Harish Harish - 1 year ago 103
Java Question

Difference in BigInteger values from a String and a byte array

Can someone please explain the difference between the below two initialisations of BigInteger.


BigInteger bi1 = new BigInteger("EF", 16);
byte[] ba = new byte[] {(byte)0xEF};
BigInteger bi2 = new BigInteger(ba);
Log.d("BIGINTEGER", "Big Integer1 = " + bi1.toString(16));
Log.d("BIGINTEGER", "Big Integer2 = " + bi2.toString(16));


Big Integer1 = ef
Big Integer2 = -11

How can I initialise a BigInteger with the value "EF" from a byte array?

Answer Source

From the BigInteger docs

Constructor and Description

BigInteger(byte[] val)

Translates a byte array containing the two's-complement binary representation of a BigInteger into a BigInteger.

The Two's-complement is the real reason.

Lets see how...

(Byte)0xef in binary = 11101111 Now convert that back to Int and you get -17 (base 10) or -11 (base 16).

Now take a look at

byte[] ba = new byte[] {0, (byte)0xEF};

This has the (Byte)0xef but prepended by 0. Which means this array has 00000000 11101111, which when converted gives the correct result.

Why was the previous case different?

Check out 2's complement rules - SO Answer, Mandatory Wikipedia link

Another way of thinking about this

0xEF in Decimal = 239

Range of Byte = -127 to 128

We have Overflow.

239 - 128 = 111

Now count this 111 from back (Numeric data types have this circular behaviour, again due to 2's complement representation).

For example: 129.toByte = -127

(129 - 128 = 1, count from back the 1st value = -127)

Shortcut to counting from back if x>128 && x<256 then x.toByte = (x - 128) - 128

Here x = 239 so x.toByte = -17

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