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#REDIRECT [[Pl CORRIG 1.DOC/fr]]
  
  +
EXERCISES PART I
  +
  +
CORRECTIONS
  +
  +
-------------------------
  +
  +
- The solutions to the exercises proposed in this file are related
  +
to exercises nr°1 in the file: EXOS_1.DOC
  +
-----------
  +
Refer to the nr° of the exercise whose solution you want to know.
  +
  +
  +
  +
------------------
  +
  +
nr° 1:
  +
------
  +
  +
The error in this program had to be found:
  +
  +
text
  +
  +
ab move.b #345,d0 ; -128<BYTE<128 !!
  +
add.b #1,d0
  +
jmp ab
  +
  +
end
  +
  +
The answer was very simple, the error is in the first instruction,
  +
indeed, we do a 'MOVE.B #345,d0' i.e., we put Byte 345 in
  +
the low weight Byte of register d0, but I remind you that the
  +
maximum value a byte can have is 128 and 345>128, this leads
  +
to an error during assembly.
  +
To correct, simply write 'MOVE.W #345,d0' or 'MOVE.L #345,d0'
  +
  +
  +
nr° 2:
  +
------
  +
The value of data register d3 after the program had to be found.
  +
  +
text
  +
  +
move.l #12,d3 ;We put the L-M=12 in the d3 register
  +
add.l #4,d3 ;We add the L-M=4 to it: d3=12+4=16
  +
add.l d3,d3 ;We add the L-M d3 to itself, so
  +
;d3=d3+d3=16+16=32 (on an L-M)
  +
move.l var,d2 ;We put the L-M pointed by 'var', i.e.
  +
;458 in the data register d2
  +
add.l d2,d3 ;we add the L-M of d2 to the L-M of d3,
  +
;so d3=d2+d3=458+32=490 (L-M)
  +
  +
data ;data segment
  +
  +
var dc.l 458
  +
  +
end
  +
  +
As the Remarks show, d3=490
  +
  +
  +
nr° 3:
  +
------
  +
It was asked what the d5 register would contain after the program.
  +
  +
text
  +
  +
move ici,d4 ;we place the word pointed by 'here' in the low
  +
;weight word of the data register d4
  +
;so d4=10 (on a word)
  +
add #5,d4 ;we add the word 5 to the low weight word of d4
  +
move d4,la ;We put the low weight word of d4 (15) at 'there'
  +
add ici,la ;We add the value pointed by 'here' to the
  +
;data pointed by 'there', 'there' therefore points to
  +
;a word (ds.w) equal to 10+15=25
  +
move la,d5 ;We put the word pointed by 'there' in the low
  +
;weight word of d5.So d5=25
  +
  +
data
  +
  +
here dc.w 10
  +
  +
bss
  +
  +
there ds.w 1
  +
  +
end
  +
  +
We then find d5=25
  +
  +
  +
nr°4:
  +
-----
  +
You had to:
  +
Find the value of a4 at the end of the program knowing that the label 'ad' is
  +
located at an address equal to $ff0 and it points (dc.l) at the number 45
  +
  +
  +
text
  +
  +
move.l #ad,a4 ;we put the address of 'ad' (L-M) in the
  +
;address register a4, so a4=$ff0
  +
move.l ad,d5 ;we put the L-M pointed by 'ad', i.e. 45 in
  +
;the data register d5
  +
  +
data
  +
  +
ad dc.l 45 ;we suppose that here PC=$ff0
  +
  +
end
  +
  +
In this specific case, a4=$ff0 (and not 45, review your lessons!)
  +
It was really very simple.
  +
  +
  +
nr°5:
  +
-----
  +
You must find the value of d5 and a5 after this program.
  +
  +
text
  +
  +
move #15,d5 ;We put the word 15 in the low weight word of d5
  +
move.l #zz,a5 ;We put the address of 'zz' in the address
  +
;register a5
  +
move zz,d5 ;We put the word pointed by 'zz' in the low
  +
;weight word of the data register d5, so d5=0.
  +
  +
data
  +
  +
zz dc.w 0
  +
  +
end
  +
  +
So d5=0 (word) and a5=address of 'zz' (L-M)
  +
  +
  +
nr°6:
  +
-----
  +
The value contained at the address 'out', the value of d0
  +
and d5 at the end of this program had to be found.
  +
  +
text
  +
  +
move #247,d0 ;we put the word 247 in the low weight word of d0
  +
move #3,d5 ;we put the word 3 in the low weight word of d5
  +
  +
bbb add #8,d0 ;we add the word 8 to the low weight word of d0
  +
move d0,out ;then we put the low weight word of d0 in 'out'
  +
dbf d5,bbb ;we come back to 'bbb' until d5=-1
  +
;i.e. 4 times because d5=3 at the beginning
  +
;at the end of the loop, d0=247+(8*4)=279 (word)
  +
bss
  +
  +
out ds.b 2 ;we reserve 2 Bytes (i.e. 1 word) in 'out'
  +
  +
end
  +
  +
In 'out' we therefore find a Word, the last value of d0 so 279, d5=-1
  +
because its register is decremented until d5=-1.
  +
  +
  +
nr°7:
  +
-----
  +
It was necessary to:
  +
Find the BINARY value of the data register d1 at the end of this program.
  +
knowing that 5438=%0001010100111110
  +
  +
text
  +
  +
move #5391,d0 ;we put the word 5391 in the low weight word of d0
  +
add #47,d0 ;we add 47, so d0=5438 (word)
  +
  +
move d0,ten ;we put the low weight word of d0 in 'ten',
  +
;as in 'ten' there is 'ds.b', only the high byte
  +
;of d0 will be in 'ten', the low weight byte
  +
;will be in 'in' (address 'ten'+1).
  +
move.b in,d1 ;we put the byte pointed by 'in' in the low weight
  +
;byte of d1
  +
;in 'in' there is the low weight byte of d0,
  +
;so d1 will contain %00111110
  +
  +
bss
  +
  +
ten ds.b 1 ;reserve one byte in 'ten'
  +
in ds.w 47 ;reserve 47 words in 'in'='ten'+1
  +
  +
data
  +
  +
null dc.w 5438 ;nothing to do here...
  +
  +
end
  +
  +
  +
We therefore have d1=%00111110 (byte)
  +
  +
  +
nr°8:
  +
-----
  +
What is going to happen in 'gag'?
  +
  +
text
  +
  +
move.l #gag,a2 ;a2 points to the address of 'gag'
  +
  +
gag move.b #100,d1 ;we put the byte 100 in the low weight
  +
;of d1
  +
jmp (a2) ;we jump to the address pointed by a2 (so
  +
;the address of 'gag')
  +
  +
end
  +
  +
  +
The program is therefore an infinite loop, we put the byte 100 in d1
  +
an infinite number of times.
  +
  +
  +
nr°9:
  +
-----
  +
What will the value of d3 be after this program?
  +
  +
text
  +
  +
move.b #12,lab ;we put the byte 12 in 'lab'
  +
move.b #14,lac ;" " " " the byte 14 in 'lac'
  +
move.b #15,lad ;" " " " the byte 15 in 'lad'
  +
  +
move.l #lab,a3 ;we put the address of 'lab' in a3
  +
move.b #1,d1 ;we put the byte 1 in the low weight byte of a1
  +
  +
fff move.b (a3)+,d3 ;we take the value pointed by a3, we increment
  +
;a3 by 1 (.b) and we put it in the low weight byte
  +
;of d3.
  +
dbf d1,fff ;we decrement d1 by 1 unit until d1=-1
  +
;if d1 is different from -1, we jump to 'fff'
  +
  +
add.b (a3),-(a3);we add the byte pointed by a3 (we decrement
  +
;a3 by 1 (.b)) to the byte pointed by the new
  +
;address of a3.
  +
move.b -(a3),d3 ;We decrement a3 by 1 (.b) and we put the value
  +
;thus pointed in the low weight byte of d3
  +
add.b (a3),d3 ;We add the value pointed by a3 to d3 (byte)
  +
  +
  +
bss
  +
  +
lab ds.w 1 ;we reserve 1 L-M
  +
lac ds.b 1 ;we " 1 byte
  +
lad ds.w 5 ;we reserve 5 words
  +
  +
end
  +
  +
  +
In 'fff', the first time: d1=1, a3 points to 'lab'.
  +
  +
With 'move.b (a3)+,d3', we take the byte in 'lab', we increment a3 by
  +
one unit (.b) so now a3 points at 'lab'+1 and we put the byte in
  +
the low weight byte of d3, so d3=12 (we have put the byte 12 in 'lab').
  +
  +
Then we have:
  +
  +
'dbf d1,fff': d1=d1-1=0 as 0 is different from -1, we jump to 'fff'.
  +
  +
again:
  +
  +
'move.b (a3)+,d3': we take the byte pointed by a3 (here a3 points to the
  +
second byte of 'lab', which is 0 since here nothing has been initialized...)and we put it
  +
in d3 after having incremented a3 by one unit: d3=0 and a3 points on
  +
'lab'+2='lac'.
  +
We arrive at 'dbf d1,fff',again, d1=d1-1=-1, this time we will no more
  +
go to 'fff' because d1=-1: we continue.
  +
  +
Then comes 'add.b (a3),-(a3)', a3 points to 'lac' ('lab'+2), we take the
  +
data (the byte) pointed by a3 (which is worth 14), we decrement a3 by one
  +
unit (.b) and we add it to the new value thus pointed by a3 (which is 0)
  +
  +
finally:
  +
  +
With 'move.b -(a3),d3' , we decrement a3 by one unit (so a3 points at
  +
'lab') and we put the data thus pointed in the low weight byte of
  +
d3, so d3=12.
  +
  +
Finally, with 'add.b (a3),d3', we add to d3 the data pointed by a3,
  +
as a3 points at 'lab', we get: d3=d3+12=12+12=24
  +
  +
Finally: d3=24. and a3 points at 'lab' ...
  +
  +
  +
  +
  +
  +
  +
  +
  +
  +
nr°10:
  +
------
  +
  +
What will be the values of d0, d1, d2, d3, a0 after this program?
  +
Easy...
  +
  +
text
  +
  +
move.b #10,a ;we put the byte 10 in a
  +
move.b #11,b ; 11 in b
  +
move.b #12,c ; 12 in c
  +
move.b #13,d ; 13 in d
  +
move.b #98,e ; 98 in e
  +
  +
move.l #begin,a0 ;we put the address of 'begin' in a0
  +
;:a0 points to 'begin'
  +
move.b (a0)+,d0 ;we put the byte pointed by a0 :0 because nothing is
  +
;initialized in 'begin',(we increment a0 by one
  +
;unit (.b)), in d0, so:
  +
;d0=0 (byte), a0=a0+1, so a0 points at 'a'.
  +
move.b (a0)+,d1 ;same for d1: d1=10, a0 points at 'a'+1='b'
  +
move.b (a0)+,d2 ;for d2: d2=11, a0 points at 'b'+1='c'
  +
move.b (a0),d3 ;We place the byte pointed by a0 in d3:
  +
;a0 points at 'c', therefore d3=12
  +
move.b -(a0),d4 ;decrement a0 by one unit and put the byte
  +
;thus pointed by a0 in the low weight byte
  +
; of d4:
  +
;as a0 was pointing to 'c', we have now:
  +
;a0='c'-1='b', so d4=11 (we have put 11 in 'b')
  +
  +
bss
  +
  +
begin ds.b 1 ;reserve 1 byte in begin
  +
a ds.b 1 ;in a
  +
b ds.b 1 ;in b
  +
c ds.b 1 ;in c
  +
d ds.b 1 ;in d
  +
e ds.b 1 ;in e
  +
f ds.l 4 ;reserve in L-M in f
  +
  +
end
  +
  +
So: d0=0, d1=10, d2=11, d3=12, d4=11, a0 points at 'b'.
  +
  +
  +
nr°11:
  +
------
  +
One had to find the value of a4 and d0, d1 after this program.
  +
  +
text
  +
  +
move.b #3,d0 ;we put the byte 3 in the register d0
  +
move.l #stack,a4 ;we put the address of 'stack' in a4
  +
  +
rrr move.b -(a4),d1 ;we decrement a4 by one unit (.b) and put the
  +
;byte pointed by a4 in the low weight byte of d1
  +
add.b #4,d1 ;we add the byte 4 to the low weight byte of d1
  +
dbf d0,rrr ;loop until d0=-1
  +
  +
data
  +
  +
dc.b 1,2,3,4,5,6,7,8,9 ;9 bytes are stored here
  +
stack dc.b 245 ;byte 245 at the address 'stack'
  +
  +
end
  +
  +
  +
at the end of the program, d0=-1, it is the loop counter...
  +
  +
the loop:
  +
  +
1st pass: d0=3, a4 points to 'stack'
  +
with 'move.b -(a4),d1', we decrement a4 by 1 unit (because .b), we take
  +
the byte pointed by a4 and we place it in the low weight byte of d1:
  +
a4='stack'-1 so a4 is pointing to the byte '9', d1=9+4=13 because we 'add.b #4, d1'
  +
We decrement d0 and go back to 'rrr' because d0=2
  +
  +
2nd pass: d0=2, a4 points to the byte '9'
  +
The same operation as before:
  +
a4 is pointing to the byte '8', d1=8+4=12 (byte), d0=2-1=1
  +
  +
3rd pass: d0=1, a4 points to the byte '8'
  +
the same, we obtain:
  +
a4 is pointing to the byte '7', d1=7+4=11 (byte), d0=1-1=0
  +
  +
4th pass: d0=0, a4 points to the byte '7'
  +
the same, we have:
  +
a4 is pointing to the byte '6', d1=6+4=10 (byte), d0=0-1=-1 so we get out of
  +
the loop and finally:
  +
  +
d0=-1, d1=10, a4 points to the byte '6'
  +
  +
  +
nr°12:
  +
------
  +
Let's find what is in 'out' and the value of d2.
  +
  +
text
  +
  +
a equ 3 ;we assign 3 to 'a'
  +
b equ 5 ; 5 to 'b'
  +
  +
move.b a,d2 ;we put the byte pointed by 'a' in the low weight
  +
;of d2, so d2=3
  +
add.b b,d2 ;we add the byte pointed by 'b', so
  +
;d2=d2+5=8 (byte)
  +
  +
move.l #g,a5 ;we put the address of 'g' in a5
  +
move.b d2,(a5) ;we put the low weight byte of d2 at
  +
;the address pointed by a5 (so in 'g')
  +
move.b d2,-(a5) ;we put the low weight byte of d2 (we decrement
  +
;a5 by one unit) at the new address of a5
  +
;(here, a5 points to 'out', so we will find d2
  +
;i.e. the byte 8 in 'out')
  +
move.b (a5)+,d2 ;we put the byte pointed by a5 (so the byte 8)
  +
;(we increment a5 by 1 unit) in the low weight byte
  +
;of d2: d2=8 and a5 points to 'g'
  +
move.b (a5),d2 ;we put the byte pointed by a5 in the low weight
  +
;of d2: a5 points to 'g', in 'g' there is
  +
;the byte 8 so d2=8 once again.
  +
  +
bss
  +
  +
out ds.b 1
  +
g ds.b 4
  +
  +
end
  +
  +
conclusion: in 'out' there is the byte 8 and d2=8 (byte)
  +
  +
  +
nr°13:
  +
------
  +
We must find the value of d1, d2, d3, d4 and a0.

Revision as of 00:57, 17 December 2023

EXERCISES PART I

CORRECTIONS

- The solutions to the exercises proposed in this file are related to exercises nr°1 in the file: EXOS_1.DOC

Refer to the nr° of the exercise whose solution you want to know.


nr° 1:

The error in this program had to be found: 

    text

ab move.b #345,d0  ; -128<BYTE<128  !! 

    add.b     #1,d0
    jmp       ab

    end

The answer was very simple, the error is in the first instruction, indeed, we do a 'MOVE.B #345,d0' i.e., we put Byte 345 in the low weight Byte of register d0, but I remind you that the maximum value a byte can have is 128 and 345>128, this leads to an error during assembly. To correct, simply write 'MOVE.W #345,d0' or 'MOVE.L #345,d0'


nr° 2:

The value of data register d3 after the program had to be found. 

    text

    move.l    #12,d3    ;We put the L-M=12 in the d3 register
    add.l     #4,d3     ;We add the L-M=4 to it:  d3=12+4=16
    add.l     d3,d3     ;We add the L-M d3 to itself, so
                        ;d3=d3+d3=16+16=32 (on an L-M)
    move.l    var,d2    ;We put the L-M pointed by 'var', i.e.
                        ;458 in the data register d2
    add.l     d2,d3     ;we add the L-M of d2 to the L-M of d3,
                        ;so d3=d2+d3=458+32=490 (L-M)

    data                ;data segment

var dc.l 458 

    end

As the Remarks show, d3=490


nr° 3:

It was asked what the d5 register would contain after the program. 

    text

    move      ici,d4    ;we place the word pointed by 'here' in the low
                        ;weight word of the data register d4
                        ;so d4=10 (on a word)
    add       #5,d4     ;we add the word 5 to the low weight word of d4
    move      d4,la     ;We put the low weight word of d4 (15) at 'there' 
    add       ici,la    ;We add the value pointed by 'here' to the
                        ;data pointed by 'there', 'there' therefore points to
                        ;a word (ds.w) equal to 10+15=25
    move      la,d5     ;We put the word pointed by 'there' in the low
                        ;weight word of d5.So d5=25

    data

here dc.w 10 

    bss

there ds.w 1 

    end

We then find d5=25


nr°4:

You had to: Find the value of a4 at the end of the program knowing that the label 'ad' is located at an address equal to $ff0 and it points (dc.l) at the number 45 


    text

    move.l    #ad,a4   ;we put the address of 'ad' (L-M) in the
                       ;address register a4, so a4=$ff0 
    move.l    ad,d5    ;we put the L-M pointed by 'ad', i.e. 45 in
                       ;the data register d5
         
    data

ad dc.l 45 ;we suppose that here PC=$ff0 

    end

In this specific case, a4=$ff0 (and not 45, review your lessons!) It was really very simple.


nr°5:

You must find the value of d5 and a5 after this program. 

    text

    move      #15,d5    ;We put the word 15 in the low weight word of d5
    move.l    #zz,a5    ;We put the address of 'zz' in the address
                        ;register a5
    move      zz,d5     ;We put the word pointed by 'zz' in the low
                        ;weight word of the data register d5, so d5=0.
    
    data

zz dc.w 0 

    end

So d5=0 (word) and a5=address of 'zz' (L-M)


nr°6:

The value contained at the address 'out', the value of d0 and d5 at the end of this program had to be found. 

    text

    move      #247,d0   ;we put the word 247 in the low weight word of d0
    move      #3,d5     ;we put the word 3 in the low weight word of d5

bbb add #8,d0 ;we add the word 8 to the low weight word of d0 

    move      d0,out    ;then we put the low weight word of d0 in 'out'
    dbf       d5,bbb    ;we come back to 'bbb' until d5=-1
                        ;i.e. 4 times because d5=3 at the beginning
                        ;at the end of the loop, d0=247+(8*4)=279 (word)
    bss

out ds.b 2 ;we reserve 2 Bytes (i.e. 1 word) in 'out' 

    end

In 'out' we therefore find a Word, the last value of d0 so 279, d5=-1 because its register is decremented until d5=-1.


nr°7:

It was necessary to: Find the BINARY value of the data register d1 at the end of this program. knowing that 5438=%0001010100111110 

    text

    move      #5391,d0  ;we put the word 5391 in the low weight word of d0
    add       #47,d0    ;we add 47, so d0=5438 (word)

    move      d0,ten    ;we put the low weight word of d0 in 'ten',
                        ;as in 'ten' there is 'ds.b', only the high byte
                        ;of d0 will be in 'ten', the low weight byte
                        ;will be in 'in' (address 'ten'+1).
    move.b    in,d1     ;we put the byte pointed by 'in' in the low weight
                        ;byte of d1
                        ;in 'in' there is the low weight byte of d0,
                        ;so d1 will contain %00111110

    bss

ten ds.b 1 ;reserve one byte in 'ten' in ds.w 47 ;reserve 47 words in 'in'='ten'+1 

    data

null dc.w 5438 ;nothing to do here... 

    end


We therefore have d1=%00111110 (byte)


nr°8:

What is going to happen in 'gag'? 

    text

    move.l    #gag,a2        ;a2 points to the address of 'gag'

gag move.b #100,d1 ;we put the byte 100 in the low weight 

                             ;of d1
    jmp       (a2)           ;we jump to the address pointed by a2 (so
                             ;the address of 'gag')

    end


The program is therefore an infinite loop, we put the byte 100 in d1 an infinite number of times.


nr°9:

What will the value of d3 be after this program? 

    text

    move.b    #12,lab   ;we put the byte 12 in 'lab'
    move.b    #14,lac   ;"  "  "  "  the byte 14 in 'lac'
    move.b    #15,lad   ;"  "  "  "  the byte 15 in 'lad'

    move.l    #lab,a3   ;we put the address of 'lab' in a3 
    move.b    #1,d1     ;we put the byte 1 in the low weight byte of a1

fff move.b (a3)+,d3 ;we take the value pointed by a3, we increment 

                        ;a3 by 1 (.b) and we put it in the low weight byte
                        ;of d3.
    dbf       d1,fff    ;we decrement d1 by 1 unit until d1=-1
                        ;if d1 is different from -1, we jump to 'fff'

    add.b     (a3),-(a3);we add the byte pointed by a3 (we decrement 
                        ;a3 by 1 (.b)) to the byte pointed by the new
                        ;address of a3.
    move.b    -(a3),d3  ;We decrement a3 by 1 (.b) and we put the value
                        ;thus pointed in the low weight byte of d3
    add.b     (a3),d3   ;We add the value pointed by a3 to d3 (byte)



    bss

lab ds.w 1 ;we reserve 1 L-M lac ds.b 1 ;we " 1 byte lad ds.w 5 ;we reserve 5 words 

    end


In 'fff', the first time: d1=1, a3 points to 'lab'.

With 'move.b (a3)+,d3', we take the byte in 'lab', we increment a3 by one unit (.b) so now a3 points at 'lab'+1 and we put the byte in the low weight byte of d3, so d3=12 (we have put the byte 12 in 'lab').

Then we have:

'dbf d1,fff': d1=d1-1=0 as 0 is different from -1, we jump to 'fff'.

again:

'move.b (a3)+,d3': we take the byte pointed by a3 (here a3 points to the second byte of 'lab', which is 0 since here nothing has been initialized...)and we put it in d3 after having incremented a3 by one unit: d3=0 and a3 points on 'lab'+2='lac'. We arrive at 'dbf d1,fff',again, d1=d1-1=-1, this time we will no more go to 'fff' because d1=-1: we continue.

Then comes 'add.b (a3),-(a3)', a3 points to 'lac' ('lab'+2), we take the data (the byte) pointed by a3 (which is worth 14), we decrement a3 by one unit (.b) and we add it to the new value thus pointed by a3 (which is 0)

finally:

With 'move.b -(a3),d3' , we decrement a3 by one unit (so a3 points at 'lab') and we put the data thus pointed in the low weight byte of d3, so d3=12.

Finally, with 'add.b (a3),d3', we add to d3 the data pointed by a3, as a3 points at 'lab', we get: d3=d3+12=12+12=24

Finally: d3=24. and a3 points at 'lab' ...





nr°10:

What will be the values of d0, d1, d2, d3, a0 after this program? Easy... 

    text

    move.b    #10,a     ;we put the byte 10 in a
    move.b    #11,b     ;            11 in b
    move.b    #12,c     ;            12 in c 
    move.b    #13,d     ;            13 in d 
    move.b    #98,e     ;            98 in e

    move.l    #begin,a0   ;we put the address of 'begin' in a0
                        ;:a0 points to 'begin'
    move.b    (a0)+,d0  ;we put the byte pointed by a0 :0 because nothing is
                        ;initialized in 'begin',(we increment a0 by one 
                        ;unit (.b)), in d0, so:
                        ;d0=0 (byte), a0=a0+1, so a0 points at 'a'.
    move.b    (a0)+,d1  ;same for d1: d1=10, a0 points at 'a'+1='b'
    move.b    (a0)+,d2  ;for d2: d2=11, a0 points at 'b'+1='c'
    move.b    (a0),d3   ;We place the byte pointed by a0 in d3:
                        ;a0 points at 'c', therefore d3=12 
    move.b    -(a0),d4  ;decrement a0 by one unit and put the byte
                        ;thus pointed by a0 in the low weight byte
                        ; of d4:
                        ;as a0 was pointing to 'c', we have now:
                        ;a0='c'-1='b', so d4=11 (we have put 11 in 'b')

    bss

begin ds.b 1 ;reserve 1 byte in begin a ds.b 1 ;in a b ds.b 1 ;in b c ds.b 1 ;in c d ds.b 1 ;in d e ds.b 1 ;in e f ds.l 4 ;reserve in L-M in f 

    end

So: d0=0, d1=10, d2=11, d3=12, d4=11, a0 points at 'b'.


nr°11:

One had to find the value of a4 and d0, d1 after this program. 

    text

    move.b    #3,d0     ;we put the byte 3 in the register d0
    move.l    #stack,a4   ;we put the address of 'stack' in a4

rrr move.b -(a4),d1 ;we decrement a4 by one unit (.b) and put the 

                        ;byte pointed by a4 in the low weight byte of d1
    add.b     #4,d1     ;we add the byte 4 to the low weight byte of d1
    dbf       d0,rrr    ;loop until d0=-1

    data

    dc.b      1,2,3,4,5,6,7,8,9   ;9 bytes are stored here

stack dc.b 245 ;byte 245 at the address 'stack' 

    end


at the end of the program, d0=-1, it is the loop counter...

the loop:

1st pass: d0=3, a4 points to 'stack' with 'move.b -(a4),d1', we decrement a4 by 1 unit (because .b), we take the byte pointed by a4 and we place it in the low weight byte of d1: a4='stack'-1 so a4 is pointing to the byte '9', d1=9+4=13 because we 'add.b #4, d1' We decrement d0 and go back to 'rrr' because d0=2

2nd pass: d0=2, a4 points to the byte '9' The same operation as before: a4 is pointing to the byte '8', d1=8+4=12 (byte), d0=2-1=1

3rd pass: d0=1, a4 points to the byte '8' the same, we obtain: a4 is pointing to the byte '7', d1=7+4=11 (byte), d0=1-1=0

4th pass: d0=0, a4 points to the byte '7' the same, we have: a4 is pointing to the byte '6', d1=6+4=10 (byte), d0=0-1=-1 so we get out of the loop and finally:

d0=-1, d1=10, a4 points to the byte '6'


nr°12:

Let's find what is in 'out' and the value of d2. 

    text

a equ 3 ;we assign 3 to 'a' b equ 5  ; 5 to 'b' 

    move.b    a,d2      ;we put the byte pointed by 'a' in the low weight
                        ;of d2, so d2=3
    add.b     b,d2      ;we add the byte pointed by 'b', so
                        ;d2=d2+5=8 (byte)

    move.l    #g,a5     ;we put the address of 'g' in a5
    move.b    d2,(a5)   ;we put the low weight byte of d2 at 
                        ;the address pointed by a5 (so in 'g') 
    move.b    d2,-(a5)  ;we put the low weight byte of d2 (we decrement
                        ;a5 by one unit) at the new address of a5
                        ;(here, a5 points to 'out', so we will find d2
                        ;i.e. the byte 8 in 'out')
    move.b    (a5)+,d2  ;we put the byte pointed by a5 (so the byte 8)
                        ;(we increment a5 by 1 unit) in the low weight byte
                        ;of d2: d2=8 and a5 points to 'g'
    move.b    (a5),d2   ;we put the byte pointed by a5 in the low weight 
                        ;of d2: a5 points to 'g', in 'g' there is
                        ;the byte 8 so d2=8 once again. 

    bss

out ds.b 1 g ds.b 4 

    end

conclusion: in 'out' there is the byte 8 and d2=8 (byte)


nr°13:

We must find the value of d1, d2, d3, d4 and a0.

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