Harish - 2 months ago 5x

Java Question

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

Input:

`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));

Output:

`Big Integer1 = ef`

Big Integer2 = -11

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

Answer

From the BigInteger docs

Constructor and Description

BigInteger(byte[] val)

Translates a byte array containing the

two's-complement binary representationof 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

Source (Stackoverflow)

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