71 operating system, OS interview questions and answers

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Operating system interview questions and answers


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Q1. We have two concurrent processes a and b and Peterson’s algorithm for mutual exclusion is applied between the two processes. Below shows the program executed by the process.

Repeat flag [a] = true;
Turn = b;
While (Q) do no-op;
Enter critical section, perform actions and then exit critical section
Flag [a] = false;
Perform other non-critical section actions until false;

What should be the value of predicated Q in the while loop for the program to guarantee mutual exclusion?

1.   Flag [b] = true and turn = b
2.   Flag [b] = true and turn = a
3.   Flag [a] = true and turn = a
4.   Flag [a] = true and turn = b
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ANSWER: Flag [b] = true and turn = b

Q2. The two processes P1 and P2 which are accessing the shared variables A and B are protected by two binary semaphores Na and Nb respectively and both initialized to 1. Let X and Y be the usual semaphore operators. X decrements the semaphore value and Y increments the semaphore value. The code of P1 and P2 are given below:

P1
While true do {
M1: ……………..
M2: ………………
A = A+1;
B = B-1;
Y (Na);
Y (Nb);

P2
While true do {
M3: ……………..
M4: ………………
B = B+1;
A = B-1;
Y (Nb);
Y (Na);

What should be the correct operators at M1, M2, M3 and M4 in order to avoid deadlock condition?

1.    X(Na), X(Nb), X(Nb), X(Na)
2.    X(Na), X(Nb), X(Na), X(Nb)
3.    X(Nb), X(Na), X(Nb), X(Na)
4.    X(Nb), X(Na), X(Na), X(Na)
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ANSWER: 2. X(Na), X(Nb), X(Na), X(Nb)

Q3. Set of processes with their respective arrival time and the CPU burst times are given in milliseconds.

Using the pre-emptive shortest remaining processing time first (SRPTF) algorithm, the average turnaround time for these processes is

1.    6.50
2.    6.00
3.    5.50
4.    5.00
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ANSWER: 3. 5.50

Q4. Consider the features given below.

A. Once the execution of the program terminates, another program is immediately scheduled for execution.
B. For execution more than one program may be loaded into main memory at the same time.
C. Another program is immediately scheduled for execution if a program waits for certain events such as IN/OUT.

The combination that will suffice to characterize an operating system as a multi-programmed operating system is

1.    A only
2.    A and B only
3.    A and C only
4.    A, B and C
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ANSWER: 4. A, B and C

For questions 5 and 6 refer to the data given below:

Consider two concurrent processes A and B whose code is given below.
Process A:
While (1) {
L:
            Print ‘ 0’; 
            Print ‘ 1’;
M:
}

Process B:
While (1) {
N:
            Print ‘1’;
            Print ‘1’;
O:
}

The two processes A and B are to be synchronized using binary semaphores P and Q and the synchronization statements can be inserted only at points L, M, N and O.

Q5. The one that will always lead to an output string ‘001100110011’ is

1.     A(P) at L, V(Q) at M, A(Q) at N, V(P) at O, P initially 1 and Q initially 0
2.     A(P) at L, V(P) at M, A(Q) at N, V(Q) at O, P and Q initially 1
3.     A(P) at L, V(Q) at M, A(Q) at N, V(Q) at O, P initially 1 and Q initially 0
4.     A(P) at L, V(Q) at M, A(Q) at N, V(P) at O, P and Q initially 1
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ANSWER: 1. A(P) at L, V(Q) at M, A(Q) at N, V(P) at O, P initially 1 and Q initially 0

Q6. The output string never contains a substring of the form 01n0 or 10n1 (n is odd) is ensured by

1.    A(P) at L, V(P) at M, A(P) at N, V(P) at O, P initially 1
2.    A(P) at L, V(P) at M, A(Q) at N, V(Q) at O, P and Q initially 1
3.    A(P) at L, V(Q) at M, A(Q) at N, V(P) at O, P and Q initially 1
4.    V(P) at L, V(Q) at M, A(P) at N, A(Q) at O, P and Q initially 1
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ANSWER: 1. A(P) at L, V(P) at M, A(P) at N, V(P) at O, P initially 1

For questions 7 and 8 refer to the data given below:

For translation from 32 bit virtual to 32 bit physical address the processor uses 2-level page tables and for both levels the page tables are stored in the main memory. The memory is byte addressable. For translation, 10 MSB’s of the virtual address are used as index into the first level page table and the next 10 bits are used as index into the second level page table. The 12 LSB’s of the virtual address are used as offset within the page. Assume that the translation look aside buffer of the processor has the hit rate of 96% and it caches recently used virtual page numbers and the corresponding physical page numbers. Also assume that the page table entries in both levels of page tables are 4 bytes wide. The physically addressed cache of the processor has a hit rate of 90%. Cache access time = TLB access time = 1ns and main memory access time = 10ns.

Q7. What is the approximate average time taken to access a virtual address if we assume that no page faults occur?

1.    4ns
2.    3ns
3.    2ns
4.    1ns
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ANSWER: 1. 4ns

Q8. What is the amount of memory required for storing the page tables of the process if a process has only following pages in its virtual address space?

Starting at virtual address 0X00000000 two contiguous code pages
Starting at virtual address 0X00400000 two contiguous data pages
Starting at virtual address 0XFFFFF000 stack page

1.    20Kbyte
2.    19Kbyte
3.    17Kbyte
4.    16Kbyte
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ANSWER: 4. 16Kbyte

Q9. Three processes with total execution time of 10, 20 and 30 units respectively arrive at time Zero. First 20% of the execution time of each process is spent in doing IN/OUT, next 70% of time is spent in doing computation and the last 10% of the time is spent in again doing IN/OUT. The algorithm which is used by the operating system is shortest remaining compute time first scheduling algorithm and it schedules new process either when the running process gets blocked in IN/OUT or when the running process finishes its compute burst. If we assume that all IN/OUT operations can be overlapped as much as possible, for what percentage of time does the CPU remains idle?

1.    80.5%
2.    30.8%
3.    10.6%
4.     0%
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ANSWER: 3. 10.6%

Q10. P1 and P2 are two processes that need to access a critical section of the code. The synchronization construct that is used by the processes is given below:

P1
While (true) {
Wants1 = true;
While (wants2 == true);
/* critical section*/
Wants1 = false;
} (/*remainder section)

P2
While (true) {
Wants2 = true;
While (wants1 == true);
/* critical section*/
Wants2 = false;
} (/*remainder section)

Wants1 and wants2 are shared variables and are initialized to false. The statement that holds true for the above construct is

1.     It does not ensure bounded waiting
2.     It does not ensure mutual exclusion
3.     It does not prevent deadlocks but ensures mutual exclusion
4.     It requires that processes enter the critical section in strict alternation.

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ANSWER: 3. It does not prevent deadlocks but ensures mutual exclusion

For questions 11 and 12 refer to the data given below:

Consider a c program barrier which is a synchronization construct where a set of processes synchronizes globally, i.e each process in the set arrives at the barrier and waits for all others to arrive and then all processes leave the barrier. Let S be the binary semaphore with P and V functions and number of processes in the set be 3.

Void barrier (void) {
1:          P(S);
2:          process_arrived++;
3:          V(S);
4:          while (process_arrived != 3);
5:          P(S);
6:          process_left++;
7:          if(process_left ==3); {
8:          process_arrived = 0;
9:          process_left = 0;
10:       }
11:        V(S);
}

process_arrived and process_left are the variables that are shared among all processes and are initialized to zero. In a concurrent program all the three processes all the barrier function when they need to synchronize globally.

Q11. The above implementation of barrier is incorrect. The statement that holds true is

1.       Lines 6 to 10 needs to be inside the critical section.
2.      The implementation of barrier is correct if there are only two processes instead of three.
3.      The above implementation of barrier may lead to deadlock if two barrier invocations are used in immediate succession.
4.      The above implementation of barrier is incorrect due to the use of binary semaphores S
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ANSWER: 3. The above implementation of barrier may lead to deadlock if two barrier invocations are used in immediate succession.

Q12. In order to rectify the problem in the implementation the statement that is true is

1.     The variable process_left is made private instead of shared.
2.     Context switch is disabled at the beginning and then re-enabled at the end
3.     Lines 6 to 10 are replaced by process_arrived.
4.     At the beginning of the barrier the first process to enter the barrier waits until process_arrived becomes zero before proceeding to execute P(S).
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ANSWER: 4. At the beginning of the barrier the first process to enter the barrier waits until process_arrived becomes zero before proceeding to execute P(S).

Q13. Consider a single processor system that has three resources types A, B and C and are shared by three processes. There are 5 units of each resources type. In the table shown below column alloc denotes the number of units of each resource type allocated to each process and the column request denotes the number of units of each resource type requested by a process in order to complete execution. From the processes P0, P1, P2 which one will finish last?

1.      P0
2.      P1
3.      P2
4.      System is in deadlock hence none of the process will finish last
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ANSWER: 3. P2

Q14. We have atomic fetch-and-set a, b instruction which unconditionally sets the memory location to 1 and it fetches the old value of a in b without allowing any intervening access to the memory location a. Implementation of function P and V on binary semaphore S are given below.

Void P (binay_semaphore*s) {
Unsigned b;
Unsigned *a = & (S-> value);
Do {

} While (b);
}
Void V (binay_semaphore *S) {
S -> value = 0;
}

The statement that holds true is

1.      The implementation of V is wrong.
2.      The implementation may not work if context switching is disabled in P.
3.      The code does not implement binary semaphore.
4.      Instead of using fetch-and-set, a pair of normal load/store can be used
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ANSWER: 2. The implementation may not work if context switching is disabled in P.

Q15. Table given below shows the arrival time and execution time for different processes. Shortest Remaining Time First (SRTF) process scheduling algorithm is used by the operating system. The total waiting time for process P2 is

1.    15
2.    20
3.    30
4.    45
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ANSWER: 1. 15

OS interview questions - June 23, 2014 at 01:25 PM by Nihal Singh

Q.1 What are the functions of operating system?

The operating system controls and coordinates the use of hardware among the different processes and applications. It provides the various functionalities to the users. The following are the main job of operating system.

  • Resource utilization
  • Resource allocation
  • Process management
  • Memory management
  • File management
  • I/O management
  • Device management

Q.2 Describe system calls and its type

System calls works as a mediator between user program and service provided by operating system. In actual situation, functions that make up an API (application program interface) typically invoke the actual system calls on behalf of the application programmer.

Types of System Calls

System calls can be grouped roughly into five major categories:

Sr No. Example
1 Process control Create process, terminate process, end, allocate and free memory etc
2 File manipulation Create file, delete file, open file, close file, read, write.
3 Device manipulation request device, release device, read, write, reposition, get device attributes, set device attributes etc.
4 Information maintenance get or set process, file, or device attributes
5 Communications Send, receive messages, transfer status information

Q.3 Explain Booting the system and Bootstrap program in operating system.

The procedure of starting a computer by loading the kernel is known as booting the system.

When a user first turn on or booted the computer, it needs some initial program to run. This initial program is known as Bootstrap Program. It is stored in read-only memory (ROM) or electrically erasable programmable read-only memory (EEPROM). Bootstrap program locates the kernel and loads it into main memory and starts its execution.

Q.4 Describe Main memory and Secondary memory storage in brief.

Main memory is also called random access memory (RAM). CPU can access Main memory directly. Data access from main memory is much faster than Secondary memory. It is implemented in a semiconductor technology, called dynamic random-access memory (DRAM).

Main memory is usually too small to store all needed programs. It is a volatile storage device that loses its contents when power is turned off. Secondary memory can stores large amount of data and programs permanently. Magnetic disk is the most common secondary storage device. If a user wants to execute any program it should come from secondary memory to main memory because CPU can access main memory directly.

Q.5 What are the advantages of multiprocessor system?

Systems which have more than one processor are called multiprocessor system. These systems are also known as parallel systems or tightly coupled systems.
Multiprocessor systems have the following advantages.

Increased Throughput: Multiprocessor systems have better performance than single processor systems. It has shorter response time and higher throughput. User gets more work in less time.

Reduced Cost: Multiprocessor systems can cost less than equivalent multiple single processor systems. They can share resources such as memory, peripherals etc.

Increased reliability: Multiprocessor systems have more than one processor, so if one processor fails, complete system will not stop. In these systems, functions are divided among the different processors.

Q.6 Is it possible to have a deadlock involving only one process? Explain your answer.

Deadlock with one process is not possible. Here is the explanation.

A deadlock situation can arise if the following four conditions hold simultaneously in a system.

  • Mutual Exclusion.
  • Hold and Wait.
  • No Preemption.
  • Circular-wait.

It is not possible to have circular wait with only one process, thus failing a necessary condition for Circular wait. There is no second process to form a circle with the first one. So it is not possible to have a deadlock involving only one process.

OS interview questions - May 31, 2013 at 12:17 PM by Kshipra Singh

1. What is an operating system?

An operating system is a collection of software programs which control the allocation and usage of various hardware resources in the system. It is the first program to be loaded in the computer and it runs in the memory till the system is shut down.

Some of the popular Operating Systems are DOS, Windows, Ubuntu, Solaris etc.

2. What are its main functions?

The main functions of an OS are:

a.) Process Management
b.) Memory Management
c.) Input/ Output Management
d.) Storage/ File system management

Operating system interview questions for freshers - Basic OS interview
Advanced operating system interview questions and answers

3. What is a Kernel?

- Kernel is the part of OS which handles all details of sharing resources and device handling.
- It can be considered as the core of OS which manages the core features of an OS.
- Its purpose is to handle the communication between software and hardware
- Its services are used through system calls.
- A layer of software called shell wraps around the Kernel.

4. What are the main functions of a Kernel?

The main functions of OS a Kernel are:

•Process management
•Device management
•Memory management
•Interrupt handling
•I/O communication
•File system management

5. What are the different types of Kernel?

Kernels are basically of two types:

a.) Monolithic Kernels - In this architecture of kernel, all the system services were packaged into a single system module which lead to poor maintainability and huge size of kernel.
b.) Microkernels - They follow the modular approach of architecture. Maintainability became easier with this model as only the concerned module is to be altered and loaded for every function. This model also keeps a tab on the ever growing code size of the kernel.

6. What are the disadvantages of Microkernels?

Following are the main disadvantages of Microkernels. Usually these disadvantages are situation based.

a.) Larger running memory footprint
b.) Performance loss due to the requirement of more software for interfacing.
c.) Difficulty in fixing the messaging bugs.
d.) Complicated process management.

7. What is a command interpreter?

It is a program that interprets the command input through key board or command batch file. It helps the user to interact with the OS and trigger the required system programs or execute some user application.

Command interpreter is also referred to as:
- Control card interpreter
- Command line interpreter
- Console command processor
- Shell

8. Explain Process.

A process is a program that is running and under execution. On batch systems, it is called as a "job" while on time sharing systems, it is called as a "task".

9. Explain the basic functions of process management.

Important functions of process management are:

- Creation and deletion of system processes.
- Creation and deletion of users.
- CPU scheduling.
- Process communication and synchronization.

10. What do you know about interrupt?

- Interrupt can be understood as a signal from a device causing context switch.
- To handle the interrupts, interrupt handlers or service routines are required.
- The address of each Interrupt service routine is provided in a list which is maintained in interrupt vector.

11. What is a daemon?

- Daemon - Disk and execution monitor, is a process that runs in the background without user’s interaction. They usually start at the booting time and terminate when the system is shut down.

12. How would you identify daemons in Unix?

- The name of daemons usually end with 'd' at the end in Unix.
- For e.g. httpd, named, lpd.

13. What do you mean by a zombie process?

- These are dead processes which are not yet removed from the process table.
- It happens when the parent process has terminated while the child process is still running. This child process now stays as a zombie.

14. What do you know about a Pipe? When is it used?

- It is an IPC mechanism used for one way communication between two processes which are related.
- A single process doesn't need to use pipe. It is used when two process wish to communicate one-way.

15. What is a named pipe?

- A traditional pipe is unnamed and can be used only for the communication of related process. If unrelated processes are required to communicate - named pipes are required.
- It is a pipe whose access point is a file available on the file system. When this file is opened for reading, a process is granted access to the reading end of the pipe. Similarly, when the file is opened for writing, the process is granted access to writing end of the pipe.
- A named pipe is also referred to as a named FIFO or just FIFO.

16. What are the various IPC mechanisms?

IPC - Inter Process Communication.

Various IPC mechanisms are:

a.) Sockets
b.) Pipes
c.) Shared memory
d.) Signals
e.) Message Queues

17. What is a semaphore?

- A semaphore is a hardware or a software tag variable whose value indicates the status of a common resource.
- Its purpose is to lock the common resource being used. A process which needs the resource will check the semaphore to determine the status of the resource followed by the decision for proceeding.
- In multitasking operating systems, the activities are synchronized by using the semaphore techniques.

18. What kind of operations are possible on a semaphore?

Two kind of operations are possible on a semaphore - 'wait' and 'signal'.

19. What is context switching?

- Context is associated with each process encompassing all the information describing the current execution state of the process
- When the OS saves the context of program that is currently running and restores the context of the next ready to run process, it is called as context switching.
- It is important for multitasking OS.

20. Tell us something about Mutex.

- Mutex - ‘Mutual Exclusion Lock’ is a lock which protects access to shared data resource.
- Threads can create and initialize a mutex to be used later.
- Before entering a critical region the mutex is locked. It is unlocked after exiting the critical region. If any thread tries to lock the mutex during this time, it can nt do so.

21. What is a critical section?

It is a section of code which can be executed only by one process at a time.

22. What is synchronization? What are the different synchronization mechanisms?

Synchronization means controlling access to a resource that is available to two or more threads or process. Different synchronization mechanisms are:

- Mutex
- Semaphores
- Monitors
- Condition variables
- Critical regions
- Read/ Write locks

23. What is the basic difference between pre-emptive and non-pre-emptive scheduling.

Pre-emptive scheduling allows interruption of a process while it is executing and taking the CPU to another process while non-pre-emptive scheduling ensures that a process keeps the CPU under control until it has completed execution.

24. Is non-pre-emptive scheduling frequently used in a computer? Why?

No, it is rarely used for the reasons mentioned below:

- It can not ensure that each user gets a share of CPU regularly.
- The idle time with this increases reducing the efficiency and overall performance of the system.
- It allows program to run indefinitely which means that other processes have to wait for very long.

25. Explain condition variable.

- These are synchronization objects which help threads wait for particular conditions to occur.
- Without condition variable, the thread has to continuously check the condition which is very costly on the resources.
- Condition variable allows the thread to sleep and wait for the condition variable to give it a signal.

26. What are read-write locks?

- Read - write locks provide simultaneous read access to many threads while the write access stays with one thread at a time. They are especially useful in protecting the data that is not frequently written but read simultaneously by many threads.
- They are slower than mutexes.

27. What is a deadlock?

- It is a condition where a group of two or more waiting for the resources currently in use by other processes of the same group.
- In this situation every process is waiting for an event to be triggered by another process of the group.
- Since no thread can free up the resource a deadlock occurs and the application hangs.

28. What are the necessary conditions for deadlock to occur?

a.) At least one resource should be occupied in a non-sharable condition.
b.) A process holding at least one resource is waiting for more resources currently in use by other processes.
c.) It is not possible to pre-empt the resource.
d.) There exists a circular wait for processes.

29. Name the functions constituting the OS's memory management.

- Memory allocation and de-allocation
- Integrity maintenance
- Swapping
- Virtual memory

30. Name the different types of memory?

a.) Main memory also called primary memory or RAM
b.) Secondary memory or backing storage
c.) Cache
d.) Internal process memory

31. Throw some light on Internal Process Memory.

- This memory consists of a set of high-speed registers. They work as temporary storage for instructions and data.

32. Explain compaction.

During the process of loading and removal of process into and out of the memory, the free memory gets broken into smaller pieces. These pieces lie scattered in the memory. Compaction means movement of these pieces close to each other to form a larger chunk of memory which works as a resource to run larger processes.

33. What are page frames?

Page frames are the fixed size contiguous areas into which the main memory is divided by the virtual memory.

34. What are pages?

- Pages are same sized pieces of logical memory of a program. Usually they range from 4 KB to 8 KB depending on the addressing hardware of the machine.
- Pages improve the overall system performance and reduces requirement of physical storage as the data is read in 'page' units.

35. Differentiate between logical and physical address.

- Physical addresses are actual addresses used for fetching and storing data in main memory when the process is under execution.
- Logical addresses are generated by user programs. During process loading, they are converted by the loader into physical address.

36. When does page fault error occur?

- It occurs when a page that has not been brought into main memory is accessed.

37. Explain thrashing.

- In virtual memory system, thrashing is a high page fault scenario. It occurs due to under-allocation of pages required by a process.
- The system become extremely slow due to thrashing leading to poor performance.

38. What are the basic functions of file management in OS?

- Creation and deletion of files/ directories.
- Support of primitives for files/ directories manipulation.
- Backing up of files on storage media.
- Mapping of files onto secondary storage.

39. Explain thread.

- It is an independent flow of control within a process.
- It consists of a context and a sequence of instructions for execution.

40. What are the advantage of using threads?

The main advantages of using threads are:

a.) No special communication mechanism is required.
b.) Readability and simplicity of program structure increases with threads.
c.) System becomes more efficient with less requirement of system resources.

41. What are the disadvantages of using threads?

The main disadvantages of using threads are:
- Threads can not be re-used as they exist within a single process.
- They corrupt the address space of their process.
- They need synchronization for concurrent read-write access to memory.

42. What is a compiler?

A compiler is a program that takes a source code as an input and converts it into an object code. During the compilation process the source code goes through lexical analysis, parsing and intermediate code generation which is then optimized to give final output as an object code.

43. What is a library?

It is a file which contains object code for subroutines and data to be used by the other program.

44. What are the advantages of distributed system?

Advantages of distributed system are:
- Resources get shared
- Load gets shared
- Reliability is improved
- Provide a support for inter-process communication

Operating system Interview questions - August 06, 2008 at 18:10 pm by Amit Satpute
1.Explain the meaning of Kernal.

The kernel is the essential center of a computer operating system, the core that........            
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2.What is a command interpreter?

The part of an Operating System that interprets commands and carries them out.........
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3.What is a daemon?

In Unix and some other operating systems, a daemon is a computer program that runs.........
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4.Explain the basic functions of process management.

The basic functions of the OS wrt the process management are :........
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5.What is a named pipe?

A connection used to transfer data between separate processes, usually on separate computers. ........
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6.What is pre-emptive and non-preemptive scheduling?

Tasks are usually assigned with priorities. At times it is necessary to run a certain.........
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7.What is a semaphore?

A semaphore is a variable. There are 2 types of semaphores:.........
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8.Explain the meaning of mutex.

A mutex and the binary semaphore are essentially the same..........
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9.What are the different types of memory?

the types of memory in a computer system are: .........
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10.Explain the meaning of virtual memory.

Virtual memory is an approach to make use of the secondary storage devices as an extension.........
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Operating system Interview questions with answers posted on May 06, 2009 at 13:10 pm by Vidya Sagar 
11.What is RTOS?

A certain capability within a specified time constraint is guaranteed by an operating system called ‘real time operating system’.......
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12.What is the difference between hard real-time and soft real-time OS?

Critical task completion on time is guaranteed by a hard real time system......
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13.What type of scheduling is there in RTOS?

The tasks of real time operating system have 3 states namely, ‘running’, ’ready’, ‘blocked’...........
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14.What is interrupt latency?

The time between a device that generates an interrupt and the servicing of the device that generated the interrupt is known as interrupt latency...........
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15.What is priority inheritance?

Priority inversion problems are eliminated by using a method called priority inheritance...........
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16.What is spin lock?

In a loop a thread waits simply (‘spins’) checks repeatedly until the lock becomes available............   
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17.What is an operating system? What are the functions of an operating system?

An operating system is an interface between hardware and software. OS is responsible for managing and co-ordinating the activities of a computer system..................
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18.What is paging? Why paging is used?

OS performs an operation for storing and retrieving data from secondary storage devices for use in main memory. Paging is one of such memory management scheme.................
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19.Difference between a process and a program

- A program is a set of instructions that are to perform a designated task, where as the process is an operation which takes the given instructions and perform the manipulations as per the code...............
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20.What is the meaning of physical memory and virtual memory?

Physical memory is the only memory that is directly accessible to the CPU. CPU reads the instructions stored in the physical memory and executes them continuously...............
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21.What is the difference between socket and pipe?

Sockets: Socket is a part of OSI layer model. Communication among different layers is performed through sockets. Application layer serves through some sockets to the presentation layer and upper application layer................
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22.What are the difference between THREAD, PROCESS and TASK?

A program in execution is known as ‘process’. A program can have any number of processes. Every process has its own address space..................
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23.Difference between NTFS and FAT32

The differences are as follows:NTFS:- Allows the access local to Windows 2000, Windows 2003, Windows NT with service pack 4 and later versions may get access for some file..................
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24.Differentiate between RAM and ROM

RAM: - Volatile memory  - Electricity needs to flow continuously - Program information is stored in RAM - RAM is read / write memory..............
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25.What is DRAM? In which form does it store data?

DRAM – Dynamic Random Access Memory. One of the read / write memory. DRAM is cheap and does the given task................
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26.What is cache memory? Explain its functions.

Cache memory is RAM. The most recently processing data is stored in cache memory. CPU can access this data more quickly than it can access data in RAM................
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27.Differentiate between Complier and Interpreter

- The program syntax is checked by the compiler; where as the keywords of the program is checked by the interpreter......................
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28.Describe different job scheduling in operating systems.

Job scheduling is an activity for deciding the time for a process to receive the resources they request.
First Come First Served: In this scheduling, the job that is waiting for a long time is served next................
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29.What is a Real-Time System?

Real-time system is the study of hardware and software, which are subject to the operational deadlines from event to the system response.................
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30.What do you mean by deadlock?

Dead lock is a situation of two or more processes waiting for each other to finish their tasks. In this situation no progress or no advancement is made...............
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31.Difference between Primary storage and secondary storage

Primary memory storages are temporary; where as the secondary storage is permanent. Primary memory is expensive and smaller, where as secondary memory is cheaper and larger................
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question

Create a file using the Vi editor with the name ‘names.txt’ which contains 10 names of yours and your friends. Write a script that will sort the names in the alphabetical order and print in on the screen. Your script should also have to test whether there are 10 names in the file and if not it should print an error message.

mphee 04-15-2014 06:19 AM

Good questions

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Neha 03-1-2014 07:46 PM

os intrvw qustn

u should provide some download link the way it is given is too bad to read.

ashish 09-15-2013 02:40 AM

difficult to copy and download

There is no question about your provided information really these are very useful for IT Exam preparation.

But it is difficult to see every question with answer in a single shot and there no download mechanism available so please make available one of these method.

vijay 02-12-2013 09:52 AM

os questions regarding

thanks.....very much useful

archana 01-16-2013 04:18 AM

operating system interview questions

nice but need more than this

Anusha 12-22-2011 02:40 AM

os

it's nice, but need much more...

sanga 11-6-2011 12:28 PM

 

 
 
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