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

user_109785 0
Viewed 154543

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.

numbers bytearray algorithm java
9 Answers

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
coobird_171720

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.

Mnementh_210050

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"));
user_2680 
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

user_148242430

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)

user_79187170
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