Q#: 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]
mov[0fff], ax
mov ax , [0010]
mov [002f] , ax
Q#: 29 ( Marks: 2 )
Write types of Devices?
Ans:
There are two types devices used in pc.
- Input devices(keyboard, mouse,)
- Output devices.(monitor, printer)
Q#: 30 ( Marks: 2 )
What dose descriptor 1st 16 bit tell?
Ans:
Each segment is describe by the descriptor like
- base,
- limit,
- and attributes,
it basically define the actual base address.
Q#: 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)
Q#: 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
Q#: 33 ( Marks: 3 )
How many Types of granularity also name them?
Ans:
There are three types of granuality :
- Data Granularity
- Business Value Granularity
- Functionality Granularity
Q#: 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)
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
Q#: 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 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;
}
Q#: 36 ( Marks: 5 )
Write the code of “break point interrupt routine”.
Ans:
[org 0x0100]
jmp start
flag: db 0 ; ………………flag whether a key pressed
oldisr: dd 0 ; ……………..space for saving old ISR
names: db 'FL =CS =IP =BP =AX =BX =CX =DX =SI =DI =DS =ES ='
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
Q#: 1 ( Marks: 1 ) :-
SP is associated with…………. By default
► SS
► DS
► CS
► ES
Q#: 2 ( Marks: 1 ) :-
Which bit of the attributes byte represents the red component of foreground color
► 5
► 4
► 3
► 2
Q#: 3 ( Marks: 1 ) :-
An 8 x 16 font is stored in ______________ bytes.
► 2
► 4
► 8
► 16
Q#: 4 ( Marks: 1 ) :-
In DOS input buffer, the number of characters actually read on return is stored in ___________ byte.
► third
► fourth
► first
► second
Q#: 5 ( Marks: 1 ) :-
Which of the following gives the more logical view of the storage medium
► BIOS
► DOS
► Both
► None
Q#: 6 ( Marks: 1 ) :-
In STOSW instruction, when DF is clear, SI is
► Incremented by 1
► Incremented by 2
► Decremented by 1
► Decremented by 2
Q#: 7 ( Marks: 1 ) :-
Which of the following interrupts is Non maskable interrupt
► INT 2
► INT 3
► INT 0
► INT 1
Q#: 8 ( Marks: 1 ) :-
Which of the following IRQs is connected to serial port COM 2?
► IRQ 0
► IRQ 1
► IRQ 2
► IRQ 3
Q#: 9 ( Marks: 1 ) :-
The time interval between two timer ticks is ?
► 40ms
► 45ms
► 50ms
► 55ms
