‫االختبار النهائي‬
1434/1435 ‫للفصل الدراسي الثاني للعام الجامعي‬
225 CS Computer Organizations & Assembly Lang. Prog.
709 :‫شعبة‬
‫ ساعتان‬:‫الزمن‬
4:‫المستوى‬
‫ درجة‬50 :‫الدرجة العظمي‬
‫جامعة سلمان بن عبد العزيز‬
‫كلية اآلداب والعلـوم بوادي الدواسر‬
‫قســـم علوم الحاسب و المعلومات‬
Note: Answer all questions.
Part A: (Knowledge Base Questions)
[10*2=20M]
1. Differentiate between computer architecture and computer organization?
Answer:
Organization is how features are implemented
– Control signals, interfaces, memory technology.
– e.g. Is there a hardware multiply unit or is it done by repeated addition?
Computer organization is a study of a Computer Architecture. E.g. Memory, Registers,
RAM, ROM, CPU, ALU, 16 bit/ 32 bit/ 64 bit architecture, what different parts makes a
computer, etc.
Computer Architecture: Refers to specification of the relationship between different
hardware components of a computer system. It defines high and low level of abstractions.
At high level CPU is presented and low level its parts like ALU, MU, CU represented
2. Define the following terms
a. Compiler:
b. Interpreter.
c. Assembler.
Compiler: This converts high level language program written in English to low level language
program written in binary 1’s & 0’s.
Interpreter: This converts high level language program to low level language.
Difference between compiler & Interpreter
Compiler converts whole program one at a time but whereas interpreter converts line by line.
Assembler: It is used to translate assembly level language into low level language. Assembly
language written in mnemonics.
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3. Write the full form of the following instructions
a. BIU: Bus Interface Unit
b. INTR: Interrupt
c. JNZ: Jump No Zero
d. SF: sign Flag
4. List few addressing modes sets available in Intel X86 PC?
Answer: Addressing modes.
Input
Output
Figure 1.3.
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

Memory
Single-bus structure.
Implied
Register
Immediate
Direct
Register indirect
Based
Indexed
Based Indexed
Based Indexed with displacement
Processor
5. Outline the structure of computer bus?
Answer: Computer Bus Structure
6. Name some word processing operations?
Answer: Word Processing Operations
Crating new document, saving the document, modifications, templates, inserting
graphics, several of data, forms reports generations.
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7. Describe about parameter passing mechanisms using stacks?
Answer: In STACK PUSH operation is made to push the contents of registers over the
top of the stack. While deleting also the contents of top of stack will be POPed in to the
registers. But in assembly languages there is no concept of parameter passing
mechanisms however to communicate the input values to be sent or receive will be done
by storing in various storage devices.
Input: Stacks which receive information
Output: This stack produces some information
Uses: Which registers to used.
8. State following statements are true or false
a. The jump instructions can be divided into conditional and unconditional
Jumps.
[True]
b. Computer bus used to transfer the data between input & output peripherals of
[True]
Computers.
c. The five logical instructions are AND, OR,NOT,EX-OR, NOR
[True]
d. Queue allows First in First out (F.I.F.O.) process.
[True]
9. Discriminate between high and low level languages?
Answer: High Level Language: High level languages are written in English language
which the user can understand. Where as the low level languages will be written in only
binary language which only computer can understand.
10. Write the format for IF ELSE, REPEAT, FOR loop used in assembly language
programming?
Answer:
END_FOR
MOV CX,80 ; number of stars to display
MOV AH,2
;Display character function
MOV DL,’*’ ;Character to display
TOP: INT 21H ;Char to display star
LOOP TOP
•
MOV AH,1
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•
REPEAT:
•
INT 21H
CMP AL,’ ‘
JNE REPEAT
MOV Dx,0
;DX counts the chars
MOV AH,1;Prepares to read
INT 21H
;chars in AL
WHILE_:
CMP Al,0DH ;CR?
JE END_WHILE; yes, exit
INC DX ;not CR increment
INT 21H
JMP
; read a character
WHILE_ ; loop back
END_WHILE
IF_ElSE LOOP
– Program to display the AL, BL values
MOV AH,2
;Display to prepare
CMP AL,BL ;If AL>=BL
JNBE ELSE_ ;NO display char if BL
MOV DL,AL
JMP DISPLAY;Go to Display
MOV DL,BL
PART B: Interpersonal Skills & Responsibility
[ 4*2.5=10M ]
11. Sketch the neat diagrams for interrupts used by the processor?
Answer:
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12. Illustrate the neat architecture of internal organization of Intel x86 PC?
Answer:
13. Show the neat architecture for Slice coding interface with languages such as
Supreme language CNN?
Answer:
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14. Evaluate the following addressing modes with a neat architecture
a. SUB (BX). [ Register Indirect addressing modes]
b. ADD AX [ Direct Addressing mode]
Answer: SUB (BX).
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C.f. indirect addressing
EA = (BX) First referencing
BX conatains the address of CX which contains the value.
The content of CX added with AX and result stores in the AX. AX AX+CX
Operand is in memory cell pointed to by contents of register R
Large address space (2n)
One fewer memory access than indirect addressing
ADD AX [ Direct Addressing mode]
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Address field contains address of operand
Effective address (EA) = address field (A)
e.g. ADD A
P.T.O. 
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Add contents of cell A to accumulator
Look in memory at address A for operand
Single memory reference to access data
No additional calculations to work out effective address
Limited address space
PART C: Cognitive skills
[ 4*2.5=10M ]
15. Interpret following instruction and write the result
a. CMP AL, BL. [ AL = 5, BL=5]. If AL>=BL NO display char if BL
a. CASE AX [AX=4] .
Answer:
– Program to display the AL, BL values
•
MOV AH,2 ;Display to prepare
•
CMP AL,BL ;If AL>=BL
•
JNBE ELSE_
•
MOV DL,AL
•
JMP DISPLAY;Go to Display
•
MOV DL,BL
;NO display char if BL
16. Design the architecture for Indirect Addressing modes?
Answer:
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17. Design CALL & RET operations for below given procedure and represent with a neat
architecture? Before CALL
Offset Address
Code Segment
MAIN PROC
Stack Segment
Offset Address
0010
I.P0012
CALL PROC1
Next Instruction
00FC
PROC 1
0200
00FE
First Instruction
0100
 S.P.
RET
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P.T.O. 
After CALL
Offset Address
Code Segment
MAIN PROC
Stack Segment
Offset Address
0010
I.P0012
CALL PROC1
Next Instruction
00FC
PROC 1
00FE
0012
 S.P.
First Instruction
0200
0100
RET
Before RET
Offset Address
Code Segment
MAIN PROC
Stack Segment
Offset Address
0010
0012
CALL PROC1
Next Instruction
00FC
PROC 1
00FE
0012
 S.P.
First Instruction
0200
0100
RET
I.P. 0300
After RET
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P.T.O. 
Offset Address
Code Segment
MAIN PROC
Stack Segment
Offset Address
0010
I.P. 0012
CALL PROC1
Next Instruction
00FC
PROC 1
00FE
First Instruction
0200
0100
 S.P.
RET
18. Predict the use of following instruction sets
a. SAL & SAR: Shift Arithmetic Left & Shift Arithmetic Right.
b. JA & JAE: Jump if Above & Jump Above or Equal.
c. CAL & RET: CALL & RETURN of procedures.
d. MOVSB & MOVSW: Move Store Byte & Move Store Word.
PART D: Communication, Information Technology Numerical skills
[ 4*2.5=10M ]
19. Calculate following number base conversions.
a. (1F0C)16 = (0011 1111 0000 1100 )8
b. (10101)2+(11001)2=(X)2
10101
21
+ 11001 + 25
101110
46
20. If the String 1 =”Hello” & String 2 =”World”, then find then length of both the strings?
Answer: String1= Hello String 2 = World Strcat= String1+ String 2 Hello+World=
HelloWorld
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21. If a magnetic disc has 100 cylinders, each containing 10 tracks of 10 sectors, and each
sector can contain 128 bytes, what is the maximum capacity of the disk in bytes show
with the help of a neat architecture?
Answer: Magnetic disc has 100 cylinders.
Each containing 10 tracks of 10 sectors.
Sector can contain 128 bytes
What is the maximum capacity of the disk in bytes?
100 * 10 * 10 * 128 =280000
22. Write the program for implementing the stack applications for reversing a given
integer?
Answer: ;
.MODEL
SMALL
.STACK
100h
.CODE
MAIN
PROC
; display user prompt
MOV
AH, 2
; prepare to display
MOV
DL, '?'
; char to display
INT
21h
; Display '?'
; initialize counter
XOR
CX,CX
; count = 0
; input a character
MOV
AH,1
; read character function
INT
21h
; character into AL
;while char is not a carriage return do
while1:
cmp
al, 0dh
;CR?
je
ewhile
; yes exit loop
; save character on the stack
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P.T.O. 
push
ax
; push it on stack
inc
cx
; count = count + 1
; read a character
int
21h
jmp
while1
ewhile:
; goto new line
mov
ah,2
mov
dl,0dh
int
21h
mov
dl, 0ah
int
21h
jcxz
exit
; CR
LF
;exit if no char read
; For count times do
TOP:
; pop a character from the stack
pop
dx
; display it
int
21h
loop
top
;end for
exit:
MOV
AH,4CH
INT
21h
MAIN
END
ENDP
MAIN
Course Instructor
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Supervisor of Department
P.T.O. 
Lecturer: Mujthaba Gulam Muqeeth
Signature: …………….
Dr. Saied M. Abd –El –atty
Signature: …………….
------Wish U Best of Luck------
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