How to convert a byte array to its numeric value (Java)?

I have an 8 byte array and I want to convert it to its corresponding numeric value.

e.g.

byte[] by = new byte[8];  // the byte array is stored in 'by'

// CONVERSION OPERATION
// return the numeric value

I want a method that will perform the above conversion operation.

One could use the Buffers that are provided as part of the java.nio package to perform the conversion.

Here, the source byte[] array has a of length 8, which is the size that corresponds with a long value.

First, the byte[] array is wrapped in a ByteBuffer, and then the ByteBuffer.getLong method is called to obtain the long value:

ByteBuffer bb = ByteBuffer.wrap(new byte[] {0, 0, 0, 0, 0, 0, 0, 4});
long l = bb.getLong();

System.out.println(l);

Result

4

I'd like to thank dfa for pointing out the ByteBuffer.getLong method in the comments.

---

Although it may not be applicable in this situation, the beauty of the Buffers come with looking at an array with multiple values.

For example, if we had a 8 byte array, and we wanted to view it as two int values, we could wrap the byte[] array in an ByteBuffer, which is viewed as a IntBuffer and obtain the values by IntBuffer.get:

ByteBuffer bb = ByteBuffer.wrap(new byte[] {0, 0, 0, 1, 0, 0, 0, 4});
IntBuffer ib = bb.asIntBuffer();
int i0 = ib.get(0);
int i1 = ib.get(1);

System.out.println(i0);
System.out.println(i1);

Result:

1
4

Assuming the first byte is the least significant byte:

long value = 0;
for (int i = 0; i < by.length; i++)
{
   value += ((long) by[i] & 0xffL) << (8 * i);
}

Is the first byte the most significant, then it is a little bit different:

long value = 0;
for (int i = 0; i < by.length; i++)
{
   value = (value << 8) + (by[i] & 0xff);
}

Replace long with BigInteger, if you have more than 8 bytes.

Thanks to Aaron Digulla for the correction of my errors.

If this is an 8-bytes numeric value, you can try:

BigInteger n = new BigInteger(byteArray);

If this is an UTF-8 character buffer, then you can try:

BigInteger n = new BigInteger(new String(byteArray, "UTF-8"));

Simply, you could use or refer to guava lib provided by google, which offers utiliy methods for conversion between long and byte array. My client code:

    long content = 212000607777l;
    byte[] numberByte = Longs.toByteArray(content);
    logger.info(Longs.fromByteArray(numberByte));

You can also use BigInteger for variable length bytes. You can convert it to Long, Integer or Short, whichever suits your needs.

new BigInteger(bytes).intValue();

or to denote polarity:

new BigInteger(1, bytes).intValue();

Complete java converter code for all primitive types to/from arrays http://www.daniweb.com/code/snippet216874.html

Each cell in the array is treated as unsigned int:

private int unsignedIntFromByteArray(byte[] bytes) {
int res = 0;
if (bytes == null)
	return res;


for (int i=0;i<bytes.length;i++){
	res = res | ((bytes[i] & 0xff) << i*8);
}
return res;
}

public static long byteArrayToLong(byte[] bytes) {
    return ((long) (bytes[0]) << 56)
            + (((long) bytes[1] & 0xFF) << 48)
            + ((long) (bytes[2] & 0xFF) << 40)
            + ((long) (bytes[3] & 0xFF) << 32)
            + ((long) (bytes[4] & 0xFF) << 24)
            + ((bytes[5] & 0xFF) << 16)
            + ((bytes[6] & 0xFF) << 8)
            + (bytes[7] & 0xFF);
}

convert bytes array (long is 8 bytes) to long

You can try use the code from this answer: https://stackoverflow.com/a/68393576/7918717

It parses bytes as a signed number of arbitrary length. A few examples:

bytesToSignedNumber(false, 0xF1, 0x01, 0x04) returns 15794436 (3 bytes as int)

bytesToSignedNumber(false, 0xF1, 0x01, 0x01, 0x04) returns -251592444 (4 bytes as int)

bytesToSignedNumber(false, 0xF1, 0x01, 0x01, 0x01, 0x01, 0x01, 0x01, 0x01, 0x04) returns -1080581331768770303 (8 of 9 bytes as long)