将字节数组转换为大整数：为什么数组输出中有这么多“ f”个十六进制值？

Question

I'm trying to compare the BigInteger version of my digest with the actual digest bytes to see if any data was lost in the BigInteger conversion. I found that it has all the same hex values in the output, except the byte array output has a lot of f 's in it, but the BigInteger output does not. Why is that?

Console Output

a7b7e9592daa0896db0517bf8ad53e56b1246923

ffffffa7
ffffffb7
ffffffe9
59
2d
ffffffaa
8
ffffff96
ffffffdb
5
17
ffffffbf
ffffff8a
ffffffd5
3e
56
ffffffb1
24
69
23

Code

import java.math.BigInteger;
import java.nio.ByteBuffer;
import java.security.MessageDigest;
import java.security.NoSuchAlgorithmException;
import java.util.Arrays;

public class Project2
{
    public static void main(String[] args)
    {
        try
        {
            ByteBuffer buffer = ByteBuffer.allocate(4);
            buffer.putInt(0xAABBCCDD); //Test value
            byte[] digest = MessageDigest.getInstance("SHA-1").digest(buffer.array());
            BigInteger bi = new BigInteger(1, digest);

            //Big Integer output:
            System.out.println(bi.toString(16));
            System.out.println("");

            //Byte array output:
            for(byte b : digest)
            {
                System.out.println(Integer.toHexString(b));
            }
        }
        catch (NoSuchAlgorithmException e)
        {
            e.printStackTrace();
        }
    }
}

Answer 1

Bytes are signed, so they're sign-extended when converted to int (for Integer.toHexString). Thus any negative byte will become a negative integer, which has high-order bits 1 (two's complement). Use

System.out.println(Integer.toHexString(b & 0xFF));

to mask out the sign-extended bits, leaving only the bottom 8 bits.