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 1st 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

Leave a Reply

Related Posts Plugin for WordPress, Blogger...