldn't matter (C, etc.). Try doing a bitwise or.

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