public static  void getIEEEBytes(double d,
    byte[] target,
    int pos) 
{
    long val = Double.doubleToLongBits(d);
    target[pos]   = (byte) (val & 0xff);
    target[pos + 1] = (byte) ((val & 0xff00)  > > 8);
    target[pos + 2] = (byte) ((val & 0xff0000)  > > 16);
    target[pos + 3] = (byte) ((val & 0xff000000)  > > 24);
    target[pos + 4] = (byte) ((val & 0xff00000000L)  > > 32);
    target[pos + 5] = (byte) ((val & 0xff0000000000L)  > > 40);
    target[pos + 6] = (byte) ((val & 0xff000000000000L)  > > 48);
    target[pos + 7] = (byte) ((val & 0xff00000000000000L)  > > 56) ;
}

Puts the IEEE representation of the double provided into the array at the designated position

 public static double getIEEEDouble(byte[] data,
    int pos) 
{
    int num1 = IntegerHelper.getInt(data[pos], data[pos + 1],
                                    data[pos + 2], data[pos + 3]);
    int num2 = IntegerHelper.getInt(data[pos + 4], data[pos + 5],
                                    data[pos + 6], data[pos + 7]);
    // Long.parseLong doesn't like the sign bit, so have to extract this
    // information and put it in at the end.  (Acknowledgment:  thanks
    // to Ruben for pointing this out)
    boolean negative = ((num2 & 0x80000000) != 0);
    // Thanks to Lyle for the following improved IEEE double processing
    long val = ((num2 & 0x7fffffff) * 0x100000000L) +
                  (num1 <  0 ? 0x100000000L + num1 : num1);
    double value = Double.longBitsToDouble(val);
    if (negative)
    {
      value = -value;
    }
    return value;
}

Gets the IEEE value from the byte array passed in