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CS609 System Programming Solved MCQs + Solved Subjective Questions For Final Term Exam

CS609 System Programming Solved MCQs + Solved Subjective Questions For Final Term Exam

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CS609 System Programming Solved MCQs + Solved Subjective Questions For Final Term Exam

What are the differences between hardware and software interrupts?

 

Software interrupts are invoked by means of some software instruction or statement and hardware interrupt is invoked by means of some hardware controller generally. The only difference between them is the method by which they are invoked

 

How would a procedure written for software interrupt will be different from that written for hardware interrupt?

 

Write a program that will turn on/off the speaker and connect it with the interval timer whenever

Ctrl+Alt+S is pressed Timer interrupt

#include <dos.h>

#include <bios.h>

void interrupt (*oldint15) ( );

void interrupt newint15 (unsigned int BP, unsigned int DI, unsigned int SI, unsigned int DS, unsigned int ES, unsigned int DX, unsigned int CX, unsigned int BX, unsigned int AX, unsigned int IP, unsigned int CS, unsigned int flags);

void main ( )

{

oldint15 = getvect (0x15); setvect (0x15, newint15); keep (0, 1000);

}

void interrupt newint15( unsigned int BP, unsigned int DI, unsigned int SI, unsigned int DS, unsigned int ES, unsigned int DX, unsigned int CX, unsigned int BX, unsigned int AX, unsigned int CS, unsigned int IP, unsigned int flags)

{

if (_AH = = 0x4F)

{

if (_AL == 0x1F)

{

outport (0x43, 0xB4); outport (0x42, 0xFF); outport (0x42, 0x21);

outport (0x61, inport(0x61) ^ 3);

}

}

else

(*oldint15) ( );

}

CS609 MCQs For Papers

What is the difference between a simple program and a TSR program? How can we stop multiple loading of a TSR program into memory?

TSR (Terminate and Stay Resident) program is a program which programs the interrupt number

65H but in this case the new interrupt 65H function remains in memory even after the termination of the program and hence the vector of int 65h does not become a dangling pointer. A TSR need to be loaded once in memory. Multiple loading will leave redundant copies in memory So we need to have some check which will load the program only once

 

 

How interrupts are processed? List down five differences between hardware and software interrupts.

Software interrupts are invoked by means of some software instruction or statement and hardware interrupt is invoked by means of some hardware controller generally.

 

5w.


hatisnulmlodem2marks

 

If data is to transferred from one computer to another through some media which can carry digital data then the modem can be eliminated and the UART on both computers can be interconnected. Such arrangement is called a NULL modem

 

 

6t.ypesofinterrupts3marks

 

Interrupt means to break the continuity of some ongoing task. When we talk of computer interrupt we mean exactly the same in terms of the processor. When an interrupt occurs the continuity of the processor is broken and the execution branches to an interrupt service routine.Two types of interrupts are:

Software interrupts

Hardware Interrupts.

DMA stands for DirecMt emoryAcce s .

 

REGS is a               

 

Size of IVT is  1024 bytes

 

Set the Interrupt vector means to change the double word sized interrupt vector within IVT.

 

T/F

 

 

 

Display device (Monitor) performs  memory mapped I/O. NMI stands for       .

CS609 MCQs For Papers

If keyboard buffer is empty the head and tail points at the same location.  T/F

 

Tail of keyboard should get to get  the start of buffer.

 

Interrupt  65H is empty.

 

Interval time is used to divide input frequency. T/F

 

An I/O device cannot be directly connected to the busses so  controller is placed between CPU

and I/O.

 

Standard PC can have             PPI.

1, 4, 8, 16

 

The BIOS int 0x1Ah can be used to configure RTC. Keyboard uses port 64H as status port.

Communication b/w keyboard and keyboard controller is                         Asynchronous serial

Synchronouserial

Parallel

None

0xF3 means typematic rate will be sent in next byte. By cascading two DMAs                                        bits can be transferred.

4, 8, 16, 32

 

Timer interrupt occurs  18.2 times in a second.

 

PPI interconnection                  bits is cleared to indicate low nibble is being sent. D1, D2, D3D,  4

 

 

 

Write TSR that use 17h to ignore spaces. 5 marks

#include <dos.h>

void interrupt (*old)( ); void interrupt newfunc ( ); main( )

{ old=getvect(0x17); setvect(0x17,newfunc); keep(0,1000);

CS609 MCQs For Papers

}

void interrupt newfunc( )

{

if (_AH==0)

{

if ( _AL != ‘ ‘ ) (*old)();

}

}

 

Write TSR to display name in center with interrupt 8 provided 1990 is center index. 10 marks

void interrupt (*1990)();

void main()

{ old=getvect(0x08); setvect(0x08,newint); keep(0,1000);

}

void interrupt newint ()

{

… (*1990)();

}

 

www.vustudents.ning.com

 

 

If CPUID instruction is not present then the processor can be a

? 486 processor

? 386 processor

? 286 processor

? All of the above

Question No. 2

On the execution of IRET instruction, number of bytes popped from stack is

? 4 bytes

? 6 bytes

? 8 bytes

? 10 bytes

Question No. 3

Keyboard Status Byte is located at the address

? 0040:0000H

? 0040:0013H

? 0040:0015H

? 0040:0017H

Question No. 4

If we use keep (0, 1000) in a TSR program, the memory allocated to it is

 

 

 

 

 

 

 

? 64000 bytes

? 32000 bytes

? 16000 bytes

? 80000 bytes

Question No. 5

CS609 MCQs For Papers

Maximum number of interrupts in a standard PC is

? 64

? 128

? 256

? 512

 

In boot block BIOS parameter block starts from

03H

05H

08H

0BH

 

File can be              viewed as collection of clusters or blocks.

Physically

Logically

Both physically and logically

None

 

What will be the value of DL Register when we are accessing C drive using undocumented service 21H/32H?

0

1

2

3

 

Operating system name contains              bytes in boot block.

3

5

8

11

 

When LSN is equal to zero (0) it means                   First block of the disk

First block of the logical drive First block of hidden blocks None of the given

 

The size of DPB data structure is              bytes

16

32

 

 

 

 

 

 

 

64

128


CS609 MCQs For Papers

 

Cluster number can also be referred as block number. True

False

 

Using the          entry and the FAT we can access the contents of file. Reserved blocks

Root Directory Number of FAT copies None of the given

 

Control information in files is maintained using

BPB DPB FCB FPB

We can access blocks for FAT using                      BPB

DPB

FCB

Both BPB and DPB

 

File system used in CD’s is                  file system

Contiguous Chained Indexed None

 

Disadvantage of FAT32 is               Large disk can be managed in FAT32

Cluster size is reduced

Internal fragmentation is reduced

Very large table

 

Practically              entries are there in FAT 32

2^26

2^28

2^30

2^32

 

NTFS volume can be accessed directly in DOS.

True

 

 

 

 

 

 

 

False


CS609 MCQs For Papers

 

What will happen if NTFS volume is accessed in DOS? Convert it to FAT volume

Nothing will happen Error of invalid media None of the given

 

Advantages of FAT32 is/are                     Large disk size can be managed in FAT32

Cluster size is reduced

Internal fragmentation is reduced

All of the given

 

FAT based file system can store file name in               form

ASCII UNICODE

Both ASCII and UNICODE

None

 

How many bytes can be used to store a file name in NTFS?

128

255

510

1024

 

IN NTFS, FAT and root directory is replaced by

FCB

MFT

Hidden blocks

Boot sector

 

LSN of FSInfo block is available at

BPB

FAT

Root Directory

None of the given

 

DOS device drivers do not understand the_          data structures. FAT12

FAT16

FAT32

NTFS

 

For supporting long file names,                       fragments can be supported.

CS609 MCQs For Papers

12

20

26

32

 

If a file is having more than one cluster then it will be managed by

FAT BPB DPB None

 

        File system keeps the backup of its boot block. FAT12

FAT16

FAT32

None

 

If a file size is 12K and the size of the cluster is 4K then            clusters are used for the file.

2

3

4

5

 

        is the first logical sector of NTFS partition. DPB

MFT

Boot sector

None

 

A file has 2 clusters and the size of cluster is 4K. What will be the size of file?

2K

8K

16K

32K

 

In FAT32        root directory entries are there

128

256

512

None

 

 

 

Block # 2 is the safest block to store the backup of boot block. True

False

CS609 MCQs For Papers

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1.  There are two main types of interrupts namely  Hardware and Software interrupts

 

2.  To set the interrupt vector means is to change the double word sized interrupt vector within the IVT. (True / False)

 

3.  The service number is usually placed in the                        register. (AL, CL, AH, AX)

 

4.  The keyboard makes use of interrupt number                    for its input operations. (9,10,11,12)

 

5.   The service                   

15H/FFH )


is called the keyboard hook service. (15H/2FH, 15H/4FH,

 

6.  The BIOS interrupt                    can be used to configure RTC.  (1AH, 2AH, 3AH, 4AH)

 

 

 

 

7.  The interval timer can operate in                    modes. (Five, Seven, Four, Six)

 

 

 

8.                is Disk interrupt. (10H,11H,13H,14H)

 

 

 

9.  PPI stands for Peripheral Programmable Interface (PPI)

 

 

 

10. Int                   is used to control the printer via the BIOS. (17H, 18H, 20H, 21H)

CS609 MCQs For Papers

 

 

11. Counter register can be used to divide clock signal. (True False) Note : it is also used to divide frequency (See #69)

 

 

 

12. DCE stands for _ Data Communication Equipment

 

 

 

13. Interrupt ID Re gister _is used to identify the cause of interrupt.

 

 

 

14. The bit #                 of the coprocessor control word is the interrupt enable flag. (7,8,9, 6)

 

 

 

15. There are                       kinds of serial communication. (2,3,4,5)

 

 

 

16.                   store the base address for LPT1. (40:00H , 40:02H ,40:08H, 40:1AH)

 

 

 

17. The amount of memory above conventional memory (extended memory) can be determined using the service                                                               .  (15H/88H, 16H/88H, 17H/88H, 21H/88H)

 

 

 

18. The output on the monitor is controller by a controller called                             within the PC. (Video controller, Bus controller, Ram controller, None of the given)

 

 

 

19. The keyboard input character scan code is received at     port. (60H,61H,62H,63H,64H )

 

 

 

20.                 is LED control byte. (0xFD,0xED, 0xFF, 0xEE )

21. In RS232C these abbreviations stands for                                       [3 marks]

DTR           DTR (data terminal ready)

DSRDSR (data set ready)

CTS                       CTS (clear to send) RTS   RTS (Request to send)

 

22. For what purpose these services are used?                                      [3 marks]

INT 1AH/01                      Set Clock Counter

INT 1AH/02                      Read Time INT 1AH/03   Set Time INT 1AH/04                      Read Date

23. How communication between modem can performed (in terms of data transfer). [5marks]

 

Modem is generally used to send /receive data to/from an analog telephone. Had the telephone line been purely digital there would have been no need of a modem in this form. If data is to transferred from one computer to another through some media which can carry digital data then the modem can be eliminated and the UART on both computers can be interconnected. Such arrangement is called a NULL modem.

BY MISHII

Function to toggle speaker between on and off

The port 61h is used to control the speaker only the least significant 2 bits are important. Bit 0 is used to connect the interval timer to the speaker and the bit #1 is used to turn the speaker on off. Rest of the bits are used by other devices.

 

Parallel communication

PPI is used to perform parallel communication. Devices like printer are generally based on parallel communication.  It’s called parallel because a number of bits are transferred from one point to another parallel on various lines simultaneously

 

Communication with parity check

The line control register contains important information about the behaviour of the line through which the data will be transferred. In it various bits signify the word size, length

of stop bits, parity check, parity type and also the a control bit to load the divisor value.

The bit 7 if set indicates that the base +0 and base + 1 will act as the divisor register otherwise if cleared will indicate that base + 0 is the data register.

 

RS232C work flow?

RS232C is a standard for physical dimension of the connector interconnecting a DTE(Data terminal equipment) and DCE (Data communication equipment).

Data is received through the RxD line. Data is send through the TxD line. DTR (data terminal ready) indicates that the data terminal is live and kicking. DSR(data set ready) indicates that the data set is live. Whenever the sender can send data it sends the signal RTS( Request to send) if as a result the receiver is free and can receive data it send the sender an acknowledge through

CTS( clear to send) indicating that its clear to send now.

 

1AH services

Clock Counter 1AH/00

Set Clock Counter 1AH/01

Read Time 1AH/02

Set Time 1AH/03

Read Date 1AH/04

CS609 MCQs For Papers

Set Date 1AH/05

Set Alarm 1AH/06

Disable Alarm 1AH/07

Read Alarm 1AH/09

 

Write function in relation to COM port and modem

The initialize () function initializes the COM port whose number is passed as parameter using BIOS services. The recievechar() function uses the COM port number to receive a byte from the COM port using BIOS services. the sendchar() function sends a character to the COM port using BIOS service whose number is passed as parameter. And the getcomstatus() function retrieves

the status of the COM port whose number has been specified and returns the modem and line status in an unsigned int.

 

Coprocesser

 

To access the block within cluster using BIOS services the cluster number should be converted into                                          .

Select correct option: CHS

LBA

LSN

None of the given

 

Question # 2 of 10 ( Start time: 12:21:55 PM ) Total Marks: 1

The practical limit of blocks per cluster is                    . Select correct option:

 

32 blocks per cluster

64 blocks per cluster

128 blocks per cluster

256 blocks per cluster

 

Question # 3 of 10 ( Start time: 12:23:05 PM ) Total Marks: 1

Maximum possible entries in FAT12 are                     . Select correct option:

1024

2048

4096

65536

 

Question # 4 of 10 ( Start time: 12:24:20 PM ) Total Marks: 1

When we talk about FAT based file system, in user data area first cluster number is                      . Select correct option:

0

CS609 MCQs For Papers

1

2

None of the given

 

Question # 5 of 10 ( Start time: 12:25:04 PM ) Total Marks: 1

If we know the cluster number, we can access the blocks within the cluster using BIOS services directly.

Select correct option:

 

True

False

Question # 6 of 10 ( Start time: 12:26:16 PM ) Total Marks: 1

Jump code part contains            bytes in boot block. Select correct option:

 

3

5

8

11

 

Question # 7 of 10 ( Start time: 12:27:22 PM ) Total Marks: 1

In dos we have limit of                   . Select correct option:

 

128 blocks per cluster

256 blocks per cluster

32 blocks per cluster

64 blocks per cluster

 

Question # 8 of 10 ( Start time: 12:28:01 PM ) Total Marks: 1

Drive parameter block is derived from                           . Select correct option:

FCB FAT BPB CPB

Question # 9 of 10 ( Start time: 12:29:26 PM ) Total Marks: 1

The directory structure of DOS is like                     . Select correct option:

 

Array Tree Linked list

None of the given

CS609 MCQs For Papers

 

Question # 10 of 10 ( Start time: 12:30:19 PM ) Total Marks: 1

             is the first block on disk. Select correct option:

 

LSN =0

LBA=0

LBA=1

Both LBA=0 and LSN=0

 

 

Question # 1 of 10 ( Start time: 12:20:29 PM )  Total Marks: 1

To access the block within cluster using BIOS services the cluster number should be converted into      .

Select correct option:

 

CHS LBA LSN

None of the given

 

Question # 2 of 10 ( Start time: 12:21:55 PM )  Total Marks: 1

The practical limit of blocks per cluster is                    . Select correct option:

 

32 blocks per cluster

64 blocks per cluster

128 blocks per cluster

256 blocks per cluster

 

Question # 3 of 10 ( Start time: 12:23:05 PM )  Total Marks: 1

If FAT entry is between FFF0H to FFF6H in FAT16 then                   . Select correct option:

 

Cluster is available

It is a Reserved cluster

It is next file  cluster

It is a last file cluster

 

Question # 4 of 10 ( Start time: 12:24:20 PM )  Total Marks: 1

What will be the value of the word located at 1Fh in DPB when number of free clusters on drive is not known?

Select correct option:

 

0000H

 

 

 

 

 

 

 

1111H FFFFH


n>

Question No. 3

Keyboard Status Byte is located at the address

? 0040:0000H

? 0040:0013H

? 0040:0015H

? 0040:0017H

Question No. 4

If we use keep (0, 1000) in a TSR program, the memory allocated to it is

 

 

 

 

 

 

 

? 64000 bytes

? 32000 bytes

? 16000 bytes

? 80000 bytes

Question No. 5

Question # 5 of 10 ( Start time: 12:25:04 PM )  Total Marks: 1

If we know the cluster number, we can access the blocks within the cluster using BIOS services directly.

Select correct option:

 

True

False

 

Question # 6 of 10 ( Start time: 12:26:16 PM )  Total Marks: 1

Jump code part contains            bytes in boot block. Select correct option:

 

3

5

8

11

 

Question # 7 of 10 ( Start time: 12:27:22 PM )  Total Marks: 1

In dos we have limit of                   . Select correct option:

 

128 blocks per cluster

256 blocks per cluster

32 blocks per cluster

64 blocks per cluster

 

Question # 8 of 10 ( Start time: 12:28:01 PM )  Total Marks: 1

We can access the contents of File by using the root directory entry and                  . Select correct option:

 

Reserved Blocks

Number of FAT copies

File Allocation Table (FAT) None of the given

 

Question # 9 of 10 ( Start time: 12:29:26 PM )  Total Marks: 1

The directory structure of DOS is like                     . Select correct option:

 

Array

CS609 MCQs For Papers

Tree

 

Linked list

 

None of the given

 

 

 

Question # 10 of 10 ( Start time: 12:30:19 PM )  Total Marks: 1

             is the first block on disk. Select correct option:

 

LSN =0

LBA=0

LBA=1

Both LBA=0 and LSN=0

 

 

Question # 1 of 10 ( Start time: 02:32:19 PM )  Total Marks: 1

Disadvantage of FAT32 is                   . Select correct option:

 

Large disk size can be managed in FAT32

Cluster size is reduced

Internal fragmentation is reduced

Very large table (not sure)

 

Question # 2 of 10 ( Start time: 02:33:47 PM )  Total Marks: 1

In FAT32, lower                 bits are used. Select correct option:

 

26

28

30

32

 

Question # 3 of 10 ( Start time: 02:34:23 PM )  Total Marks: 1

File system used in CD’s is                       file system. Select correct option:

 

Contiguous

Chained

Indexed (guess) None of the given

CS609 MCQs For Papers Question # 4 of 10 ( Start time: 02:35:53 PM )  Total Marks: 1

If a file size is 12K and the size of the cluster is 4K then                   clusters are used for the file. Select correct option:

 

2

 3

4

5

 

Question # 5 of 10 ( Start time: 02:37:23 PM )  Total Marks: 1

To store a UNICODE character           is/are needed. Select correct option:

 

Nibble

Byte

2 Bytes

4 Bytes

 

 

 

Cluster size is reduced in                        . Select correct option:

 

FAT12

FAT16

 FAT32

None of the given

 

Question # 7 of 10 ( Start time: 02:40:18 PM )  Total Marks: 1

For supporting long file names,                        fragments can be supported. Select correct option:

 

12

20

26

32

 

Question # 8 of 10 ( Start time: 02:41:28 PM )  Total Marks: 1

A file has 2 clusters and the size of cluster is 4K. What will be the size of file? Select correct option:

 

2K

8K

16K

32K

CS609 MCQs For Papers

 

In NTFS, Backup of boot block is stored at block #                        . Select correct option:

 

2

6

8

10

 

 

 

Question # 10 of 10 ( Start time: 02:44:18 PM )  Total Marks: 1

When we mark a file as deleted by placing 0xE5 then the chain of clusters in FAT is also replaced by    .

Select correct option:

 

E5

1

0

N

 

Quiz: CS609                                                                         Total Marks 10

Instructions:

1)  Upload this Quiz after solving with in 24 hours No Extra Time will be given or Quiz will not be accepted via email.

2)  Do No mark more then one choices.

3)  It is better you high light the choice with Read like   a) correct choice

 

 

 

1)  The interval timer can operate in           modes.

a)  Three b)  Four c)   Five d)   Six

 

2)  Highest capacity physical capacity of the disk according to the IDE interface is

                  . a)  127GB b)   100 GB

c)  80 GB

d)  300 GB

 

3)  Partition Table can be read using the extended                             Services.

a)  13 H

CS609 MCQs For Papers

b)  14 H

c)  15 H

d)  None of given

 

4)  File control block (FCB) is                       byte long.

a)

32

b)

64

c)

16

d)

128

 

5)  On the execution of IRET instruction, number of bytes popped from stack is

 

 

a)  4 bytes b)   6bytes c)  8 bytes d)  10 bytes

 

6)   If CPUID instruction is not present then the processor can be a                               

a)  486 processor b)   386 processor c)   286 processor

d)  All of the given choices

 

7)   Extended memory is available if the processor is of the type                                

a)  AT

b)  XT

c)  All of the given choices d)   None of them

 

8)   The built in mechanism within the UART for error detection is                             

a)  hamming code b)   parity

c)  CRC16 (cyclic redundancy check 16 bit )

d)  CRC32 (cyclic redundancy check 32 bit )

 

9)    In Protected Mode, the segment registers are used as                              

a)  Descriptor b)   Selector

c)  All of the given choices

d)  None of the given choices

 

10) If three Programmable interrupt controllers are cascaded then how many interrupt driven hardware IO devices can be attached                                                              

CS609 MCQs For Papers

a)

12

b)

18

c)

23

d)

24

 

 

 

 

 

 

 

 

 

CS609 MCQs For Papers

 

 

There are two main types of interrupts, namely

 

PC based and Window based

Hardware based and Kernal based

Hardware interrupts and Software interrupts

None of the given

 

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

Standard PC operates in two modes in terms of memory which are

 

Real mode and Extended Mode Base mode and Memory Mode None of the given

Real mode and protected mode

 

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

BPB stands for                   .

 

BIOS parameter block BIOS processing block Base processing block BIOS partition block

 

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

The Function of I/O controller is to provide                          .

 

I/O control signals

Buffering

Error Correction and Detection

All of given

 

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

IVT is a table containing               byte entries each of which is a far address of an interrupt service routine.

 

2

4

8

16

 

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

Each paragraph in keep function is           bytes in size.

CS609 MCQs For Papers

 

8

16

32

 

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

Interrupt 9 usually reads the                    from keyboard.

 

ASCII code

Scan code

Both ASCII and Scan code

None of the above

 

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

A software interrupt does not require EOI (End of interrupt).

 

True

False

 

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

To store each character in keyboard buffer             bytes are required.

 

2

4

6

8

 

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

Interrupt           is empty; we can use its vector as a flag.

 

9H

13H

15H

65H

 

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

Command register is an            bit register

4

8

16

32

 

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

The interval timer can operate in           modes.

 

Three Four Five Six

 

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

The following command “outportb (0x61,inportb(0x61) | 3);” will                     .

 

Turn on the speaker Turn off the speaker Toggle the speaker None of the above

 

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

The PPI acts as an interface between the CPU and a parallel                   .

 

I/O device

CPU BUS

None of Given

 

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

DTE is                         .

 

Data terminal equipment Data transmitting equipment Dual terminal equipment None of the given.

 

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

DSR stands for                     .

 

Data set ready Data service ready Data stock ready None of the given

 

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

In self test mode the output of the UART is routed to its input.

 

True

False

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

Interrupt             is used to get or set the time.

 

0AH

1AH

2AH

3AH

 

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

            is used to set time.

 

1A/02H

1A/03H

1A/04H

1A/05H

 

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

Communication between keyboard and keyboard controller is                       .

 

Asynchronous serial Synchronous serial Parallel communication None of the given

)

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

Following is not a method of I/O

 

Programmed I/O Input driven I/O Hardware Based I/O None of given

 

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

The Function of I/O controller is to provide                          .

 

I/O control signals

Buffering

Error Correction and Detection

All of given

 

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

Which of the following are types of ISR                        .

 

BIOS (Basic I/O service ) ISR DOS ISR

ISR provided by third party device drivers

All of the given

 

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

Interrupt service number is usually placed in                            register.

CH CL AH AL

 

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

NMI Stand for

 

Non Maskable Interrupt Non Multitude Interrupt Non Maskable Instruction None of Given

 

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

A single interrupt controller can arbitrate among             different devices.

 

4

6

8

10

 

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

Hardware Interrupts are                    .

 

Preemptive

Non-Preemptive

Both Preemptive and Non-Preemptive

None of Given

 

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

The microprocessor package has many signals for data. Below are some in Correct priority order (Higher to Lower).

 

Reset,Hold,NMI,INTR

NMI, INTR,Hold,Reset INTR,NMI,Reset,Hold None of the Given

 

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

The interval timer can operate in           modes.

 

Three Four Five Six

 

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

PPI stands for                           .

 

Parallel Programmable interface Peripheral Programmable interface Port Programmable interface

None of the given

 

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

The following command “outportb (0x61,inportb(0x61) & 0xFC);” will

 

Turn on the speaker Turn off the speaker Toggle the speaker None of the given

 

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

The PPI acts as an interface between the CPU and a parallel                   .

 

I/O device

CPU BUS

None of Given

 

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

BIOS DO NOT support            .

 

LPT1

LPT2

LPT3

LPT4

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

          bit is cleared to indicate the low nibble is being sent.

D1

D2

D3

D4

 

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

The bit              of Line control register in UART, if cleared will indicate that DLL is the data register.

 

1

3

5

7

 

 

 

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

                 used to determine the amount of conventional memory interfaced with the processor in kilobytes.

 

 

 

INT 10 H INT 11 H INT 12 H INT 13 H

 

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

Bit number                of coprocessor control word is the Interrupt Enable Flag.

 

7

8

9

10

 

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

To distinguish 486 with Pentium CPUID Test is used.

 

True

False

 

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

            is LED control byte.

 

 

 

 

 

 

 

0xF3

0xED

0xE5

0xFF


 

 

2. Int                            service 0 can be used to set the line parameter of the UART or COM port.

14H

15H

13H

None of the given option

3. In case of synchronous communication a timing signal is required to identify the start and end of a bit.

True

False

 

4. In self test mode the output of the UART is routed to its input

True

False

6. The                  fuction uses the COM port number to receive a byte from the COM port using

BIOS services. recievebyte() receive() recievechar()

 

 

 

 

10. The                            function initialize the COM port whose number is passed as parameter using BIOS services.

Initializecom() Initialize() Recievechar()

None of these option

 

 

 

11.  XON whenever received indicates the start of communication and  XOFF whenever received indicates a temporary pause in the communication.

 

14.                   is a device incorporated into the PC to update time even if the computer is off.

Clock counter

ROM Clock

Real time clock

What is the location of timer count in BIOS data area?

Sol.

0040:006C

 

Question No: 22           ( Marks: 3 )

Write a function which fills the whole screen with blanks (space). Sol.

unsigned char far *scr= (unsigned char far*)0xb8000000 void main()

{

int i;

for (i=0; i<2000; i++)

{

*scr=0x20;

*(scr+1)=0x07;

scr=scr+2;

}

}

 

Question No: 23           ( Marks: 5 )

For what purpose the port 61H is used? Sol

The port 61h is used to control the speaker only the least significant 2 bits are important. Bit 0 is used to connect the interval timer to the speaker and the bit #1 is used to turn the speaker on off. Rest of the bits are used by other devices.

 

Question No: 24           ( Marks: 10 )

Write down a C program that will protect the boot block to be written by other application.

Hint.

Use interrupt 13 for accessing the boot block information. Sol.

#pragma inline

#include <dos.h>

#include <bios.h>

void interrupt (*oldtsr) ( );

void interrupt newtsr (unsigned int BP, …, flags);

//must provide all the arguments

void main ( )

{

oldtsr = getvect (0x13); setvect(0x13, newtsr); //corrected keep (0, 1000);

}

void interrupt newtsr(unsigned int BP, unsigned int DI,

unsigned int SI, unsigned int DS, unsigned int ES, unsigned

int DX, unsigned int CX, unsigned int BX, unsigned int AX, unsigned int IP, unsigned int CS,

unsigned int flags) //corrected

{

if ( _AH == 0x03)

if(( _DH == 1 && _CH == 0 && _CL == 1)&& _DL >= 0x80)

{

asm clc;

asm pushf;

asm pop flags;

return;

}

_ ES = ES; _DX = DX;

_CX = CX; _BX = BX;

_AX = AX;

*oldtsr;

asm pushf;

asm pop flags;

AX = _AX; BX = _BX; CX = _CX; DX = _DX; ES = _ES;

}

MIDTERM    EXAMINATION

Spring 2009

CS609- System Programming (Session - 1)

Question No: 21           ( Marks: 2 ) ICW and OCW stand for?

Sol.

Initialize control words, operation control words

 

Question No: 22           ( Marks: 3 )

What is the usage of Interrupt ID Register within the UART?

Sol.

 

Once an interrupt occurs it may be required to identify the case of the interrupt. This register is used to identify the cause of the interrupt.

Interrupt ID Register

0 bit for Trigger Triggered

2 1 bits for Modem/Line

00 =Change in Modem Status

01 = THR is Empty

10 = Data is Ready

11 =Error in Data

 

Question No: 23           ( Marks: 5 )

Write a program that surpasses the input of ‘s’ from keyboard. Scan code of ‘s’ is 0x1F.

#include <dos.h>

void interrupt(*old)(); void interrupt newint9(); void main()

{

old=getvect(0x09); setvect(0x09,newint9); keep(0,1000);

}

void interrupt newint9()

{

if (inportb(0x60)==0x1F) //corrected

{

outportb(0X20,0x20);

return;

} (*old)();

}

 

Question No: 24           ( Marks: 10 ) Explain FIFO Queue in UART.                         [5]

This feature is available in the newer version of the UART numbered 16500. A queue or a buffer of the input or output bytes is maintained within the UART in order to facilitate

more efficient I/O. The size of the queue can be controlled through interrupt ID register.  slide.

 

Write down the structure of Modem control register.  [5]

26 question thy or 6 subjective

jis mein aik hi coding i or wo b handout se hi

or last question tha k keyboard cpu se kesy interact kar k interupt generate karta hai ... diagram k sath

or abrevation sub yad karin

objective mein bth hi question thy aberivation se mutaliq or ziada concetration subjective pe rakihin

coding ka itna nahi aea paper mein

 

Total 26 Questions..

20 mcsqs are from that file which i sent u yesterday...almost all mcqs..

 

Then 21st question is the Write the base adresses of LPTs. Sol.

LPT1=40:08, LPT2=40:0A, LPT3=40:0C, LPT4=40:0E

 

22nd is the abriviations of the internal registers of UARTS e.g THR,RBR etc

 

23rd What is the value of LEDs (from last lecture)

 

24th  i forgot same from the last lecture about LED.

25th is a 5 marks programm.

26th is the diagram and explaination of the STATUS REGISTER A of the RTL.

 

Q (5 marks)

Explain the usage of XON and XOFF is software based flow control

Ans (Page # 158 , lec # 17)

 

Q

Write a function that will read the status of COM Port and return the modem line status is a unsigned int. COM port number will be passed as parameter.

 

 

 

Q Write down the purpose of int 12H and int 15H/88H Ans :  int 12H : used for memory interface

int 15H/88H  return = No. of kb above 1 MB

 

 

 

Q

Write down the detail of the Service (00,01,02) of int 17H which is used in printer.

 

Ans :   00 : Display characters

01 : Initialze Printer

02 :  Request printer.

 

 

 

some questions of Another Paper. Q (2 Marks )

write stands for

THR transmitter holding register RBR receiver buffer register DLL (band rate divisor) low DLM  (band rate divisor) high

 

 

 

Q

write stands for. DSR Data set ready

DTR Date terminal ready

RTS ready to send

CTS  clear to send

 

 

 

Q#4:-

a. Write down three differences between Logical Sector Number (LSN) and Logical Block

Addressing (LBA).

b. What is meant by polling mode in communication between software and UART and what is its disadvantage as compared to interrupt mode. [max 5 line answer]

 

Q#5:-In IRQ2 and IRQ3 which one has the highest priority?

Can’t be determined

Both have same priority

IRQ3

IRQ2

 

Q#6:-Extended memory is available if the processor is of the type

None of the given choices

All of the given choices

XT

• AT

 

Q#7:-In NTFS, boot sector is stored at

First and 6th sector

First and Last sector

Only at Last sector

Only at First sector

 

Q#8:-DMA driven IO is the technique used for performing IO

All of the given choices

By transferring the data directly from IO port to processor and vice versa

By using the processor to perform an IO routine only when data for IO operation is available.

By keeping the processor tied up either checking for the possibility of an IO operation or

performing the IO operation

 

Q#9:-A full duplex communication system is to be implementing using a PPI (peripheral programmable interface). By the virtue of the printer interface provided by the standard PCs the unit of data transfer for such communication will be

A double word

A word

A byte

A nibble

what is a Descriptor 2marks

waht is a segment memory and write its attribute 5marks what size of enteries of GDT,LDT and IDT 3marks

what is the contiguous Memory Management waht is FAT12 and 32 entries size 2marks

IN NTFS,the FAT and root directory has been replaced by ........ 2marks write the aotonmy of NTFS 5marks

redundant

Q # 1  [Markes:5]

Write down the structure of segment register in protected modes? Q # 2  [Markes:5]

Write down the anatomy of NTFS based file system? Q # 3 [Markes:5]

Find the root directory sector where reserved sectors = 1 and sector per FAT = 9. Use appropriate assumption where needed.

Q # 4   [Markes:3]

How many maximum root directory entries are possible in FAT 12 & FAT 16? Q # 5   [Markes:3]

What is the CHS and LBA address of MBR? Q # 6   [Markes:3]

Write down the names of three different states of viruses. Q # 7   [Markes:3]

How contents of small and large files are managed in MFT ? Q # 8   [Markes:2]

What is the segmentation is context of  non contiguous memory management system? Q # 9   [Markes:2]

Write down the major enhancement of FAT32 comparing to FAT12 and 16. Q # 10  [Markes:2]

Write down the purpose of interrupt 12H and interrupt 15H/88H. Q # 11 [Markes:2]

For what he purpose services are used

Interrupt

1AH /06

1AH /07

1AH /09

Single Transfer in DMA (2 marks)

what kind of data send to the keyboard (2 marks)

Undocumented Services (2marks) Highest IDE capacity biosdisk() (2marks)

In NTFS, FAT32 we replace the root directories with                            ?(2marks)

hand out Page#64 (3marks)

Multiple loading will leave redundant copies in memory

answer given below:- int flag;

flag =1; keep(0,1000); if (flag==1) Make TSR else

exit Program

 

 

 

logical to physical address translation? (3marks)

What is MFT? (3marks)

what is the contiguous Memory Management (3marks)

what size of enteries of GDT,LDT and IDT (3marks)

 

Differenciate b/W LSN and LBA (5marks)

anatomy of an NTFS (5 marks)

write the step of Virus Detection (5marks

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