Do the dirty work of decoding; made a private static method to
facilitate testing the algorithm
long raw_number = number;
// mask off the two low-order bits, 'cause they're not part of
// the number
raw_number = raw_number >> 2;
double rvalue = 0;
if ((number & 0x02) == 0x02)
{
// ok, it's just a plain ol' int; we can handle this
// trivially by casting
rvalue = ( double ) (raw_number);
}
else
{
// also trivial, but not as obvious ... left shift the
// bits high and use that clever static method in Double
// to convert the resulting bit image to a double
rvalue = Double.longBitsToDouble(raw_number << 34);
}
if ((number & 0x01) == 0x01)
{
// low-order bit says divide by 100, and so we do. Why?
// 'cause that's what the algorithm says. Can't fight city
// hall, especially if it's the city of Redmond
rvalue /= 100;
}
return rvalue;
