在java中，大的负十六进制值转换为long

Question

the result should have a negative value, but its positive.

How can I fix this?

Thanks!!

BigInteger b = new BigInteger("80000000000000004308000000000000", 16);

System.out.println("long value: "+b.longValue());

--> long value: 4830110600354856960

Answer 1

http://download.oracle.com/javase/1,5,0/docs/api/java/math/BigInteger.html#longValue%28%29

See the above page to understand why it won't be negative. It returns the low 64bits, so your last 64 bits must be higher than Long.MAX_VALUE to cause a negative value.

Answer 2

Your string representation is a signed long but is being presented to BigInteger as an unsigned string (sign is denoted by using a "-" at the start of your string).

The String representation consists of an optional minus sign followed by a sequence of one or more digits in the specified radix.

Bit shifting or correcting your string is needed to make this work from a string instantiation.

I think the best answer is to convert your string to a byte array and use the BigInteger(byte[] val) instantiation which will recognize a negative or positive number based on two's complement. Many options exist for that string to byte array conversion. Take your pick.

... oh, and your number is too large to fit into a long, so that's going to be an issue too; you get the least significant bits.

Answer 3

If you always have 128-bit numbers and assume the highest bit is your sign then you can use the following lines:

BigInteger neg = BigInteger.ONE.shiftLeft(127);
BigInteger b = new BigInteger("80000000000000004308000000000000", 16);
if(b.compareTo(neg) >= 0) { 
    b = neg.subtract(b); 
}

Note: b.longValue() will only be appropriate if the number of bits fits into a long which may not be the case for such large numbers.