I would be delighted, actually, to learn that I'm wrong and haveQuote:>mantissa = comp_t & 0x1FFF
>exponent = comp_t & 0xE000
>double ticks = ldexp( (double)mantissa, 3*exponent)
TryQuote:>someone provide a more sensible explanation.
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
is equivalent to multiplication by 2 .