CS401 Current Final Term Paper 2021 - VU Answers

 

CS401 Current Final Term Paper 2021


VU Answers Providing CS401 Current Final Term Paper 2021. Most Important and Useful CS401 Current Paper 2021. Easy to See and Download CS401 Paper.

CS401 CURRENT FINALTERM PAPER FALL 2020 

Provide by VU Answer

 

Cs401

Mcqs past papers sa tha

3masks questions:

Diff between serial and parallel port communication.

How to speed up multitasking.

Ik program code aya tha for cant remember the condition.

Subjectives Question and its  Answers of my Final Tearm CS401 on 12-Sep-2020


Q) Format of DOS input buffer:

Answer:- 

Offset Size Description

00 1 maximum characters buffer can hold

01 1 number of chars from last input which may be recalled OR number of characters actually read, excluding CR

02 n actual characters read, including the final carriage return

Q) Describe MOVS instructions

Answer:- (Page 92) MOVS MOVS transfers a byte or word from the source location DS:SI to the destination ES:DI and updates SI and DI to point to the next locations. MOVS is used to move a block of memory. The DF is important in the case of overlapping blocks.

Q) How interrupts are handled in protected mode & IDTR.

Answer:- (Page 182)

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.

Q)  VESA service "INT 10 – VESA – Get SuperVGA Information" uses which registers to return the result? 

Answer:- (Page 180) To return the result, 

"INT 10 – VESA – Get SuperVGA Information” uses: Return: AL = 4Fh if function supported AH = status 

INT 10 – VESA – Get SuperVGA Infromation

AX = 4F00h

ES:DI -> buffer for SuperVGA information

Return:

AL = 4Fh if function supported

AH = status

Q) define interrupt INT 0*80

Answer:- 

int 0x80 ; multitasking kernel interrupt

Q)Define HOOKING And Unhooking

HOOKING AN INTERRUPT

To hook an interrupt we change the vector corresponding to that interrupt. As soon as the interrupt vector changes, that interrupt will be routed to the new handler. By hooking this interrupt a debugger can get control after every instruction and display the registers etc 

Unhooking Interrupt

We now add the interrupt restoring part to our program. This code resets the interrupt vector to the value it had before the start of our program.

Q) How interrupts are handled in protected mode.

Answer:- 

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.

Q) INT 14 READ OPERATION from SERIAL PORT

Answer:- 

The BIOS INT 14 provides serial port services. We will use a mix of BIOS services and direct port access for our example. A major limitation in using BIOS is that it does not allows interrupt driven data transfer, i.e. we are interrupted whenever a byte is ready to be read or a byte can be transferred since the previous transmission has completed. To achieve this we have to resort to direct port access. Important BIOS services regarding the serial port 

INT 14 - SERIAL - READ CHARACTER FROM PORT

AH = 02h

DX = port number (00h-03h)

Return:

AH = line status

AL = received character if AH bit 7 clear


Q) What is EAX

AX register is extended 32 bit register is renamed as EAX

Q) Define multitasking, its environment and conditions

Answer:- 

Multitasking: 

*A multitasking environment allows applications to be constructed as a set of independent tasks, each with a separate thread of execution and its own set of system resources. The inter-task communication facilities allow these tasks to synchronize and coordinate their activity. 

*Multitasking provides the fundamental mechanism for an application to control and react to multiple, discrete real-world events and is therefore essential for many real-time applications. 

*Multitasking creates the appearance of many threads of execution running concurrently when, in fact, the kernel interleaves their execution on the basis of a scheduling algorithm. This also leads to efficient utilization of the CPU time and is essential for many embedded applications where processors are limited in computing speed due to cost, power, silicon area and other constraints. 


To realize such a system, the following major functional conditions are to be carried out.


A.Process Management

• interrupt handling 

• Task scheduling and dispatch 

• create/delete, suspend/resume task 

• manage scheduling information – priority, scheduling policy, etc


B.Inter-process Communication and Synchronization

• Code, data and device sharing 

• Synchronization, coordination and data exchange mechanisms 

• Deadlock and Live-lock detection


C.Memory Management 

• Dynamic memory allocation 

• Memory locking 

• Services for file creation, deletion, reposition and protection


D.Input/Output Management

• Handles request and release functions and read, write functions for a variety of peripherals


Q) write a program add 10 numbers using register + offset addressing mode in assembly language

Cs401

50 questions

40 mcqs 80% from past papers

1. In truncated file service which register read service and file attributes

2.SACS instruction works?

3.which interrupt is called scheduler.which main purpose of that.

4.which interrupt interrupt and control back to dos.write name of interrupt and service number.

5.serial port intiziltaion service number and also write attributes of it with service number those initialize them.

6.physical address btna tha

7.data movment instruction de v thi un ka btna tha k kon sy processor me use hoti hain.

8.code dia hua tha btna tha k instruction kya kam kr rhi us me.

9.code likhna tha apna name on top of screen show krny ka.

10.code likhna tha aik array ka jis me 128 elements ho har element 4 byte ka ho or har element ki value 0 ho.

Overall paper bhut easy tha..


#CS401 Final Term Current Paper FALL 2020-21

27-FEB-2021 

9:30 am paper

Esc by hooking code.. 

Attribute of changing video service to information about font color... 

Which service is used with int 13 for rea data write data and gatting data... 

Difference between 16bit and 32bit DIV

Mcqs past papers or handouts sy ay thy


Cs401 2:30 06-03-2021

Mcqs past se thy or ak or  file se..

Short long question

1 two name system operation computer strt krny k ly..

2 real mode ko protected mode me kasy krty is trha ka tha question.

3 code dia tha( int x int y)  c me is ko assembly me likhna tha..

4 ak so acha us me error btana tha just..

5 detail from likhna thi 5:6:5 ki video me use hota wo btana tha..5 red 6 green 5 blue tha ans iska..

6 ak code ki line skip thi w/o likhna thi..

7 ak program bnana tha k physically output btani thi us ki..

8 is me code dia tha k segment line ko physically  kasy krty AH =01h AH =09h ye tha...


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Current Paper and Important thoughts are recommended there will be higher chance to ask for exams in repeated questions. 


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