《unix操作系统设计》英文版习题答案1

S.1Uni x Internals (Apri l/May-2012, Set-1) JNTU-AnantapurB.T ech. III-Year II-Sem.( JNTU-Anantapur)Code No.: 9A05602/R09B.Tech. III Year II Semester Regular ExaminationsApril/May - 2012UNIX INTERNALS( Computer Science and Engineering )Time: 3 HoursMax. Marks: 70Answer any FIVE Questions All Questions carry equal marks- - -1.(a)Draw and explain the architecture of UNIX systems. (Unit-I, Topic No. 1.5.1)(b)Discuss in detail the sleep and wakeup procedures. (Unit-V, Topic No. 5.6)2.(a)If several processes contend for a buffer, the kernel guarantees that none of them sleep forever, but it does not guarantee that a process will not be starved out from use of a buffer. Redesign getblk so that a process is guaranteed eventual use of a buffer. (Unit-II, Topic No. 2.2)(b)What are the advantages of buffer cache? (Unit-II, Topic No. 2.5)3.(a)Give brief description about the accessing of inodes. (Unit-III, Topic No. 3.1)(b)Explain in detail the continuous file allocation system. (Unit-III, Topic No. 3.7)4.(a)What is the purpose of link system call? Write and explain the algorithm for the same. (Unit-IV, Topic No. 4.12)(b)Write short notes on maintenance of the file system. (Unit-IV, Topic No. 4.15)5.(a)What are regions? Explain their role in the creation of processes. (Unit-V, Topic No. 5.2)(b)Describe in detail about the context of a process. (Unit-V, Topic No. 5.3)6.(a)Explain the security problems that exists if a setuid program is not write protected. (Unit-VI, Topic No. 6.6)(b)Write a program that a parent and child shares a file access. (Unit-VI, Topic No. 6.1)7.(a)Discuss in detail about the scheduling parameters. (Unit-VII, Topic No. 7.1)(b)What is profiling? Explain it in detail. (Unit-VII, Topic No. 7.3)8.(a)Explain clists in detail. (Unit-VIII, Topic No. 8.3)(b)Write a program for reading data from risk by using block and raw interfaces.(Unit-VIII, Topic No. 8.3)S.2Spectrum ALL-IN-ONE Journal for Engineering Students, 2013B.T ech. III-Year II-Sem.( JNTU-Anantapur)(a)Draw and explain the architecture of UNIX systems.Answer :April/May-12, Set-1, Q1(a)For answer refer Unit-I, Q10.(b)Discuss in detail the sleep and wakeupprocedures.Answer :April/May-12, Set-1, Q1(b)For answer refer Unit-V , Q17.(a)If several processes contend for a buffer,the kernel guarantees that none of them sleep forever, but it does not guarantee that a process will not be starved out from use of a buffer. Redesign getblk so that a process is guaranteed eventual use of a buffer.Answer :April/May-12, Set-1, Q2(a)Algorithm getblkInput: file system numberblock numberOutput: locked buffer that can now be used for block {While (buffer not found){If (block is not in hash queue)If (there are no buffers on free list){sleep (event any buffer becomes free);continue;}}}}(b)What are the advantages of buffercache?Answer :April/May-12, Set-1, Q2(b)For answer refer Unit-II, Q12, Topic: Advantages of Buffer Cache.(b)Explain in detail the continuous file allocation system.Answer :April/May-12, Set-1, Q3(b)Note: In given question, the word contiguous is misprinted as continous.In a contiguous allocation method all the files are arranged in a sequential blocks of memory. Therefore,according to this technique, if a file size is k blocks and it starts from a block s then it spans till stk-1 block numbers.With this approach file access is much faster as all files occupy contiguous blocks. For sequential access of files the physical address of the last referred block is noted, so that file access can start from the next block to avoid repeated accesse of the previous blocks. Direct access is also supported since only the starting address and the block numbers are required.Though access time is minimal, contiguous allocation does not make efficient use of the disk space. This means that the allocation of space for a new file should be efficient enough so that disk space is not wasted.01234567891011IMP 1213141516171819USER 2021222324252627CHAT28293031DISKFOO DirectoryFilename Starting locationSize Foo Imp User Chat0614263764S.3Uni x Internals (Apri l/May-2012, Set-1) JNTU-AnantapurB.T ech. III-Year II-Sem.( JNTU-Anantapur)common strategies used for this purposes are best fit, worst fit and first fit. First fit is considered to usually be faster than other two strategies. However, best fit is also considered to be efficient than worse fit in terms of both time and disk space utilization.Even if any of these strategies are used all of them suffer with external fragmentation. External fragmentation occurs when a request for a new file cannot be satisfied because the largest available continous chunk of blocks is not sufficient enough for the file. Therefore, even though there is enough space available for the file, but it is not utilized because it is not contiguous.External fragmentation can be resolved with the help of a scheme called “compaction”. Compaction involves rearranging all free memory locations in a sequential order so that contiguous disk space can be allocated for new files instead of space getting wasted. Compaction is carried out in both off line and online modes. In offline mode, all operations are suspended and the file system is unmounted and finally compaction is performed. In this strategy a lot of time is wasted and hence it is avoided. In online mode,compaction is performed alongwith other system operations,but it effects business performance and reduces time wastages.However, contiguous allocation scheme also requires to know the size of the new file for storage allocation. This requirement looks simple but it might be difficult to determine the size of an output file after its execution. Also if the size of the file is determined in advance and the file takes a lot of time to reach its final size, then a lot of space is wasted when the file has not reached its final size. Therefore, a modified contiguous allocation scheme is used that allocates some fixed amount of space for the first time and if the file requires more space then another chunk of contiguous blocks is allocated. The chunks of free blocks allocated after the first allocation is called “infant” and these two chunks of contiguous blocks are linked to one another.Q4.(a)What is the purpose of link system call?Write and explain the algorithm for the same.Answer :April/May-12, Set-1, Q4(a)Link ( ) System CallLink system call creates a new directory for an inode that already exists. It does this by linking a file to a new name in the file system directory. It has the following syntax,Link (existing filename, new filename)The file system consists of pathnames for each link.Processes use the pathnames to access files. All the filenames are given equal priorities since the Kernel is unaware of the original filename.Figure: Files Linked in the File System Tree After executing the system callsLink(“/usr/foo/sys/src”,“/usr/lib/src'');Link(“/usr/lib/file3.h” ,“/usr/Foo/src/file2h”);The paths–“/usr/Foo/sys/file2.h”,“/usr/lib/file 3.h” and “/usr/lib/src/file2-h” belong to the same file.The Kernel gives the authority of linking directories only to the superuses. This makes the coding part easy for the program that traverses the file system tree. Superusers have to be careful enough while linking directories. The ability of linking directories must be supported on early versions of the system. This is because, the mkdir command used for creating new directories depends on the ability of linking directories. Algorithm of link system call, the following algorithm shows the linking of files.1.Obtain the inode for existing filename using the“namei” algorithm.2.If there are many links on file/linking directorywithout superuser permission.(a)Release the inode using “iput” algorithm.(b)Return error.3.Increment the link count.4.Update the disk copy of inode.5.Unlock inode.6.Obtain the parent for the new filename inodeusing “namei” algorithm.7.If the new file already exists,(a)Undo the update done in step (4)(b)Return error.8.Make a new directory in the parent directory ofthe new filename.9.Add the new filename and the inode number ofthe existing filename.10.Release parent directory inode using “iput”algorithm.11.Release the inode of the existing file using “iput”algorithm.S.4Spectrum ALL-IN-ONE Journal for Engineering Students, 2013 The Kernel uses “namei” algorithm to look up for the inode for the source file. It increments the link count and updates the disk copy of the inode. Then it unlocks the inode and searches for the target file. The “Link” call will fail if the file is not available. The Kernel decrements the link count which was previously incremented. If the file is not available then it finds a free slot in the parent directory of the target file. It writes the name of the target file and inode number of the source file into it. Using the “iput” algorithm, it releases the inode of the target file parent directory. The target file did not exist in actual. Hence no more inodes are there to be released.Finally, the Kernel releases the source file inode which is accessed by another file in the file system. The link count increases by 1. It keeps track of the count of directory entries belonging to the file. This differentiates it from the inode reference count. If the file is not accessed by any other process then inode reference count of the file becomes 0 and the link count will be atleast 2.(b)Write short notes on maintenance of the file system.Answer :April/May-12, Set-1, Q4(b) For answer refer Unit-IV, Q17.Q5.(a)What are regions? Explain their role in the creation of processes.Answer :April/May-12, Set-1, Q5(a) For answer refer Unit-V, Q3, Topic: Regions.(b)Describe in detail about the context of a process.Answer :April/May-12, Set-1, Q5(b) For answer refer Unit-V, Q7.Q6.(a)Explain the security problems that exists if a setuid program is not write protected. Answer :April/May-12, Set-1, Q6(a) The security problems that exists if a setuid program is not write protected are as follows,1.The first problem is associated with executing a sequence of commands defined by the attacker.2.The second problem is associated with substituting data of the attacker’s choosing for data created by aprogram.In the first problem, an attacker takes advantage of the setuid program’s running with special privileges to force it to execute whatever commands the attacker needs. For instance, assume that an attacker found a copy of the Bourne shell that was setuid to root. The attacker could then execute the shell since, the shell is setuid to root these commands would be executed as though root had typed them. Hence, the attacker could do anything he wants, since root is the most highly privileged user on the system. Even if the shell were changed to read from a command file rather than accept commands interactively, the attacker could file simply create his own script and run the shell using it. This situation must be avoided and it is easy to avoid but it occurs frequently.Several problems are possible with world-writable files. Occasionally programs will use temporary files for various purposes, the function of the program depending on what is in the file. If the program closes the temporary file at any point and then reopens it later, an attacker can replace the temporary file with a file with other data that will cause the program to act as the attacker desires. If the replacement file has the same owner and group as the temporary file, it can be very difficult for the program to determine if it is being spoofed.(b)Write a program that a parent and child shares a file access.Answer :April/May-12, Set-1, Q6(b) For answer refer Unit-VI, Q3.Q7.(a)Discuss in detail about the scheduling parameters.Answer :April/May-12, Set-1, Q7(a) Scheduling parameters are used for process scheduling each entry in the process table contains a priority field that can be used for process scheduling.For remaining answer refer Unit-VII, Q2.B.T ech. III-Year II-Sem.( JNTU-Anantapur)S.5Uni x Internals (Apri l/May-2012, Set-1) JNTU-AnantapurB.T ech. III-Year II-Sem.( JNTU-Anantapur)(b)What is profiling? Explain it in detail.Answer :April/May-12, Set-1, Q7(b)For answer refer Unit-VII, Q10.Q8.(a)Explain clists in detail.Answer :April/May-12, Set-1, Q8(a)For answer refer Unit-VIII, Q11.(b)Write a program for reading data from risk by using block and raw interfaces.Answer :April/May-12, Set-1, Q8(b)#include “Fcntl.h”main ( ){char buffer1 [4096], buffer2 [4096];int Fd1, Fd2, i ;if (((Fd1 = open(“/dev/dsk5”, 0_RDONLY))= = –1)|| ((Fd2 = open(“/dev/rdsk5”, 0_RDONLY))= = –1))){printf (“read operation failed\n”);exit( );}for (i = 0; i < size of (buffer1); i + +)if (buffer 1[i]! = buffer2 [i]){printf(''different at offset %d\n'', i);exit ( );}printf(“reads successfully matched\n”);}。