http://java.sun.com/j2se/1.4.2/docs/api/java/math/BigInteger.html#BigInteger(jav
a.lang.String,%20int)

http://java.sun.com/j2se/1.4.2/docs/api/java/math/BigInteger.html#doubleValue()

For example by just coding the conversion by hand:

    long result = 0;
    String s = "fbcd000000000000"
    for(int i = 0; i < s.length(); i++) {
        result = (result << 4) | Character.digit(s.charAt(i), 16);
    }

You need to add error handling if s can be to long or s not necessarily
includes hex digits only.

/Thomas

The problem is parseLong is expecting a signed number. The unsigned
one you give it is too big.

You can see the problem if you run this code fragment
     String toConvert = "bfdf000000000000";
     long temp = Long.parseLong( toConvert, 16 );
     System.out.println( temp );

Your code contains another error, masked by boxing. This is closer to
what you meant:

     long temp = 0xbfdf000000000000L;
     double d = Double.longBitsToDouble( temp );
     System.out.println( d );

Double.longBitsToDouble returns a double, not a Double.

I see three ways to solve your unsigned problem:

1. arrange to get get the bits in binary rather than hex.

2. construct the low and high 32 bits then shift then OR.

3. Convert hex chars from unsigned to signed, perhaps when generated
(see http://mindprod.com/jgloss/hex.html) or by checking for  high
char being >= 8 , replace it with equivalent char <8 with high bit
off, then mask the sign bit back in afterwards. Conceptually simple,
but in practice long winded.

For hints on how to do the details, so the at the code in
http://mindprod.com/products1.html#LEDATASTREAM
for readDouble