Re: A89: Addx
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Re: A89: Addx
Scott Noveck writes:
> That's the free one from Motorola, right? I've been trying to figure out how
> to order it. . .
Two ways to do it. One is that you download it from their website and
read the PDF file on-line or print it out (with their newer CPUs
that's your only option - they decided not to print 1000+ page manuals).
The other possibility is to visit the following URL:
http://www.mot-sps.com/home/lit_ord.html
and order the book(s) you need (it is a free service). You must know the
book's code number, which may or may not be easy to guess. If you are
about programming info, try the M68000PM/AD which encodes the book
titled "Programmer's Reference Manual" and includes all the 68k
processors except the 68060 and the CPU32+ and it includes the
68881/882 FPUs and the 68851 PMMU as well. It is SW info only but a
quite detailed one.
> So let's say that I have a "decimal" number being represented in a word, so
> that the upper byte is the integer portion and the lower byte is the decimal
> portion - therefore:
>
> <00000010|10000000> represents 2.5 (the upper byte is 1*2^1, the lower byte
> _represents_ 2^-1, or .5)
>
> If I have this word in d1 want to round it off the get rid of the decimal
> portion, will this work?
>
> round:
> clr.w d3
> lsr.w #8,d1 ;integer part remains, x flag is set if the
> ;MSB in the decimal portion is set
> addx.w d3,d1 ;A68k won't allow addx.w #0,d1 - they must both be dx's
>
Yes, that will work. If you really want to save d3, though, you can do
this:
lsr.w #7,d1 ; MSB of the fractional is in bit 0
add.w #1,d1 ; Round it up
lsr.w #1,d1 ; Get rid of the bottom bit
Note that this is longer (and slower) than the addx method but it
saves you a register.
Regards,
Zoltan
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