Skip to main content

What is the purpose of the command interpreter? Why is it usually separate from the kernel? Would it be possible for the user to develop a new command interpreter using the system‐call interface provided by the operating system?

 What is the purpose of the command interpreter? Why is it usually separate from the kernel? Would it be possible for the user to develop a new command interpreter using the system‐call interface provided by the operating system?


Answer: It reads commands from the user or a file of commands and executes them, usually by turning them into one or more system calls. It is usually not part of the kernel since the command interpreter is subject to changes. A user should be able to develop a new command interpreter using the system‐call interface provided by the operating system. The command interpreter allows a user to create and manage processes and also determine ways by which they communicate (such as through pipes and files). As all of this functionality could be accessed by a user‐level program using the system calls, it should be possible for the user to develop a new command‐line interpreter.


Why is the separation of mechanism and policy desirable in an operating system?


Answer: Mechanism and policy must be separate to ensure that systems are easy to modify. No two system installations are the same, so each installation may want to tune the operating system to suit its needs. With mechanism and policy separate, the policy may be changed at will while the mechanism stays unchanged. This arrangement provides a more flexible system. 


What are the two models of interprocess communication? What are the strengths and weaknesses of the two approaches?


Answer: The two models of interprocess communication are the message-passing Model and the shared‐memory model.


In what ways is the modular kernel approach similar to the layered approach? In what ways does it differ from the layered approach?


Answer: The modular kernel approach requires subsystems to interact with each other through carefully constructed interfaces that are typically narrow (in terms of the functionality that is exposed to external modules). The layered kernel approach is similar in that respect. However, the layered kernel imposes a strict ordering of subsystems such that subsystems at the lower layers are not allowed to invoke operations corresponding to the upper‐layer subsystems. There are no such restrictions in the modular‐kernel approach, wherein modules are free to invoke each other without any constraints.


What is the main advantage for an operating‐system designer of using a virtual‐machine architecture? What is the main advantage for a user?


Answer: The system is easy to debug, and security problems are easy to solve. virtual machines also provide a good platform for operating system research since many different operating systems may run on one physical system.


What is the purpose of interrupts? What are the differences between a trap and an interrupt? Can traps be generated intentionally by a user program? If so, for what purpose?


Answer: An interrupt is a hardware‐generated change‐of‐flow within the system. An interrupt handler is summoned to deal with the cause of the interrupt; control is then returned to the interrupted context and instruction. A trap is a software‐generated interrupt. An interrupt can be used to signal the completion of an I/O to obviate the need for device polling. A trap can be used to call operating system routines or to catch arithmetic errors.

Labels:

Comments

Post a Comment

Popular posts from this blog

40 Redis interview questions for 2021 - 2022

  Redis interview questions 1.What is Redis?. 2. What is the data type of Redis? 3. What are the benefits of using Redis? 4. What are the advantages of Redis over Memcached? 5. What are the differences between Memcache and Redis? 6. Is Redis single-process and single-threaded? 7. What is the maximum storage capacity of a string type value? 8. What is the persistence mechanism of Redis? Their advantages and disadvantages? 9. Redis common performance problems and solutions: 10. What is the deletion strategy of redis expired keys? 11. Redis recycling strategy (elimination strategy)? 12. Why does edis need to put all data in memory? 13. Do you understand the synchronization mechanism of Redis? 14. What are the benefits of Pipeline? Why use pipeline? 15. Have you used Redis cluster? What is the principle of cluster? 16. Under what circumstances will the Redis cluster solution cause the entire cluster to be unavailable? 17. What are the Java clients supp...
Post a Comment
Read more

8 common methods for server performance optimization

  1. Use an in-memory database In-memory database is actually a database that puts data in memory and operates directly. Compared with the disk, the data read and write speed of the memory is several orders of magnitude higher. Saving the data in the memory can greatly improve the performance of the application compared to accessing it from the disk. The memory database abandoned the traditional way of disk data management, redesigned the architecture based on all data in memory, and made corresponding improvements in data caching, fast algorithms, and parallel operations, so the data processing speed is faster than that of traditional databases. Data processing speed is much faster.       But the problem of security can be said to be the biggest flaw in the memory database. Because the memory itself has the natural defect of power loss, when we use the memory database, we usually need to take some protection mechanisms for the data on the memory in advance, such...
Post a Comment
Read more

Recursion-maze problem - Rat in the Maze - Game

  package com.bei.Demo01_recursion; public class MiGong {     public static void main(String[] args)  {         //First create a two-dimensional array to simulate the maze         int [][]map=new int[8][7];         //Use 1 for wall         for (int i = 0; i <7 ; i++) {             map[0][i]=1;             map[7][i]=1;         }         for (int i = 0; i <8 ; i++) {             map[i][0]=1;             map[i][6]=1;         }         //Set the bezel         map[3][1]=1;         map[3][2]=1;         //Output         for (int i = 0; i <8 ; i++) {             for (int j = 0; j ...
Post a Comment
Read more