CS401 Finalterm Paper 2

FINALTERM EXAMINATION

Spring 2010

CS401- Computer Architecture and Assembly Language Programming (Session - 2)

Time: 90 min
Marks: 58
Question No: 1 ( Marks: 1 ) - Please choose one
Suppose AL contains 5 decimal then after two left shifts produces the value as
5
10
15
20
Question No: 2 ( Marks: 1 ) - Please choose one
In graphics mode a location in video memory corresponds to a _____________ on the screen.
line
dot
circle
rectangle
Question No: 3 ( Marks: 1 ) - Please choose one
Creation of threads can be
static
dynamic
easy
difficult
Question No: 4 ( Marks: 1 ) - Please choose one
The thread registration code initializes the PCB and adds it to the linked list so that the
__________ will give it a turn.
assembler
scheduler
linker
debugger
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Question No: 5 ( Marks: 1 ) - Please choose one
VESA VBE 2.0 is a standard for
High resolution Mode
Low resolution Mode
Medium resolution Mode
Very High resolution Mode
Question No: 6 ( Marks: 1 ) - Please choose one
Which of the following gives the more logical view of the storage medium
BIOS
DOS
Both
None
Question No: 7 ( Marks: 1 ) - Please choose one
Which of the following IRQs is derived by a key board?
IRQ 0
IRQ 1
IRQ 2
IRQ 3
Question No: 8 ( Marks: 1 ) - Please choose one
Which of the following IRQs is used for Floppy disk derive?
IRQ 4
IRQ 5
IRQ 6
IRQ 7
Question No: 9 ( Marks: 1 ) - Please choose one
Which of the following pins of a parallel port connector are grounded?
10-18
18-25
25-32
32-39
Question No: 10 ( Marks: 1 ) - Please choose one
The physical address of IDT( Interrupt Descriptor Table) is stored in _______
GDTR
IDTR
IVT
IDTT
Question No: 11 ( Marks: 1 ) - Please choose one
In NASM an imported symbol is declared with the ............................ while and exported
symbol is declared with the ............................
Global directive, External directive
External directive, Global directive
Home Directive, Foreign Directive
Foreign Directive, Home Directive
Question No: 12 ( Marks: 1 ) - Please choose one
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In 68K processors there is a ........................ program counter (PC) that holds the address of
currently executing instruction
8bit
16bit
32bit
64bit
Question No: 13 ( Marks: 1 ) - Please choose one
To reserve 8-bits in memory ___ directive is used.
db
dw
dn
dd
Question No: 14 ( Marks: 1 ) - Please choose one
In the “mov ax, 5” 5 is the __________ operand.
source
destination
memory
register
Question No: 15 ( Marks: 1 ) - Please choose one
RETF will pop the segment address in the
CS register
DS register
SS register
ES register
Question No: 16 ( Marks: 1 ) - Please choose one
For the execution of the instruction “DIV BL”, the implied dividend will be stored in
AX
BX
CX
DX
Question No: 17 ( Marks: 1 ) - Please choose one
When a number is divided by zero ”A Division by 0” interrupt is generated. Which
instruction is used for this purpose
INT 0
INT 1
INT 2
This interrupt is generated automatically
Question No: 18 ( Marks: 1 ) - Please choose one
INT 21 service 01H is used to read character from standard input with echo. It returns the
result in ______ register.
AL
BL
CL
BH
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Question No: 19 ( Marks: 1 ) - Please choose one
BIOS sees the disks as
logical storage
raw storage
in the form of sectors only
in the form of tracks only
Question No: 20 ( Marks: 1 ) - Please choose one
In 9pin DB 9, which pin number is assigned to CD (Carrier Detect) ?
1
2
3
4
Question No: 21 ( Marks: 1 ) - Please choose one
In 9pin DB 9, Signal ground is assigned on pin number
4
5
6
3
Question No: 22 ( Marks: 1 ) - Please choose one
In 9pin DB 9, RI (Ring Indicator) is assigned on pin number
6
7
8
9
Question No: 23 ( Marks: 1 ) - Please choose one
Motorola 68K processors have ....................... 23bit general purpose registers.
4
8
16
32
Question No: 24 ( Marks: 1 ) - Please choose one
When two devices in the system want to use the same IRQ line then what will happen?
An IRQ Collision
An IRQ Conflict
An IRQ Crash
An IRQ Blockage
Question No: 25 ( Marks: 1 ) - Please choose one
In the instruction MOV AX, 5 the number of operands are
1
2
3
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4
Question No: 26 ( Marks: 1 ) - Please choose one
Which flags are NOT used for mathematical operations ?
Carry, Interrupt and Trap flag.
Direction, Interrupt and Trap flag.
Direction, Overflow and Trap flag.
Direction, Interrupt and Sign flag.
Question No: 27 ( Marks: 2 )
How can we improve the speed of multitasking?
Ans:
We can iµprove the speed of µultitasking by changing the frequency of tiµer interrupt
.
Question No: 28 ( Marks: 2 )
Write instructions to do the following. Copy contents of memory location with offset 0025
in the current data segment into AX.
Ans:
€
Mov ax , [0025]
€
µov[0fff], ax
€
µov€ ax , [0010]
€€€µov [002f] , ax
Question No: 29 ( Marks: 2 )
Write types of Devices?
Ans:
There are two types devices used in pc.
1. Input devices(keyboard, µouse,)
2. Output devices.(µonitor, printer)
Question No: 30 ( Marks: 2 )
What dose descriptor 1st 16 bit tell?
Ans:
Each segµent is describe by the descriptor like
1. base,
2. liµit,
3. and attributes,
it basically define the actual base address.
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Question No: 31 ( Marks: 3 )
List down any three common video services for INT 10 used in text mode.
Ans:
INT 10 - VIDEO - SET TEXT-MODE CURSOR SHAPE
AH = 01h
CH = cursor start and options
CL = bottom scan line containing cursor (bits 0-4)
Question No: 32 ( Marks: 3 )
How to create or Truncate File using INT 21 Service?
Ans:
INT 21 - TRUNCATE FILE
AH = 3Ch
CX = file attributes
DS:DX -> cs401 filename
Return:
CF = error flag
AX = file handle or error code
Question No: 33 ( Marks: 3 )
How many Types of granularity also name them?
Ans:
There are three types of granuality :
1. Data Granularity
2. Business Value Granularity
3. Functionality Granularity
Question No: 34 ( Marks: 5 )
How to read disk sector into memory using INT 13 service?
Ans:
INT 13 - DISK - READ SECTOR(S) INTO MEMORY :
AH = 02h
AL = number of sectors to read (must be nonzero)
CH = low eight bits of cylinder number
CL = sector number 1-63 (bits 0-5)
high two bits of cylinder (bits 6-7, hard disk only)
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DH = head number
DL = drive number (bit 7 set for hard disk)
ES:BX -> data buffer
Return:
CF = error flag
AH = error code
AL = number of sectors transferred
Question No: 35 ( Marks: 5 )
The program given below is written in assembly language. Write a program in C to
call this assembly routine.
[section .text]
global swap
swap: mov ecx,[esp+4] ; copy parameter p1 to ecx
mov edx,[esp+8] ; copy parameter p2 to edx
mov eax,[ecx] ; copy *p1 into eax
xchg eax,[edx] ; exchange eax with *p2
mov [ecx],eax ; copy eax into *p1
ret ; return from this function
Ans:
The above code will asseµble in c through this coµµand. Other aurwise er
ror will occur.
Nasµ−f win32 swap .asµ
This coµµand will generate swap.obj file.
The code for given prograµ will be as follow.
#include <stdio.h>
Void swap(int* pl, int* p2);
Int µain()
{
Int a=10,
Int b= 20;
Print f (“ a=%d b=%d\n” , a ,b);
Swap (&a ,&b);
Print f (“ a=%d b=%d\n” , a ,b);
Systeµ ( “ pause” );
Return 0;
}
Question No: 36 ( Marks: 5 )
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Write the code of “break point interrupt routine”.
Ans:
Breakpoint interrupts service routine :
debugISR: push bp
mov bp, sp ; …………….to read cs, ip and flags
push ax
push bx
push cx
push dx
push si
push di
push ds
push es
sti ;…………………….. waiting for keyboard interrupt
push cs
pop ds ;…………………… initialize ds to data segment
mov ax, [bp+4]
mov es, ax ; ………………….load interrupted segment in es
dec word [bp+2] ; ……………….decrement the return address
mov di, [bp+2] ;………………… read the return address in di
mov word [opcodepos], di ;…………. remember the return position
mov al, [opcode] ; …………..load the original opcode
mov [es:di], al ;………….. restore original opcode there
mov byte [flag], 0 ; …………set flag to wait for key
call clrscr ;……………. clear the screen
mov si, 6 ; …………..first register is at bp+6
mov cx, 12 ;………… total 12 registers to print
mov ax, 0 ; …………..start from row 0
mov bx, 5 ; ………….print at column 5
push ax ; ………………..row number
push bx ;………………. column number
mov dx, [bp+si]
push dx ;………………. number to be printed
call printnum ;…………….. print the number
sub si, 2 ; ……………….point to next register
inc ax ; ………………..next row number
loop l3 ; ……………….repeat for the 12 registers
mov ax, 0 ; ………………..start from row 0
mov bx, 0 ; ………………..start from column 0
mov cx, 12 ; …………………..total 12 register names
mov si, 4 ;……………………. each name length is 4 chars
mov dx, names ; …………………..offset of first name in dx
push ax ;………………………. row number
push bx ; ………………………column number
push dx ; ……………………….offset of string
push si ; ………………………….length of string
call printstr ; ………………………….print the string
add dx, 4 ;………………………….. point to start of next string
inc ax ; ……………………………new row number
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loop l1 ;…………………………….. repeat for 12 register names
or word [bp+6], 0x0100 ; ……………………set TF in flags image on stack
keywait: cmp byte [flag], 0 ;……………………. has a key been pressed
je keywait ; ………………….. no, check again
pop es
pop ds
pop di
pop si
pop dx
pop cx
pop bx
pop ax
pop bp
iret
start: xor ax, ax
mov es, ax ; ……………………point es to IVT base
mov word [es:1*4], trapisr ;…………………. store offset at n*4
mov [es:1*4+2], cs ; …………………...store segment at n*4+2
mov word [es:3*4], …………………..debugisr ; store offset at n*4
mov [es:3*4+2], cs ; …………………..store segment at n*4+2
cli ; ………………….disable interrupts
mov word [es:9*4], kbisr ; ………………….store offset at n*4
mov [es:9*4+2], cs ; ……………………...store segment at n*4+2
sti ; ………………………enable interrupts

By : ADEEL ABBAS 

www.allvupastpapers.blogspot.com 

AdeelAbbasbk@gmail.com