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Posts Tagged ‘VU CS401- Computer Architecture and Assembly Language Programming FinalTerm solved unsolved past papers Spring 2010’

VU CS401- Computer Architecture and Assembly Language Programming FinalTerm solved unsolved past papers Spring 2010

VU CS401- Computer Architecture And Assembly Language Programming FinalTerm Solved Unsolved Past Papers Spring 2010

 

FINALTERM  EXAMINATION

Spring 2010

CS401- Computer Architecture and Assembly Language Programming

(Session – 3)

Time: 90 min

Marks: 58

 

Question No: 1   ( Marks: 1 )   – Please choose one

 

SP is associated with…………. By default

 

SS

 

DS CS ES

 

Question No: 2   ( Marks: 1 )   – Please choose one

 

Which bit of the attributes byte represents the red component of foreground color

 

5 4 3

2

 

Question No: 3   ( Marks: 1 )   – Please choose one

 

An 8 x 16 font is stored in ______________ bytes.

 

2 4 8

16

 

Question No: 4   ( Marks: 1 )   – Please choose one

 

In DOS input buffer, the number of characters actually read on return is stored in

___________ byte.

 

third fourth first

 

second

 

Question No: 5   ( Marks: 1 )   – Please choose one

 

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

BIOS

 

DOS

 

Both None

 

Question No: 6   ( Marks: 1 )   – Please choose one

 

In STOSW instruction, when DF is clear, SI is

 

Incremented by 1

 

Incremented by 2

 

Decremented by 1

 

 

 

 

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Decremented by 2

 

 

Question No: 7   ( Marks: 1 )   – Please choose one

 

Which of the following interrupts is Non maskable interrupt

 

INT 2

 

INT 3 INT 0 INT 1

 

Question No: 8   ( Marks: 1 )   – Please choose one

 

Which of the following IRQs is connected to serial port COM 2?

 

IRQ 0 IRQ 1 IRQ 2

 

IRQ 3

 

Question No: 9   ( Marks: 1 )   – Please choose one

 

The time interval between two timer ticks is ?

 

40ms 45ms 50ms

 

55ms

 

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
In 68K processors there is a 32bit …………………. that holds the address of currently
executing instruction
Program counter
Stack pointer
Register
Stack
Question No: 13 ( Marks: 1 ) - Please choose one

 

 

 

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Single step interrupt is

Hardware interrupt

 

Like divide by zero interrupt

 

Like divide by 1 interrupt Software interrupt

 

Question No: 14   ( Marks: 1 )   – Please choose one

 

Which of the following is NOT true about registers:

Their operation is very much like memory

 

Intermediate results may also be stored in registers They are also called scratch pad ram

 

None of given options

 

Question No: 15   ( Marks: 1 )   – Please choose one

 

Types of jump are:

short, near

 

short, near, far

 

near, far short, far

 

Question No: 16   ( Marks: 1 )   – Please choose one

 

MS DOS uses ____ display mode.

 

Character based

 

Graphics based Numeric based Console based

 

Question No: 17   ( Marks: 1 )   – Please choose one

 

Which of the following IRQs is derived by a timer device?

 

IRQ 0

 

IRQ 1 IRQ 2 IRQ 3

 

Question No: 18   ( Marks: 1 )   – Please choose one

 

In programmable interrupt controller, which of the following ports is referred as a control port.

 

19

 

20

 

21 22

 

Question No: 19   ( 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

 

Question No: 20   ( Marks: 1 )   – Please choose one

 

 

 

 

 

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In 9pin DB 9, which pin number is assigned to DSR (DataSet Ready) ?

 

4 5

6

7

 

Question No: 21   ( Marks: 1 )   – Please choose one

 

In 9pin DB 9, which pin number is assigned to TD (Transmitted Data) ?

 

1 2

3

4

 

Question No: 22   ( Marks: 1 )   – Please choose one

 

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

4

 

5

 

6 3

 

Question No: 23   ( Marks: 1 )   – Please choose one

 

8088 is a ………………………

 

16 bit processor

 

32 bit processor 64 bit processor 128 bit processor

 

Question No: 24   ( Marks: 1 )   – Please choose one

 

The table index (TI) is set to _____ to access the GDT (Global Descriptor Table).

 

1

 

0

 

-1 -2

 

Question No: 25   ( Marks: 1 )   – Please choose one

 

VESA(Video Electronics Standards Association) organizes 16 color bits for every pixel in

 

5:5:5 format

 

5:6:5 format

 

6:5:6 format 5:6:7 format

 

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 )

 

Write instruction to allocate space for 32 PCBs.

 

 

 

 

 

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“INT 10 – VESA – Get SuperVGA Information” uses:
VESA service

eagle_eye

 

Ans:

 

multitasking kernel as a TSR [org 0x0100]

 

jmp start

 

PCB layout: ax,bx,cx,dx,si,di,bp,sp,ip,cs,ds,ss,es,flags,next,dummy 0, 2, 4, 6, 8,10,12,14,16,18,20,22,24, 26 , 28 , 30

 

 

Question No: 28   ( Marks: 2 )

 

Define short jump

 

Ans;

 

The jump is called a short jump, If the offset is stored in a single byte as in 75F2 with the opcode 75 and operand F2, the jump is called a short jump. F2 is added to IP as a signed byte

 

Question No: 29   ( Marks: 2 )

 

 

INT 14 – SERIAL – READ CHARACTER FROM PORT uses which two 8bit registers to return the results ?

 

 

Ans;

 

14 – SERIAL – READ CHARACTER FROM PORT uses these two 8bit registers to return the results:

 

AH = line status

 

AL = received character if AH bit 7 clear

 

Question No: 30   ( Marks: 2 )

 

Which registers are uses as scratch when we call a function?

 

Ans:

Following registers are uses as scratch when we call a function

 

  • EAX

 

  • ECX

 

  • EDX

 

 

 

 

 

Question No: 31   ( Marks: 3 )

 

“INT 10 – VESA – Get SuperVGA Information” uses which registers to

 

return the result?

 

To return the result,

Return:

 

AL = 4Fh if function supported

 

AH = status

 

Question No: 32   ( Marks: 3 )

 

Define the protected mode.

 

 

 

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When the processor switches into 32bit mode it is called protected mode. It can be accessed by turning on least significant bit of a register called CR0

 

(Control Register 0) and the processor switches into 32bit mode.

 

All registers in 386 have been extended to 32bits. The new names are EAX,

 

EBX,

ECX,

EDX,

 

ESI,

EDI,

ESP,

 

EBP, EIP, and EFLAGS.

 

The original names refer to the lower 16bits of these registers. A 32bit address register can access upto 4GB of memory so memory access has increased a lot.

 

 

 

 

Question No: 33   ( Marks: 3 )

 

Describe briefly INT 3 functionality.

 

The functionality of INT 3 is this , its Debug Interrupt. The special thing about this interrupt is that it has a single byte opcode and not a two byte combination where the second byte tells the interrupt number which allows it to replace any instruction what soever. It is also used by the debugger.

 

 

Question No: 34   ( Marks: 5 )

 

Read the passage carefully and choose proper word for each blank space from the list given below .

 

In descriptors the 32bit base is scattered into different places because of compatibility reasons. The limit is stored in 20 bits but the ……………defines that the limit is in terms of bytes of 4K pages therefore a maximum of 4GB size is possible.

 

The …………….. must be set to signal that this segment is present in memory. DPL

is the descriptor privilege level again related to the protection levels in 386.

……………… defines that this segment is to execute code is 16bit mode or 32bit

mode. ……………… is conforming bit that we will not be using. ………………signals

that the segment is readable. A bit is automatically set whenever the

segment is accessed.

 

(A bit, C bit, G bit, D bit, P bit , R bit, B bit)

 

SOLUTION:

 

In descriptors the 32bit base is scattered into different places because of compatibility reasons. The limit is stored in 20 bits but the …….G bit……..defines that the limit is in terms of bytes of 4K pages therefore a maximum of 4GB size is possible. The …….P bit………. must be set to signal that this segment is present in memory. DPL is the descriptor privilege level again related to the protection levels

 

in 386. ……..D bit ………. defines that this segment is to execute code is 16bit mode
or 32bit mode. ……… C  ……… is conforming bit that we will not be using. ……. R
bit……….. signals that the segment is readable. A bit is automatically set whenever
the segment is accessed.

 

 

 

 

 

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Question No: 35   ( Marks: 5 )

 

Answer the following:

 

§ What is a device driver? Ans:

 

These are operating system extensions which become part of the operating system and extend its services to new devices. Device drivers in

 

DOS are very simple. They just have their services exposed through the file system interface.

 

Device driver file starts with a header containing a link to the next driver in the first four bytes followed by a device attribute word. The most important bit in the device attribute word is bit 15 which dictates if it is a character device or a block device.

 

If the bit is zero the device is a character device and otherwise a block device.

 

Next word in the header is the offset of a strategy routine, and then is the offset of the interrupt routine and then in one byte, the number of units supported is stored. This information is padded with seven zeroes.

 

  • Strategy routine is called whenever the device is needed

 

  • it is passed a request header. Request header stores the unit requested, the command

 

  • code, space for return value and buffer pointers etc. Important command codes include

 

  1. 0 to initialize,
  2. 1 to check media,
  3. 2 to build a BIOS parameter block,
  4. 4 and 8 for read and write respectively.

 

For every command the first 13 bytes of request header are same.

 

  • Why are device drivers necessary, given that the BIOS already has code that communicates with the computer’s hardware?

 

Ans:

 

These are used for the reason of fast programming execution. device driver takes some RAM and expresses it as a secondary storage device to the operating system. Therefore a new drive is added and that can be browsed to, filed copied to and from just like ordinary drives expect that this drive is very fast as it is located in the RAM. This program cannot be directly executed since it is not a user program. This must be loaded by adding the line “device=filename.sys” in the

 

“config.sys” file in the root directory.

 

 

Question No: 36   ( Marks: 5 )

 

Write the code of “break point interrupt routine”.

 

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
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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

 

 

 

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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

 

 

 

 

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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eagle_eye

 

 

 

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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eagle_eye

 

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?

 

Ανσ:

 

Ωε χαν ιµπροϖε τηε σπεεδ οφ µυλτιτασκινγ βψ χηανγινγ τηε φρεθυενχψ οφ τιµερ ιντερρυπτ

 

.

 

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.

 

Ανσ:

 

 

Μοϖ αξ , [0025]

 

 

µοϖ[0φφφ], αξ

 

 

µοϖ€ αξ , [0010]

 

€€€µοϖ [002φ] , αξ

 

 

 

 

 

Question No: 29   ( Marks: 2 )

 

Write types of Devices?

 

Ανσ:

 

Τηερε αρε τωο τψπεσ δεϖιχεσ υσεδ  ιν πχ.

 

  1. Ινπυτ δεϖιχεσ(κεψβοαρδ, µουσε,)

 

  1. Ουτπυτ δεϖιχεσ.(µονιτορ, πριντερ)

 

 

 

 

Question No: 30   ( Marks: 2 )

 

What dose descriptor 1st 16 bit tell?

 

 

Ανσ:

 

Εαχη σεγµεντ ισ δεσχριβε βψ τηε δεσχριπτορ λικε

 

  1. βασε,

 

  1. λιµιτ,

 

  1. ανδ αττριβυτεσ,

 

ιτ  βασιχαλλψ δεφινε τηε αχτυαλ βασε αδδρεσσ.

 

 

 

 

 

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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?

Ανσ:

Τηερε αρε τηρεε τψπεσ οφ γρανυαλιτψ :

  1. Δατα  ρανυλαριτψ
  2. Βυσινεσσ ςαλυε  ρανυλαριτψ
  3. Φυνχτιοναλιτψ  ρανυλαριτψ

 

 

 

 

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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eagle_eye

 

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

 

 

Ανσ:

 

Τηε αβοϖε χοδε ωιλλ ασσεµβλε ιν χ τηρουγη τηισ χοµµανδ. Οτηερ αυρωισε ερ ρορ ωιλλ οχχυρ.

Νασµ−φ ωιν32 σωαπ .ασµ

 

Τηισ χοµµανδ ωιλλ γενερατε σωαπ.οβϕ φιλε. Τηε χοδε φορ γιϖεν προγραµ ωιλλ βε ασ φολλοω.

 

#ινχλυδε <στδιο.η> ςοιδ σωαπ(ιντ∗ πλ, ιντ∗ π2); Ιντ µαιν()

 

{

 

Ιντ α=10, Ιντ β= 20;

Πριντ φ (  α=%δ β=%δ∴ν  , α ,β);

 

Σωαπ (&α ,&β);

 

Πριντ φ (  α=%δ β=%δ∴ν  , α ,β);

 

Σψστεµ (  παυσε  );

 

 

Ρετυρν 0;

 

 

 

}

 

 

 

 

 

Question No: 36   ( Marks: 5 )

 

 

 

 

 

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eagle_eye

 

 

 

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

 

 

Cs401  2010

Question No: 1   ( Marks: 1 )   – Please choose one

 

Which feature of database provides conversion from inconsistent state of DB to a consistent state ensuring minimum data loss?

 

User accessible catalog

 

Data processing

 

Authorization service

 

Recovery service

 

 

Question No: 2   ( Marks: 1 )   – Please choose one

 

Which of the following statements is true about the views?

 

view is always a complete set of all the tables in a database View can not be used for retrieving data

 

The results of using a view are not permanently stored in the database.

 

Rows can not be updated or deleted in the view

 

 

Question No: 3   ( Marks: 1 )   – Please choose one

 

 

 

 

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Which of the following is true about TRUNCATE?

 

Can be Rolled back.

 

Activates Triggers.

 

is DML Command.

 

Resets identity of the table.

 

 

Question No: 4   ( Marks: 1 )   – Please choose one

 

Which of the following is the correct way to find out the size of cartesian product incase of CROSS JOIN?

 

the number of columns in the first table multiplied by the number of columns in the second table.

 

the number of columns in the first table multiplied by the number of rows in the second table.

 

the number of rows in the first table multiplied by the number of columns in the first table.

 

the number of rows in the first table multiplied by the number of rows in the second table.

 

 

Question No: 5   ( Marks: 1 )   – Please choose one

 

Suppose there are 8 rows and 4 columns in TABLE1 and 3 rows and 4 coulmns in TABLE2; what is the size of the cartesian product incase of CROSS JOIN between these two tables?

 

24 32 12 16

 

Question No: 6   ( Marks: 1 )   – Please choose one

 

Which of the following is not one of the properties of Transaction?

 

atomicity consistency redundancy durability

 

Question No: 7   ( Marks: 1 )   – Please choose one

 

Which of the following is INCORRECT about VIEWS?

It is not possible to left out the data which is not required for a specific view.

 

A database view displays one or more database records on the same page.

 

Views can be used as security mechanisms

 

Views are generally used to focus the perception each user

has of the database.

 

 

Question No: 8   ( Marks: 1 )   – Please choose one

 

 

 

 

 

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Each course section is assigned a particular faculty member, and each course section corresponds to a particular course. Conceptually, what is the relationship between faculty and course (not course section).

 

1:1 1:M M:M

 

Ternary

 

Question No: 9   ( Marks: 1 )   – Please choose one

 

Which of the following is used to add or drop columns in an existing table?

 

 

ALTER HAVING SELECT THEN

 

Question No: 10   ( Marks: 1 )   – Please choose one

 

Which of the following is a correct way of selecting all the columns from a table called PERSONS?

 

SELECT FROM * Persons

 

SELECT * FROM Persons

 

SELECT * WHERE Persons

 

SELECT WHERE * Persons

 

 

Question No: 11   ( Marks: 1 )   – Please choose one

 

Which of the following is NOT a feature of Indexed sequential files?

 

Records are stored in sequence and index is maintained.

 

Dense and nondense types of indexes are maintained.

 

Track overflows and file overflow areas can not be ensured.

 

Cylinder index increases the efficiency

 

 

Question No: 12   ( Marks: 1 )   – Please choose one

 

Consider the given relations Student and Instructor as given below. Please note that Fname and Lname also denote the First Name and Last Name respectively.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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Which of the following statements is correct with respect to the two relations given above?

 

The two relations are not union-compatible since their attribute names differ.

 

The two relations are union-compatible since they have the same type of tuples.

 

The set operations such as CARTESIAN PRODUCT and DIVISION can be applied on these

 

two relations.

 

To find out the students who are not instructors, it is necessary to perform the operation

 

Student ÷ Instructor.

 

 

Question No: 13   ( Marks: 1 )   – Please choose one

 

Which of the following serves as a milestone or reference point in the log file?

 

Constraints

 

Relations

 

Check points

 

Transactions identities

 

 

Question No: 14   ( Marks: 1 )   – Please choose one

 

Which of the following is not true regarding DB transactions?

 

A set of database operations that are processed partly

A database transaction is a logical unit of database operations

 

A database transaction must be atomic

 

A database transaction must contains the ACID property

 

 

Question No: 15   ( Marks: 1 )   – Please choose one

 

Which of the following are the general activities, which are performed during the development of application programs?

 

Data input programs Editing

 

Display

 

All of given

 

 

Question No: 16   ( Marks: 1 )   – Please choose one

 

Browser based forms are developed in the following tools EXCEPT

 

HTML

 

Scripting language

 

 

 

 

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Front Page

 

Web-based Forms

 

 

Question No: 17   ( Marks: 1 )   – Please choose one

 

Which of the following is not a form of optical disk?

 

CD ROM

 

WORM

 

Erasable Optical

 

EEPROM

 

 

Question No: 18   ( Marks: 1 )   – Please choose one

 

Which of the following is the correct description of cache hit?

 

When data is found in the cache

 

When data is removed in the cache

 

The number of times the cache is accessed directly by the processor

 

When data is lost from the cache

 

Question No: 19   ( Marks: 1 )   – Please choose one

 

In which of the following situations, Clustering is suitable:

 

Frequently updating Relatively static

 

Relatively deletion Relatively dynamic

 

Question No: 20   ( Marks: 1 )   – Please choose one

 

Only one type of constraint can be enforced in any table by CREATE command

 

True False

 

Question No: 21   ( Marks: 1 )   – Please choose one

 

Which of the following is disadvantage of chaining technique to handle the collisions?

 

Unlimited Number of elements Fast re-hashing

 

Overhead of multiple linked lists

Maximum number of elements must be known

 

Question No: 22   ( Marks: 1 )   – Please choose one

 

Consider the following relation R and its sample data. (Consider that these are the only tuples for the given relation)

 

 

 

 

 

 

 

 

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Which of the following statements is NOT correct?

 

The functional dependency ProjNo -> DeptNo holds over R.

 

The functional dependency (EmpNo, ProjNo) -> DeptNo holds over R. The functional dependency DeptNo -> ProjNo holds over R.

 

The functional dependency EmpNo -> DeptNo holds over R.

 

Question No: 23   ( Marks: 1 )   – Please choose one

 

An entity type is

 

defined when the database is actually constructed

a specific type such as an integer, text, date, logical etc

 

a coherent set of similar objects that we want to store data on (e.g. STUDENT, COURSE, CAR)

 

defined by the database designer

 

Question No: 24   ( Marks: 1 )   – Please choose one

 

An entity can be logically connected to another by defining a ____.

hyperlink

 

common attribute primary key

 

superkey

 

Question No: 25   ( Marks: 1 )   – Please choose one

 

You can’t modify more than one table at a time through a view.

 

True False

 

Question No: 26   ( Marks: 1 )   – Please choose one

 

Which of the following is one of the purposes of using DML commands?

 

 

 

 

Creating databases Destroying databases Inserting data in tables Non of the above

 

Question No: 27   ( Marks: 2 )

 

 

 

 

 

Question No: 28   ( Marks: 2 )

 

Define domain of an attribute.

 

 

 

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Ans:

 

Domain is the set of possible values that an attribute can have, that is, we specify a set of values either in the form of a range or some discrete values, and then attribute can have value out of those values. Domain is a form of a check or a constraint on attribute that it cannot have a value outside this set.

 

 

Question No: 29   ( Marks: 2 )

 

Write the main feature of volatile storage media?

 

storage that is lost when the power is turned off

 

Question No: 30   ( Marks: 2 )

 

Suppose you want to delete a table row by row and record an entry in the transaction log for each deleted row. Which DML command will you use?

 

DELETE * FROM student WHERE name=”Abrar”;

 

Question No: 31   ( Marks: 3 )

 

Write three benefits of using VIEWS.

 

Views are generally used to focus, simplify, and customize the perception each user has of the database. Views can be used as security mechanisms by allowing users to access data through the view, without granting the users

 

permissions to directly access the underlying base tables of the view.Views allow users to focus on specific data that interests them and on the specific tasks for which they are responsible. Unnecessary data can be left out of the view. This also increases the security of the data because users

 

 

Question No: 32   ( Marks: 3 )

 

SELECT * FROM Persons

 

WHERE FirstName LIKE ‘%da%';

 

what does the above statement return?

 

Ans:

 

Question No: 33   ( Marks: 3 )

 

What is the difference between a primary key and a unique key with reference to clustered and nonclustered indexes?

 

 

Question No: 34   ( Marks: 5 )

 

Consider a table named COMPANY with fields COMPANY_NAME,

 

DESCRIPTION, ORDER_NUMBER. Write an SQL statement to display company names in reverse alphabetical order.

 

SELECT COMPANY_NAME FROM COMPANY ORDER BY COMPANY_NAME

DESC;

 

Question No: 35   ( Marks: 5 )

 

Name the five main components of Database management systems software.

 

 

 

Question No: 36   ( Marks: 5 )

 

Give 4 similarities between Materialized views and indexes.

 

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  1. They consume storage space.

 

  1. They must be refreshed when the data in their master tables changes.

 

III. They improve the performance of SQL execution when they are used for IV. query rewrites.

 

  1. Their existence is transparent to SQL applications and users.

 

Assembly Language Paper – CS401  Paper attempted : 22 Feb 2010 at 05:00 PM

 

***********************************************************

1. BL contains 5 decimal then after right shift , BL will become

 

  • 3

 

  • 2.5

 

  • 5

 

  • 10

 

2. 8 * 16 font is stored in ________ bytes.

 

  • 3

 

  • 4

 

  • 8

 

  • 16

 

3. In DOS input buffer , number of characters actually read on return is stored in

 

  • First byte

 

  • Second byte

 

  • Third byte

 

  • Fourth byte

 

4. IRQ 0 has priority

 

  • Low

 

  • High

 

  • Highest

 

  • Medium

 

5. Thread registration code initialize PCB and add to linked list so that _____ will give it turn.

 

  • Assembler

 

  • Linker

 

  • Scheduler

 

  • Debugger

 

6. Traditional calling conventions are in ______ number

 

  • 1

 

  • 2

 

  • 3

 

  • 4

 

7. VESA VEB 2.0 is standard for

 

  • High Resolution Mode

 

  • Low Resolution Mode

 

  • Very High Resolution Mode

 

  • Medium Resolution Mode

 

8. To clear direction flag which instruction is used

 

  • Cld

 

  • Clrd

 

 

 

 

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  • Cl df

 

  • Clr df

 

9. In STOSW instruction , When DI is cleared , SI is

 

  • Incremented by 1

 

  • Incremented by 2

 

  • Decremented by 1

 

  • Decremented by 2

 

10. Interrupt that is used in debugging with help of trap flag is

 

  • INT 0

 

  • INT 1

 

  • INT 2

 

  • INT 3

 

11. INT for arithmetic overflow is

 

  • INT 1

 

  • INT 2

 

  • INT 3

 

  • INT 4

 

12. IRQ referred as

 

  • Eight Input signals

 

  • One Input signal

 

  • Eight Output signals

 

  • One output signal

 

 

13. IRQ for keyboard is ____1_____

 

14. IRQ for sound card is ______5_______

 

15. IRQ for floppy disk is ______6_______

 

16. IRQ with highest priority is

 

  • Keyboard IRQ

 

  • Timer IRQ

 

  • Sound Card

 

  • Floppy Disk

 

 

17. Pin for parallel port ground is

 

  • 10-18

 

  • 18-25

 

  • 25-32

 

  • 32-39

 

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

 

  • GDTR

 

  • IDTR

 

  • IVT

 

  • IDTT

 

19. Execution of “RET 2” results in?

 

 

 

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20. CX register is

 

  • Count register

 

  • Data register

 

  • Index register

 

  • Base register

 

21. OUT instruction uses __AX_____ as source register.

 

22. IN DB-9 connector the Data Set ready pin is at

 

  • 5

 

  • 6

 

  • 7

 

  • 8

 

23. If two devices uses same IRQ then there is

 

  • IRQ collision

 

  • IRQ conflict

 

  • IRQ drop

 

24. VESA organizes 16 bit color for every pixel in ratio

 

  • 5:5:5

 

  • 5:6:5

 

  • 6:5:6

 

  • 5:6:7

 

25. Division by zero is done by which interrupt.

Interrupt 0.

 

26. Define Hardware Interrupt & I/O ports (5 marks)

 

27. Five BIOS video services used in text mode ( 3 marks)

INT 10 – VIDEO – SET TEXT-MODE CURSOR SHAPE

 

AH = 01h

CH = cursor start and options

CL = bottom scan line containing cursor (bits 0-4)

INT 10 – VIDEO – SET CURSOR POSITION

 

AH = 02h

 

BH = page number 0-3 in modes 2&3 0-7 in modes 0&1 0 in graphics modes

 

DH = row (00h is top) DL = column (00h is left)

INT 10 – VIDEO – SCROLL UP WINDOW

 

AH = 06h

 

AL = number of lines by which to scroll up (00h = clear entire window) BH = attribute used to write blank lines at bottom of window

 

CH, CL = row, column of window’s upper left corner DH, DL = row, column of window’s lower right corner

INT 10 – VIDEO – SCROLL DOWN WINDOW

 

AH = 07h

 

AL = number of lines by which to scroll down (00h=clear entire window) BH = attribute used to write blank lines at top of window

 

CH, CL = row, column of window’s upper left corner DH, DL = row, column of window’s lower right corner

 

INT 10 – VIDEO – WRITE CHARACTER AND ATTRIBUTE AT CURSOR

 

 

 

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POSITION

 

AH = 09h

AL = character to display

BH = page number

BL = attribute (text mode) or color (graphics mode)

CX = number of times to write character

 

 

 

28. DOS allocate memory for program execution and then de-allocate , explain memory management in DOS (10 marks)

 

An important point to understand here is that whenever a program is executed in DOS all available memory is allocated to it. No memory is available to execute any new programs. Therefore memory must be freed using explicit calls to DOS for this purpose before a program is executed.

 

Important services in this regard are listed below.

INT 21 – ALLOCATE MEMORY

 

AH = 48h

 

BX = number of paragraphs to allocate Return:

 

CF = error flag

 

AX = segment of allocated block or error code in case of error BX = size of largest available block in case of error

INT 21 – FREE MEMORY

 

AH = 49h

 

ES = segment of block to free Return:

 

CF = error flag AX = error code

INT 21 – RESIZE MEMORY BLOCK

 

AH = 4Ah

 

BX = new size in paragraphs ES = segment of block to resize Return:

 

CF = error flag AX = error code

 

BX = maximum paragraphs available for specified memory block

INT 21 – LOAD AND/OR EXECUTE PROGRAM

 

AH = 4Bh

AL = type of load (0 = load and execute)

 

DS:DX -> ASCIZ program name (must include extension) ES:BX -> parameter block

 

Return:

 

CF = error flag AX = error code

 

The format of parameter block is as follows. Offset Size Description

 

00h WORD segment of environment to copy for child process (copy caller’s environment if 0000h)

 

02h DWORD pointer to command tail to be copied into child’s PSP 06h DWORD pointer to first FCB to be copied into child’s PSP 0Ah DWORD pointer to second FCB to be copied into child’s PSP

 

0Eh DWORD (AL=01h) will hold subprogram’s initial SS:SP on return 12h DWORD (AL=01h) will hold entry point (CS:IP) on return

 

 

There was fill in blanks question with 10 marks. The choice was given at bottom.

 

29. Serial Port is also accessible via I/Oports , COM 1    is accessible via ports
3F8-3FF while   COM 2    is accessible via 2F8 -2FF.

 

 

 

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The first register at 3F8 is the Transmitter holding register if written to and the receiver buffer register if read from.

 

Other register of our interest include 3F9 whose           Bit  0   must be set to enable received

 

data available interrupt and Bit 1 must be set to enable transmitter holding register empty interrupt.

 

( Transmitter , COM 1, I/O ports , COM2. bit 0 , Buffer , 3FA)

 

 

 

 

 

 

FINAL TERM EXAMINATION SPRING 2010 CS401 COMPUTER ARCHITECTURE AND ASSEMBLY

 

LANGUAGE PROGRAMMING

 

9 AUG 2010

 

 

Question No: 1 ( Marks: 1 ) – Please choose one

 

When a 32 bit number is divided by a 16 bit number, the quotient is of

 

  • 32 bits

 

  • 16 bits

 

  • 8 bits

 

  • 4 bits

 

Question No: 2 ( Marks: 1 ) – Please choose one

 

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

  • 1

 

  • 2

 

  • 3

 

  • 4

 

Question No: 3 ( Marks: 1 ) – Please choose one

3. In DOS input buffer , number of characters actually read on return is stored in

 

  • First byte
  • Second byte

 

  • Third byte
  • Fourth byte

 

Question No: 4 ( Marks: 1 ) – Please choose one
7. VESA VEB 2.0 is standard for
· High Resolution Mode
· Low Resolution Mode
·   Very High Resolution Mode
· Medium Resolution Mode
Question No: 5 ( Marks: 1 ) – Please choose one
22. IN DB-9 connector the Data Set ready pin is at
· 5
· 6
· 7
· 8
Question No: 6 ( Marks: 1 ) – Please choose one
Threads can have function calls, parameters and variables.
· global
· local

 

 

 

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  • legal
  • illegal

 

Question No: 7 ( Marks: 1 ) – Please choose one How many prevalent calling conventions do exist

  • 1

 

  • 2
  • 3
  • 4

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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Question No: 8 ( Marks: 1 ) – Please choose one

 

In 9pin DB 9 DSR is assigned on pin number

·

4

·

5

·

6

·

7

Question No: 9 ( Marks: 1 ) – Please choose one
In 9pin DB 9 CTS is assigned on pin number

·

6

·

7

·

8

·

9

Question No: 10 ( Marks: 1 ) – Please choose one
In 9pin DB 9 CD is assigned on pin number

·

1

·

2

·

3

·

4

 

Question No: 11 ( Marks: 1 ) – Please choose one

 

A 32bit address register can access upto ……………………….of memory so memory

 

access has increased a lot.

 

·     2GB

 

·     4GB

 

·     6GB

 

·     8GB

 

Question No: 12 ( Marks: 1 ) – Please choose one

 

in device attribute word which of the following bit decides whether it is a charater device or a block device

  • Bit 12
  • Bit 13

 

  • Bit 14
  • Bit 15

 

Question No: 13 ( Marks: 1 ) – Please choose one

 

9. Which of the following IRQ is cascading interrupt

 

  • IRQ 0

 

  • IRQ 1

 

  • IRQ 2

 

  • IRQ 3

 

Question No: 14 ( Marks: 1 ) – Please choose one

Which of the following interrupts is used for Arithmetic overflow

 

  • INT 1

 

  • INT 2

 

  • INT 3
  • INT 4

 

Question No: 15 ( Marks: 1 ) – Please choose one

 

An End of Interrupt (EOI) signal is sent by

  • Handler

 

 

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  • Processor

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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  • IRQ
  • PIC

 

Question No: 16 ( Marks: 1 ) – Please choose one

 

The number of pins in a parallel port connector are?

  • 20

 

  • 25
  • 30
  • 35

 

Question No: 17 ( 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: 18 ( Marks: 1 ) – Please choose one

 

A 32bit address register can access upto …………………….. of memory so memory

access has increased a lot.

 

·     2GB

·     4GB

 

·     6GB

·     8GB

 

Question No: 19 ( Marks: 1 ) – Please choose one 9 Pin Serial connector is called

  • DB-7

 

  • DB-9
  • DB-25

 

  • 9DB-5

 

Question No: 20 ( 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: 21 ( Marks: 2 )

Write brief about INT 13 – Extended READ SERVICES

 

Question No: 22 ( Marks: 2 )

 

What is Interrupt flag?

 

Question No: 23 ( Marks: 3 )

Give the name of any two descriptors

 

Question No: 24 ( Marks: 3 )

 

It is the part of Multitasking TSR caller, what will do these instructions comment against them (3)

 

Mov al, [chars+bx] Mov [es:40],al

 

Inc bx

 

 

 

 

 

 

 

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Question No: 25 ( Marks: 5 )

Write Data Movement and Arithmetic Instructions of Motorola 68 K Processor.

 

Question No: 26 ( Marks: 5 )

 

Write assembly program for “Break Interrupt Service Routine”

 

 

NOTE

 

MUST PREPARE LAST 10 LESSONS WELL. MOSTLY THE PAPER WAS FROM THEIR. ESPECIALLY LAST THREE LESSONS.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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Today’s CS401 Exam

 

final 2010 spring

 

 

 

REPLIED BY: MALIK RIZWAN ALI Question No: 1 ( Marks: 1 ) – Please choose one The physical address of the stack is obtained by

 

SS:SI combination

 

 

 

 

SS:SP combination

 

 

 

 

ES:BP combination

 

 

 

 

ES:SP combination

 

 

 

 

 

 

 

Question No: 2 ( Marks: 1 ) – Please choose one

 

 

 

 

 

 

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Value of AH in the write Graphics pixel service is

 

0Ch

 

0Bh

 

1Ch

 

2Ch

 

 

 

 

Question No: 3 ( Marks: 1 ) – Please choose one

 

Threads can have function calls, parameters and __________ variables.

 

global

 

local

 

legal

 

illegal

 

 

 

 

Question No: 4 ( Marks: 1 ) – Please choose one

 

Creation of threads can be

 

static

 

dynamic

 

easy

 

difficult

 

 

 

 

Question No: 5 ( Marks: 1 ) – Please choose one

 

How many prevalent calling conventions do exist

 

1

 

 

 

 

 

 

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2

 

3

 

4

 

 

 

 

Question No: 6 ( 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: 7 ( Marks: 1 ) – Please choose one

 

The serial port connection is a —————— connector

 

9pin DB 9

 

8pin DB 9

 

3pin DB 9

 

9pin DB 5

 

 

 

 

Question No: 8 ( Marks: 1 ) – Please choose one

 

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

 

BIOS

 

DOS

 

Both

 

 

 

 

 

 

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None

 

 

 

 

Question No: 9 ( Marks: 1 ) – Please choose one In STOSB instruction, when DF is clear, SI is

 

 

Incremented by 1

 

 

 

 

Incremented by 2

 

 

 

 

Decremented by 1

 

 

 

 

Decremented by 2

 

 

 

 

 

 

 

Question No: 10 ( Marks: 1 ) – Please choose one After the execution of STOSW the CX will be

 

 

Decremented by 1

 

 

 

 

Decremented by 2

 

 

 

 

Incremented by 1

 

 

 

 

 

 

 

 

 

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Incremented by 2

 

 

 

 

 

 

 

Question No: 11 ( Marks: 1 ) – Please choose one IRQ is referred to

 

 

 

Eight input signals

 

 

 

 

One output signal

 

 

 

 

One input signals

 

 

 

 

Eight output signals

 

 

 

 

 

 

 

Question No: 12 ( 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: 13 ( Marks: 1 ) – Please choose one

 

 

 

 

 

 

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Which of the following IRQs is connected to serial port COM 1?

 

IRQ 4

 

IRQ 5

 

IRQ 6

 

IRQ 7

 

 

 

 

Question No: 14 ( Marks: 1 ) – Please choose one

 

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

 

GDTR

 

IDTR

 

IVT

 

IDTT

 

 

Question No: 15 ( Marks: 1 ) – Please choose one

 

Assembly language is:

 

Low-level programming language

 

High-level programming language

 

Also known as machine language

 

Not considered closer to the computer

 

 

 

 

Question No: 16 ( Marks: 1 ) – Please choose one

 

The number of bits required to access 1MB of memory are

 

16 bits

 

32 bits

 

 

 

 

 

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Depends on the processor architecture

 

20 bits

 

 

 

 

Question No: 17 ( Marks: 1 ) – Please choose one

 

In STOSB instruction, SI is decremented or incremented by

 

 

 

 

3

 

2

 

1

 

4

 

 

 

 

Question No: 18 ( Marks: 1 ) – Please choose one

 

In programmable interrupt controller, which of the following ports is referred as a control port.

 

19

 

20

 

21

 

22

 

 

 

 

Question No: 19 ( 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

 

 

 

 

 

 

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CL

 

BH

 

 

 

 

Question No: 20 ( Marks: 1 ) – Please choose one

 

In device attribute word, which of the following bit decides whether it is a character device or a block device

 

Bit 12

 

Bit 13

 

Bit 14

 

Bit 15

 

 

 

 

Question No: 21 ( Marks: 1 ) – Please choose one

 

In 9pin DB 9, which pin number is assigned to CTS (Clear To Send) ?

 

6

 

7

 

8

 

9

 

 

 

 

Question No: 22 ( Marks: 1 ) – Please choose one

 

In 9pin DB 9, which pin number is assigned to RD (Received Data) ?

 

1

 

2

 

3

 

4

 

 

 

 

 

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Question No: 23 ( Marks: 1 ) – Please choose one

 

VESA(Video Electronics Standards Association) organizes 16 color bits for every pixel in

 

5:5:5 format

 

5:6:5 format

 

6:5:6 format

 

5:6:7 format

 

 

 

 

Question No: 24 ( Marks: 1 ) – Please choose one

 

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

 

4

 

8

 

16

 

32

 

 

 

 

Question No: 25 ( Marks: 1 ) – Please choose one

 

Programmable Interrupt Controller (PIC) has

 

 

 

 

One input signals and eight output signals

 

 

 

 

One input signal and one output signal

 

 

 

 

Eight input signals and one output signals

 

 

 

 

 

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Eight input signals and eight output signals

 

 

 

 

 

 

Question No: 26 ( Marks: 1 ) – Please choose one

 

Video services are classified into…………………. broad categories.

 

5

 

4

 

3

 

2

 

 

 

 

Question No: 27 ( Marks: 2 )

 

What are device drivers? give your answer in two to three lines. Device drivers are operating system extensions that become part of the

 

operating system and extend its services to new devices. Device drivers in DOS are very simple. They just have their services exposed through the file system interface.

 

Device driver file starts with a header containing a link to the next driver in the first four bytes followed by a device attribute word. The most important bit in the device attribute word is bit 15 which dictates if it is a character device or a block device. If the bit is zero the device is a character device and otherwise a block device. Next word in the header is the offset of a strategy routine, and then is the offset of the interrupt routine and then in one byte, the number of units supported is stored. This information is padded with seven zeroes.

 

Strategy routine is called whenever the device is needed and it is passed a request header. Request header stores the unit requested, the command code, space for return value and buffer pointers etc. Important command codes include 0 to initialize, 1 to check media, 2 to build a BIOS parameter block, 4 and 8 for read and write respectively. For every command the first 13 bytes of request header are same.

 

 

 

 

 

 

 

 

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Question No: 28 ( Marks: 2 )

 

For what purpose “INT 1″ is reserved ? INT 1 vector occupies location 4,

 

5, 6, and 7 INT 1, Trap, Single step Interrupt

 

This interrupt is used in debugging with the trap flag. If the trap flag is set the Single Step Interrupt is generated after every instruction. By hooking this interrupt a debugger can get control after every instruction and display the registers etc. 8088 was the first processor that has this ability to support debugging.

 

 

 

 

 

 

Question No: 29 ( Marks: 2 )

 

How interrupts are handled in protected mode.

Switching processor in the newer 32bit mode is a very easy

 

task. Just turn on the least significant bit of a new register called CR0 (Control Register 0) and the processor switches into 32bit mode called protected mode. However manipulations in the protected mode are very

 

different from those in the read mode. Handling interrupts in protected mode is also different. Instead of the IVT

 

at physical address 0 there is the IDT (interrupt descriptor table) located at physical address stored in IDTR, a special purpose register. The IDTR is also a 48bit register similar in structure to the GDTR and loaded with another special instruction LGDT.

 

 

 

 

 

 

Question No: 30 ( Marks: 2 )

 

Which bit of acknowledge is used to generate IRQ7 Pin 10,

 

the ACK pin, is normally used by the printer to acknowledge the receipt of data and show the willingness to receive more data. Signaling this pin generates IRQ 7 if enabled in the PIC and in the parallel port controller. Pin 18-25 are ground and must be connected to the external circuit ground to provide the common reference point otherwise they won’t understand each other voltage levels.

 

 

 

 

 

 

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Question No: 31 ( Marks: 3 )

 

Write the name three flags which are not used for mathematical operations. The three flags not used

 

for mathematical operations are the direction flag, the interrupt flag and the trap flag.

 

 

 

 

Question No: 32 ( Marks: 3 )

 

“INT 13 – DISK – GET DRIVE PARAMETERS ” uses which registers to return error flag and error number.

 

INT 13 – DISK – GET DRIVE PARAMETERS AH = 08h

 

DL = drive (bit 7 set for hard disk) Return:

 

CF = error flag AH = error code

 

 

 

 

Question No: 33 ( Marks: 3 )

 

Who is responsible for removing the parameter from the stack when we call a function in C and Pascal?

 

In C the caller removes the parameter while in Pascal the callee removes them. The C scheme has reasons pertaining to its provision for variable number of arguments.

 

 

 

 

 

Question No: 34 ( Marks: 5 )

 

Read the passage carefully and choose proper word for each blank space from the list given below .

 

 

 

 

In descriptors the 32bit base is scattered into different places because of compatibility reasons.

The limit is stored in 20 bits but the …………… defines that the limit is in terms of bytes of 4K
pages therefore a maximum of 4GB size is possible. The …………….. must be set to signal that

 

 

 

 

 

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this segment is present in memory. DPL is the descriptor privilege level again related to the

 

protection levels in 386. ……………… defines that this segment is to execute code is 16bit mode or
32bit mode. ……………… is conforming bit that we will not be using. ……………… signals that the
segment is readable. A bit is automatically set whenever the

 

segment is accessed.

 

 

 

 

(A bit, C bit, G bit, D bit, P bit , R bit, B bit)

 

 

 

 

Question No: 35 ( Marks: 5 )

 

Write assembly language instructions to set the timer interrupt frequency at 1 ms.

 

 

 

 

Question No: 36 ( Marks: 5 )

 

In the context of ” INT 13 – DISK – WRITE DISK SECTOR(S)” fill the blanks by choosing the correct answer against each blank space from the list given at the bottom.

 

 

 

 

AH = ………………..

 

AL = ………………………….

 

CH = …………………………

 

CL = sector number 1-63 (bits 0-5)

 

high two bits of cylinder (bits 6-7, hard disk only)

 

DH = ……………………………….

 

DL = drive number (bit 7 set for hard disk)

 

ES:BX -> …………………………………

 

 

 

 

 

 

 

 

 

 

 

 

 

VU CS401- Computer Architecture And Assembly Language Programming FinalTerm Solved Unsolved Past Papers Spring 2010

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