Re: A89: Addx
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Re: A89: Addx
>Of course it adds what's there *before* (as described in the 68k
>User's Manual, by the way ...).
That's the free one from Motorola, right? I've been trying to figure out how
to order it. . .
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
At this point, we should be left with d1 containing THREE (2.5 rounds up),
right? That was the whole point of the original question - I'm working on
some trig functions for a 360 degree tilemap engine and need to be able to
round =)
-Scott
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