Why does BigInteger parse "80"(hex) to two bytes?

Question

I want to convert a hex-string to a byte-array. I thought using BigInteger is a good idea. But for values greater than 7F it produces unexpected results.

My code:

    var bytes = new BigInteger("80", 16).toByteArray();
    for (var b : bytes) System.out.println(b);

It outputs:

0
-128

Why does this produce two bytes?

I would have expected 00 to FF to produce one byte, 0100 to FFFF to produce two bytes, and so forth.

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Side note: The first byte seems to actually matter:

new BigInteger(new byte[]{      (byte)0x80}); // produces -128 (negative!)
new BigInteger(new byte[]{   0, (byte)0x80}); // produces  128
new BigInteger(new byte[]{0, 0, (byte)0x80}); // produces  128

Answer 1

Thanks for all Your contributions, comments and answers, finally I understand what is going on.

Numbers from 0 to Hex 0x7F (decimal 127) work like this:

First Bit=sign (0=+,1=-)
|    Other bits=number
v    vvv vvvv
0    111 1111    = 0x7F = 127

Numbers from 128 to Hex 0x7FFF (decimal 32.767) work like this:

First Bit (of first byte!)=sign
|    Other bits and other bytes=number
v    vvv vvvv vvvv vvvv
0    000 0000 1000 0000    = 0x80 = 128

Long story short:

• Only the first bit of the first byte is determining the sign
• All other bits of all other bytes determine the absolute value
  • That is 7 bits of the first byte
  • And 8 bits of every other byte