Very Computer

Quote:>mantissa = comp_t & 0x1FFF
>exponent = comp_t & 0xE000

>double ticks = ldexp( (double)mantissa, 3*exponent)

I would be delighted, actually, to learn that I'm wrong and have

Quote:>someone provide a more sensible explanation.

Try

ticks = (double)((comp_t & 0x1fff) << 3 * (comp_t & 0xe000));

The reason your ldexp call is correct is that it is

mantissa times (8 to the power exponent)
or more literally
mantissa times (2 to the power (3 times exponent))

This can be done quicker with a left shift.  As K&R2 says,

        The value of E1<<E2 is E1 (interpreted as a bit pattern)
        left-shifted E2 bits; in the absence of overflow, this
                                            E2
        is equivalent to multiplication by 2  .

Chris