CS401 Computer Architecture and Assembly Language Programming Assignment No. 01

[Vuhelper]

CS401 Computer Architecture and Assembly Language Programming Assignment No. 01 Discussion and Solution Fall 2014 Due Date Nov 24, 2014


We have the following register values:

AX = 420AH,

BX = A12EH,

DS = 22EAH,

CS = 0100H,

ES = 0FEAH,

SS = 0FE0H,

SI = 0010,

BP = 02ACH and

DI = 0300H.



You are required to calculate the physical address generated by each of the following instructions?



mov [bx+si+di],ax
ADD AX,[BP+12]


Note: Every instruction is independent of others.





Question No. 2: (10 marks)



Assemble the given program using NASM.
[ORG 0X0100]

MOV CX, 5

XOR AX, AX

MOV BX, NUM



L1: ADD AX, [BX]

ADD BX, 2

SUB CX, 1

JNZ L1



MOV [SUM], AX

MOV AX, 0X4C00

INT 0X21



NUM: DW 0, 1, 2, 3, 4.

SUM: DW 0

[ATTACH=CONFIG]24151[/ATTACH]







After that load the Debugger by typing AFD in the command prompt like in Fig 1.




Fig 1: Loading debugger.



The AFD window will be opened. The complete description is given below in fig 2
Fig 2: AFD description.



When the program is loaded in the debugger, it is loaded at offset 0100, which displaces all memory accesses in our program. Execute the program step by step and examine how the memory is read and the registers are updated, how the instruction pointer moves forward.


Now we execute the program with F2 key (instruction will move one by one) and write down the contents of the specified registers in the given table





Instruction
Register current value after the instruction execution
IP

(Instruction Pointer)
Flags

OF
CF
ZF
PF
SF
AF
1


[ORG 0X0100]
AX









BX


CX


DX


2


MOV CX, 10


AX









BX


CX


DX


3


XOR AX, AX


AX









BX


CX


DX


4


MOV BX, NUM


AX









BX


CX


DX


5


ADD AX, [BX]


AX











BX


CX


DX


6


ADD BX, 2


AX









BX


CX


DX


7


SUB CX, 1


AX









BX


CX


DX


8


JNZ L1


AX









BX


CX


DX


9


ADD AX, [BX]


AX











BX


CX


DX


10


ADD BX, 2


AX









BX


CX


DX


11


SUB CX, 1


AX









BX


CX


DX


12


JNZ L1


AX









BX


CX


DX


13


ADD AX, [BX]


AX











BX


CX


DX


14


ADD BX, 2


AX









BX


CX


DX


15


SUB CX, 1


AX









BX


CX


DX


16


JNZ L1


AX









BX


CX


DX


17


ADD AX, [BX]


AX











BX


CX


DX


18


ADD BX, 2


AX









BX


CX


DX


19


SUB CX, 1


AX









BX


CX


DX


20


JNZ L1


AX









BX


CX


DX


21


ADD AX, [BX]


AX











BX


CX


DX


22


ADD BX, 2


AX









BX


CX


DX


23


SUB CX, 1


AX









BX


CX


DX


24


JNZ L1


AX



BX


CX


DX


25


MOV [SUM], AX


AX





BX

CX

DX

26


MOV AX, 0X4C00


AX





BX

CX

DX

27


INT 0X21


AX






BX

CX

DX


Table: Register contents after the execution of each instruction.

[education-2014]

We have the following register values:
AX = 420AH,
BX = A12EH,
DS = 22EAH,
CS = 0100H,
ES = 0FEAH,
SS = 0FE0H,
SI = 0010,
BP = 02ACH and
DI = 0300H.

You are required to calculate the physical address generated by each of the following instructions?

mov [bx+si+di],ax
ADD AX,[BP+12]

Solution:
1.mov [bx+si+di],ax
EA= A12EH+0010+ 0300
=AF24H
Physical address = segment*0x10+EA
= DS +EA
=22EAH + AF24H
=0x321934



ADD AX,[BP+12]
EA = 420AH , [02ACH+12]
= ADD 420AH, 02424
= 44434

Physical address = segment*0x10+EA
=SS+EA
= 0FE0H+44434
=0x45948
is it rite?