I need to convert 2 byte array ( byte[2] ) to integer value in java. How can I do that?
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5This question could use some clarification. As it stands now, we will only see what you mean by which answer you select.Victor Zamanian– Victor Zamanian2010-12-20 18:45:15 +00:00Commented Dec 20, 2010 at 18:45
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2Does this answer your question? Read two bytes into an integer?David Krell– David Krell2023-05-24 11:00:28 +00:00Commented May 24, 2023 at 11:00
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6 Answers
You can use ByteBuffer for this:
ByteBuffer buffer = ByteBuffer.wrap(myArray);
buffer.order(ByteOrder.LITTLE_ENDIAN); // if you want little-endian
int result = buffer.getShort();
See also Convert 4 bytes to int.
8 Comments
keshav
Thank you this worked. Will ByteBuffer takes care of signed and unsigned bytes?
Matt Ball
@keshav: what do you mean? There is no such thing as a "signed" or "unsigned" byte. A byte is just a group of
0 s and 1 s; the only meaning is in how you interpret those numbers.keshav
@Matt: Once the conversion is done, if the converted number is negative ( say -100 ), will I see it as -100?
Chris Dodd
This won't work, as
+ has higher precedence than <<, so a << 8 + b is the same as a << (8 + b) which is not at all what you want. In addition, byte in java is signed, so bad things will happen if the low-order byte is negative. |
In Java, Bytes are signed, which means a byte's value can be negative, and when that happens, @MattBall's original solution won't work.
For example, if the binary form of the bytes array are like this:
1000 1101 1000 1101
then myArray[0] is 1000 1101 and myArray[1] is 1000 1101, the decimal value of byte 1000 1101 is -115 instead of 141(= 2^7 + 2^3 + 2^2 + 2^0)
if we use
int result = (myArray[0] << 8) + myArray[1]
the value would be -16191 which is WRONG.
The reason why its wrong is that when we interpret a 2-byte array into integer, all the bytes are unsigned, so when translating, we should map the signed bytes to unsigned integer:
((myArray[0] & 0xff) << 8) + (myArray[1] & 0xff)
the result is 36237, use a calculator or ByteBuffer to check if its correct(I have done it, and yes, it's correct).
Comments
Well, each byte is an integer in the range -128..127, so you need a way to map a pair of integers to a single integer. There are many ways of doing that, depending on what you have encoded in the pair of bytes. The most common will be storing a 16-bit signed integer as a pair of bytes. Converting that back to an integer depends on whether you store it big-endian form:
(byte_array[0]<<8) + (byte_array[1] & 0xff)
or little endian:
(byte_array[1]<<8) + (byte_array[0] & 0xff)
Comments
Also, if you can use the Guava library:
Ints.fromByteArray(0, 0, myArray[1], myArray[0]);
It was worth mentioning since a lot of projects use it anyway.
Comments
Simply do this:
return new BigInteger(byte[] yourByteArray).intValue();
Works great on Bluetooth command conversions etc. No need to worry about signed vs. unsigned conversion.
1 Comment
brunch875
Note there's a constructor in this class where you can manually specify you wish to handle unsigned values.
import java.io.*;
public class ByteArray {
public static void main(String[] args) throws IOException {
File f=new File("c:/users/sample.txt");
byte[]b={1,2,3,4,5};
ByteArrayInputStream is=new ByteArrayInputStream(b);
int i;
while((i=is.read())!=-1) {
System.out.println((int)i);
FileOutputStream f1=new FileOutputStream(f);
FileOutputStream f2=new FileOutputStream(f);
ByteArrayOutputStream b1=new ByteArrayOutputStream();
b1.write(6545);
b1.writeTo(f1);
b1.writeTo(f2);
b1.close();
}
