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

Question

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.

Answer 1

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 

Answer 2

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.