CS401 Final Term Paper Solved 2010 - 3

 FINALTERM  EXAMINATION

Spring 2010

CS401- Computer Architecture and Assembly Language Programming

(Session - 2)

Time: 90 min

M a r k s: 58

    

Question No: 1    ( M a r k s: 1 )    http://vuzs.net

 Suppose AL contains 5 decimal then after two left shifts produces the value as

       ► 5         

       ► 10

       ► 15

       ► 20

   

Question No: 2    ( M a r k s: 1 )    http://vuzs.net

 In graphics mode a location in video memory corresponds to a _____________ on the screen.

       ► line

       ► dot

       ► circle

       ► rectangle

   

Question No: 3    ( M a r k s: 1 )    http://vuzs.net

 Creation of threads can be

       ► static

       ► dynamic

       ► easy

       ► difficult

   

Question No: 4    ( M a r k s: 1 )    http://vuzs.net

 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

   

Question No: 5    ( M a r k s: 1 )    http://vuzs.net

 VESA VBE 2.0 is a standard for

       ► High resolution Mode

       ► Low resolution Mode

       ► Medium resolution Mode

       ► Very High resolution Mode

   

Question No: 6    ( M a r k s: 1 )    http://vuzs.net

 Which of the following gives the more logical view of the storage medium

       ► BIOS

       ► DOS

       ► Both

       ► None

   

Question No: 7    ( M a r k s: 1 )    http://vuzs.net

 Which of the following IRQs is derived by a key board?

       ► IRQ 0

       ► IRQ 1

       ► IRQ 2

       ► IRQ 3

   

Question No: 8    ( M a r k s: 1 )    http://vuzs.net

 Which of the following IRQs is used for Floppy disk derive?

       ► IRQ 4

       ► IRQ 5

       ► IRQ 6

       ► IRQ 7

   

Question No: 9    ( M a r k s: 1 )    http://vuzs.net

 Which of the following pins of a parallel port connector are grounded?

       ► 10-18      

       ► 18-25     

       ► 25-32   

       ► 32-39

   

Question No: 10    ( M a r k s: 1 )    http://vuzs.net

 The physical address of IDT( Interrupt Descriptor Table) is stored in _______

       ► GDTR

       ► IDTR

       ► IVT

       ► IDTT

   

Question No: 11    ( M a r k s: 1 )    http://vuzs.net

 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    ( M a r k s: 1 )    http://vuzs.net

 In 68K processors there is a ........................ program counter (PC) that holds the address of currently executing instruction

       ► 8bit

       ► 16bit

       ► 32bit

       ► 64bit

   

Question No: 13    ( M a r k s: 1 )    http://vuzs.net

 To reserve 8-bits in memory ___ directive is used.

       ► db

       ► dw

       ► dn

       ► dd

   

Question No: 14    ( M a r k s: 1 )    http://vuzs.net

 In the “mov ax, 5”     5 is the __________ operand.

       ► source

       ► destination

       ► memory

       ► register

   

Question No: 15    ( M a r k s: 1 )    http://vuzs.net

 RETF will pop the segment address in the

       ► CS register

       ► DS register

       ► SS register

       ► ES register

   

Question No: 16    ( M a r k s: 1 )    http://vuzs.net

 For the execution of the instruction “DIV  BL”, the implied dividend will be stored in

       ► AX 

       ► BX

       ► CX 

       ► DX

   

Question No: 17    ( M a r k s: 1 )    http://vuzs.net

 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    ( M a r k s: 1 )    http://vuzs.net

 INT 21 service 01H is used to read character from standard input with echo. It returns the result in  ______ register.

       ► AL

       ► BL

       ► CL

       ► BH

   

Question No: 19    ( M a r k s: 1 )    http://vuzs.net

 BIOS sees the disks as

       ► logical storage

       ► raw storage

       ► in the form of sectors only

       ► in the form of tracks only

   

Question No: 20    ( M a r k s: 1 )    http://vuzs.net

 In 9pin DB 9, which pin number is assigned to CD (Carrier Detect) ?

       ► 1

       ► 2

       ► 3

       ► 4

   

Question No: 21    ( M a r k s: 1 )    http://vuzs.net

 In 9pin DB 9, Signal ground is assigned on pin number

       ► 4

       ► 5

       ► 6

       ► 3

   

Question No: 22    ( M a r k s: 1 )    http://vuzs.net

 In 9pin DB 9, RI (Ring Indicator) is assigned on pin number

       ► 6

       ► 7

       ► 8

       ► 9

   

Question No: 23    ( M a r k s: 1 )    http://vuzs.net

 Motorola 68K processors have ....................... 23bit general purpose registers.

       ► 4

       ► 8

       ► 16

       ► 32

   

Question No: 24    ( M a r k s: 1 )    http://vuzs.net

 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    ( M a r k s: 1 )    http://vuzs.net

 In the instruction  MOV AX, 5 the number of operands are

       ► 1

       ► 2

       ► 3

       ► 4

   

Question No: 26    ( M a r k s: 1 )    http://vuzs.net

 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    ( M a r k s: 2 )

 How can we improve the speed of multitasking?

Ans:

We can improve the speed of multitasking by changing the frequency of timer interrupt.

   

Question No: 28    ( M a r k s: 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]

mov[0fff], ax

mov  ax , [0010]

   mov [002f] , ax

     

Question No: 29    ( M a r k s: 2 )

 Write types of Devices?

Ans:

There are two types devices used  in pc.

1. Input devices(keyboard, mouse,)
2. Output devices.(monitor, printer)

    

Question No: 30    ( M a r k s: 2 )

 What dose descriptor 1^st 16 bit tell?

Ans:

Each segment is describe by the descriptor like

1. base,
2. limit,
3. and attributes,

it  basically define the actual base address.

    

Question No: 31    ( M a r k s: 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    ( M a r k s: 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    ( M a r k s: 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    ( M a r k s: 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)

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    ( M a r k s: 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 assemble in c through this command. Other aurwise error will occur.

Nasm-f win32 swap .asm

This command will generate swap.obj file.

The code for given program will be as follow.

#include <stdio.h>

Void swap(int* pl, int* p2);

Int main()

{

      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);

System ( “pause”);

Return 0;

}

    

Question No: 36    ( M a r k s: 5 )

 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

              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