comp_t

[pefv700 at perv.pe.utexas.edu]

In article <9105060921.aa11316 at art-sy.detroit.mi.us>, chap at art-sy.detroit.mi.us (j chapman flack) writes...
>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
>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