西安电子科技大学操作系统期末试题及答案解析

西安电子科技大学操作系统期末试题及答案解析work Information Technology Company.2020YEAR西安电子科技大学考试时间 120 分钟试题题号一二三四五六七八九十总分分数1.考试形式：闭卷2.考试日期：年月日3.本试卷共四大题，满分100分班级学号姓名任课老师Part 1: Select one answer (A through D) for each question ( Total 20, each 2 )1.A computer system consists of , system programsand application programs.A.Control busB. data busC.Address busD.hardware2.Thread（线程）can be implemented in .A.Kernel SpaceB. UserSpaceC. Kernel Space or User SpaceD. Noneof the above3.In OS, short term schedule（调度）means .4.A. Job scheduling B.Process scheduling5.C. Thread scheduling D. CPUscheduling6.Which one of the following item is not shared by allthreads in a process7. .8.A. Address space B.RegisterC. Open filesD.Accounting information9.In the paged memory management system, the address iscomposed of page number and the offset within the page.In the address structure shown in the following figure,.31 10 9 0A. page size is 512, 2M pages at mostB. page size is 1k, 4M pages at mostC. page size is 2k, 8M pages at mostD. page size is 4k, 16M pages at most10.Virtual memory is based on The Principle of Locality.Which of the following statements about The Principle of Locality is correct11.12. A. Program executes codes in orderB.Program assesses（访问）memory in a non-uniform mannerC.Program accesses a lot of variables continuouslyD.Program accesses a relatively small portion of theaddress space at any instant of time13.In UNIX i-node scheme, a directory entry contains onlytwo fields: the14.file name (14 bytes) and the number of the i-node forthat file (2 bytes). These parameters（参数）limit the number of files per file system to15. .16. A. 64kB. 32k17. C. 16kD. 4k18.The time required to read or write a disk block isdetermined（决定）by three factors. Which one dominates （主导，占优势）the other two times19. .20. A. Seek time B.Rotational delay time21. C. Data transfer time D.None of above22.The chmod command（命令）is used to change thepermission（许可）of file in Linux. To use it, you specify the desired permission setting and the file or files that you wish to modify. The permission settings are usually a series of bits. Which of the following bits pre sent that the file’s owner may read, write, and execute the file, while all others may only read the file23. .24. A. 755 B. 744 C. 644D. 61125. The method listed below doesn’t needCPU to participate in the transfer of data block.A.Interrupt-Driven I/OB. DMAC. Programmed I/OD. None ofabovePart2: Fill Blanks (Total 20, each 2)1.Operating System is an extended machine and .2.Parallel Systems include Symmetric（对称）multiprocessing and .3.A semaphore（信号量）S is an integer variable that isaccessed only through two standard atomic operations: and .4.Address binding of instructions and data to memoryaddresses can happen at three different stages, Compiletime, Load time and .5.Four Conditions for Deadlock: , Hold and wait,No preemption and .6.Sector 0 of the disk is called the .7.In Linux, the file metadata（元数据）is storedin .8.The security goals include data confidentiality（机密性），data integrity and .Part3: Essay Questions (Total 20, each 4)1.What is system call Use an example to illustrate（举例说明）the steps of system call.2.Please describe the diagram（图）of Process State.3.What is Monitor Can you use Monitor to implement a datastructure4.5.What is TLB What role does it play in memory management6.7.The difference of programming I/O and interrupt-drivenI/O.Part4: Integrate Questions (Total 40, each 10)1.The code below is an example program of producer-consumer. The product produce numbers from 1 to MAX and the consumer will read it. Please fill blanks in the code#include <stdio.h>#include <pthread.h>#define MAX 1000000000 /* how many numbers to produce */pthread_mutex_t the_mutex;pthread_cond_t condc, condp;int buffer = 0; /* buffer used between producer and consumer*/void *producer(void *ptr) /*produce data*/{int i;for (i=1; i<=MaX; i++) {pthread_mutex_lock(&the_mutex);while ( ① )pthread_cond_wait (&condp, &the_mutex);buffer = i;②③}pthread_exit (0);}void *consumer (void *ptr) /*consume data*/{int i, res;for (i = 1; i<=MAX; i++) {pthread_mutex_lock (&the_mutex);while ( ④ )pthread_cond_wait ( ⑤ , &the_mutex);res = buffer;buffer = 0;pthread_cond_signal(&condp);⑥printf(“buffer=%d\n” , res);}pthread_exit (0);}int main (int argc, char **argv){pthread_t pro,con;pthread_mutex_init (&the_mutex, 0);pthread_cond_init (&condc, 0);pthread_cond_init (&condp, 0);pthread_create (&con, 0, consumer, 0);pthread_create (&pro, 0, producer, 0);pthread_join (pro, 0);pthread_join (con, 0);pthread_cond_destroy (&condc);pthread_cond_destroy (&condp);pthread_mutex_destroy (&the_mutex);}2.Consider the following snapshot of a system :Answer the following questions using the banker’s algorithm:(1)What is the content of the matrix Need(4points)(2)Is the system in safe state Why (3 points)(3)If a request from process P1 arrives for(0,4,2,0), can the request be granted（允许）immediately(4)(3 points)3.Disk requests come in to the disk driver for cylinders10,22,20,2,40,6, and 38, in that order. In all cases, the arm is initially at cylinder 20. A seek takes 6 msec per cylinder moved. How much seek times is needed for(1)First Come First Served (FCFS). (3points)(2)Shorted Seek Time First (SSTF). (3 points)(3)Elevator algorithm (SCAN, initially moving upward)(4 points)4.Consider a system where the virtual memory page size is2K (2048 bytes), and main memory consists of 4 page frames. Now consider a process which requires 8 pages of storage. At some point during its execution, the page table is as shown below:Answer the following questions:(1)List the virtual address ranges for each virtualpage. (3 points)(2)List the virtual address ranges that will resultin a page fault.(3 points)(3)Give the main memory (physical) addresses foreach of the following virtual addresses (all numbers decimal): (i) 8500, (ii) 1400, (iii) 5000, (iv) 2100.(4 points)AnswersPart 1: Select one answer (A through D) for each question1. D计算机系统由硬件和软件（系统程序+应用程序）组成。