Re: A92: serious help here please


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Re: A92: serious help here please




{{I recognize these notations from the Motorola Programmer's guide!}
    ADD    0(D0,A0),D1   ; Using what seems to be the A68K syntax
These commands would appear to do the same thing--Add the value at the
location specified by A0 offset by D0 to the value in D1.}}


I thank you and the person before you for the answer to my question.  
It's sort of funny though:  After reading the answers from both of you I 
stumbled upon the disassembler in my computer, I disassembled the code 
that I had no idea what it was (by putting the hex value into it) and 
came out with your answer.  Now my question is, how does it work, so I 
reread both of your answers and came up confused.

A) both of you said something similar to "the indirect of a0+d0 and 
place that in d1 when you have 0(a0,d0),d1" or something, but
B) what if you had 16(a0,d0),d1...for example:

if you had the following:
a0=$0002
d0=$0004
d1=$0006
$0006=#8
$0016=#10
would "ADD 16(a0,d0),d1" be:
a) d1=30    (16+indirect(a0+d0)+d1)
b) d1=16    (indirect(a0+d0+16)+d1)
and if the answer is (b) then why is 16 outside parentheses?

-Rob

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