Jerome,
Interesting question - and code !
harpo equ $160
chico equ $140
elem_size equ $0c
clr.l a1 A1 = 0
moveq.l #4,d2
lea harpo(a1),a1 A1 = A1 + $160 = $160
myloop:
adda.w elem_size,a1 A1 = A1 + (5 * $0c) = $19c
dbra d2,myloop
lea chico-4*elem_size(a1),a1 A1 = $140 - $19c + (4 * $0c)
= $140 - $1cc
>> Question: what is the value of a1 at the end ?
Well it could be $ffffff74 ......
>> (I also have my idea, but I do not want to influence yet!)
BUT it depends upong how good your assembler is, does it do expressions
evaluation correctly so that the multiplication is done before the addition,
or does it do it in-line ?
My answer above shows what it should be (!) assuming correct precedence,
but if the assembler is in-line, then it will be :
(($140 - 4) * $0c) + $19c
($13c * $0c) + $19c
$ed0 + $19c
$106c
So, which answer did you think ?
By the way, you loop around FIVE times because the dbra instruction stops
when the counter reaches -1, so we pass through the loop with d2 = 4,3,2,1
and then 0, adding $0c to a1 each time, so a1 ends up at $160 + (5 * $0c)
which is (reaches for hex calculator ...) $19c.
How did I do ?
Cheers,
Norman.
-------------------------------------
Norman Dunbar
Database/Unix administrator
Lynx Financial Systems Ltd.
mailto:[EMAIL PROTECTED]
Tel: 0113 289 6265
Fax: 0113 289 3146
URL: http://www.Lynx-FS.com
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