现代操作系统(第三版)答案.pdf

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1、MODERNOPERATINGSYSTEMSSECOND EDITIONPROBLEM SOLUTIONSANDREW S.TANENBAUMVrije UniversiteitAmsterdam,The NetherlandsPRENTICE HALLUPPER SADDLE RIVER,NJ 07458课后答案网 w w w.k h d a w.c o mSOLUTIONS TO CHAPTER 1 PROBLEMS1.An operating system must provide the users with an extended(i.e.,virtual)machine,and i

2、t must manage the I/O devices and other system resources.2.Multiprogramming is the rapid switching of the CPU between multipleprocesses in memory.It is commonly used to keep the CPU busy while oneor more processes are doing I/O.3.Input spooling is the technique of reading in jobs,for example,from ca

3、rds,onto the disk,so that when the currently executing processes are finished,there will be work waiting for the CPU.Output spooling consists of firstcopying printable files to disk before printing them,rather than printingdirectly as the output is generated.Input spooling on a personal computer isn

4、ot very likely,but output spooling is.4.The prime reason for multiprogramming is to give the CPU something to dowhile waiting for I/O to complete.If there is no DMA,the CPU is fully occu-pied doing I/O,so there is nothing to be gained(at least in terms of CPU utili-zation)by multiprogramming.No matt

5、er how much I/O a program does,theCPU will be 100 percent busy.This of course assumes the major delay is thewait while data are copied.A CPU could do other work if the I/O were slowfor other reasons(arriving on a serial line,for instance).5.Second generation computers did not have the necessary hard

6、ware to protectthe operating system from malicious user programs.6.It is still alive.For example,Intel makes Pentium I,II,and III,and 4 CPUswith a variety of different properties including speed and power consumption.All of these machines are architecturally compatible.They differ only inprice and p

7、erformance,which is the essence of the family idea.7.A 25 80 character monochrome text screen requires a 2000-byte buffer.The1024 768 pixel 24-bit color bitmap requires 2,359,296 bytes.In 1980 thesetwo options would have cost$10 and$11,520,respectively.For currentprices,check on how much RAM current

8、ly costs,probably less than$1/MB.8.Choices(a),(c),and(d)should be restricted to kernel mode.9.Personal computer systems are always interactive,often with only a singleuser.Mainframe systems nearly always emphasize batch or timesharing withmany users.Protection is much more of an issue on mainframe s

9、ystems,as isefficient use of all resources.10.Every nanosecond one instruction emerges from the pipeline.This means themachine is executing 1 billion instructions per second.It does not matter atall how many stages the pipeline has.A 10-stage pipeline with 1 nsec per课后答案网 w w w.k h d a w.c o m2PROBL

10、EM SOLUTIONS FOR CHAPTER 1stage would also execute 1 billion instructions per second.All that matters ishow often a finished instructions pops out the end of the pipeline.11.The manuscript contains 80 50 700=2.8 million characters.This is,ofcourse,impossible to fit into the registers of any currentl

11、y available CPU andis too big for a 1-MB cache,but if such hardware were available,themanuscript could be scanned in 2.8 msec from the registers or 5.8 msec fromthe cache.There are approximately 2700 1024-byte blocks of data,so scan-ning from the disk would require about 27 seconds,and from tape 2 m

12、inutes 7seconds.Of course,these times are just to read the data.Processing andrewriting the data would increase the time.12.Logically,it does not matter if the limit register uses a virtual address or aphysical address.However,the performance of the former is better.If virtualaddresses are used,the

13、addition of the virtual address and the base registercan start simultaneously with the comparison and then can run in parallel.Ifphysical addresses are used,the comparison cannot start until the addition iscomplete,increasing the access time.13.Maybe.If the caller gets control back and immediately o

14、verwrites the data,when the write finally occurs,the wrong data will be written.However,if thedriver first copies the data to a private buffer before returning,then the callercan be allowed to continue immediately.Another possibility is to allow thecaller to continue and give it a signal when the bu

15、ffer may be reused,but thisis tricky and error prone.14.A trap is caused by the program and is synchronous with it.If the program isrun again and again,the trap will always occur at exactly the same position inthe instruction stream.An interrupt is caused by an external event and itstiming is not re

16、producible.15.Base=40,000 and limit=10,000.An answer of limit=50,000 is incorrectfor the way the system was described in this book.It could have been imple-mented that way,but doing so would have required waiting until the address+base calculation was completed before starting the limit check,thus s

17、low-ing down the computer.16.The process table is needed to store the state of a process that is currentlysuspended,either ready or blocked.It is not needed in a single process sys-tem because the single process is never suspended.17.Mounting a file system makes any files already in the mount point

18、directoryinaccessible,so mount points are normally empty.However,a systemadministrator might want to copy some of the most important files normallylocated in the mounted directory to the mount point so they could be found intheir normal path in an emergency when the mounted device was beingchecked o

19、r repaired.课后答案网 w w w.k h d a w.c o mPROBLEM SOLUTIONS FOR CHAPTER 1318.Forkcan fail if there are no free slots left in the process table(and possibly ifthere is no memory or swap space left).Execcan fail if the file name givendoes not exist or is not a valid executable file.Unlinkcan fail if the f

20、ile to beunlinked does not exist or the calling process does not have the authority tounlink it.19.If the call fails,for example because fd is incorrect,it can return 1.It canalso fail because the disk is full and it is not possible to write the number ofbytes requested.On a correct termination,it a

21、lways returns nbytes.20.It contains the bytes:1,5,9,2.21.Block special files consist of numbered blocks,each of which can be read orwritten independently of all the other ones.It is possible to seek to any blockand start reading or writing.This is not possible with character special files.22.System

22、calls do not really have names,other than in a documentation sense.When the library procedure read traps to the kernel,it puts the number of thesystem call in a register or on the stack.This number is used to index into atable.There is really no name used anywhere.On the other hand,the nameof the li

23、brary procedure is very important,since that is what appears in theprogram.23.Yes it can,especially if the kernel is a message-passing system.24.As far as program logic is concerned it does not matter whether a call to alibrary procedure results in a system call.But if performance is an issue,if ata

24、sk can be accomplished without a system call the program will run faster.Every system call involves overhead time in switching from the user contextto the kernel context.Furthermore,on a multiuser system the operating sys-tem may schedule another process to run when a system call completes,further s

25、lowing the progress in real time of a calling process.25.SeveralUNIXcalls have no counterpart in the Win32 API:Link:a Win32 program cannot refer to a file by an alternate name or see it inmore than one directory.Also,attempting to create a link is a convenient wayto test for and create a lock on a f

26、ile.Mount and umount:a Windows program cannot make assumptions aboutstandard path names because on systems with multiple disk drives the drivename part of the path may be different.Chmod:Windows programmers have to assume that every user can accessevery file.Kill:Windows programmers cannot kill a mi

27、sbehaving program that is notcooperating.课后答案网 w w w.k h d a w.c o m4PROBLEM SOLUTIONS FOR CHAPTER 126.The conversions are straightforward:(a)A micro year is 106 365 24 3600=31.536 sec.(b)1000 meters or 1 km.(c)There are 240bytes,which is 1,099,511,627,776 bytes.(d)It is 6 1024kg.SOLUTIONS TO CHAPTE

28、R 2 PROBLEMS1.The transition from blocked to running is conceivable.Suppose that a proc-ess is blocked on I/O and the I/O finishes.If the CPU is otherwise idle,theprocess could go directly from blocked to running.The other missing transi-tion,from ready to blocked,is impossible.A ready process canno

29、t do I/O oranything else that might block it.Only a running process can block.2.You could have a register containing a pointer to the current process tableentry.When I/O completed,the CPU would store the current machine statein the current process table entry.Then it would go to the interrupt vector

30、 forthe interrupting device and fetch a pointer to another process table entry(theservice procedure).This process would then be started up.3.Generally,high-level languages do not allow one the kind of access to CPUhardware that is required.For instance,an interrupt handler may be requiredto enable a

31、nd disable the interrupt servicing a particular device,or to manipu-late data within a process stack area.Also,interrupt service routines mustexecute as rapidly as possible.4.There are several reasons for using a separate stack for the kernel.Two ofthem are as follows.First,you do not want the opera

32、ting system to crashbecause a poorly written user program does not allow for enough stack space.Second,if the kernel leaves stack data in a user programs memory spaceupon return from a system call,a malicious user might be able to use this datato find out information about other processes.5.It would

33、 be difficult,if not impossible,to keep the file system consistent.Suppose that a client process sends a request to server process 1 to update afile.This process updates the cache entry in its memory.Shortly thereafter,another client process sends a request to server 2 to read that file.Unfor-tunate

34、ly,if the file is also cached there,server 2,in its innocence,will returnobsolete data.If the first process writes the file through to the disk after cach-ing it,and server 2 checks the disk on every read to see if its cached copy isup-to-date,the system can be made to work,but it is precisely all t

35、hese diskaccesses that the caching system is trying to avoid.课后答案网 w w w.k h d a w.c o mPROBLEM SOLUTIONS FOR CHAPTER 256.When a thread is stopped,it has values in the registers.They must be saved,just as when the process is stopped the registers must be saved.Timesharingthreads is no different than

36、 timesharing processes,so each thread needs itsown register save area.7.No.If a single-threaded process is blocked on the keyboard,it cannot fork.8.A worker thread will block when it has to read a Web page from the disk.Ifuser-level threads are being used,this action will block the entire process,de

37、stroying the value of multithreading.Thus it is essential that kernel threadsare used to permit some threads to block without affecting the others.9.Threads in a process cooperate.They are not hostile to one another.If yield-ing is needed for the good of the application,then a thread will yield.Afte

38、rall,it is usually the same programmer who writes the code for all of them.10.User-level threads cannot be preempted by the clock uless the whole processquantum has been used up.Kernel-level threads can be preempted individu-ally.In the latter case,if a thread runs too long,the clock will interrupt

39、thecurrent process and thus the current thread.The kernel is free to pick a dif-ferent thread from the same process to run next if it so desires.11.In the single-threaded case,the cache hits take 15 msec and cache misses take90 msec.The weighted average is 2/3 15+1/3 90.Thus the meanrequest takes 40

40、 msec and the server can do 25 per second.For a mul-tithreaded server,all the waiting for the disk is overlapped,so every requesttakes 15 msec,and the server can handle 66 2/3 requests per second.12.Yes.If the server is entirely CPU bound,there is no need to have multiplethreads.It just adds unneces

41、sary complexity.As an example,consider a tele-phone directory assistance number(like 555-1212)for an area with 1 millionpeople.If each(name,telephone number)record is,say,64 characters,theentire database takes 64 megabytes,and can easily be kept in the serversmemory to provide fast lookup.13.The poi

42、nters are really necessary because the size of the global variable isunknown.It could be anything from a character to an array of floating-pointnumbers.If the value were stored,one would have to give the size tocreate?global,which is all right,but what type should the second parameterof set?global b

43、e,and what type should the value of read?global be?14.It could happen that the runtime system is precisely at the point of blocking orunblocking a thread,and is busy manipulating the scheduling queues.Thiswould be a very inopportune moment for the clock interrupt handler to begininspecting those que

44、ues to see if it was time to do thread switching,since theymight be in an inconsistent state.One solution is to set a flag when the run-time system is entered.The clock handler would see this and set its own flag,课后答案网 w w w.k h d a w.c o m6PROBLEM SOLUTIONS FOR CHAPTER 2then return.When the runtime

45、 system finished,it would check the clock flag,see that a clock interrupt occurred,and now run the clock handler.15.Yes it is possible,but inefficient.A thread wanting to do a system call firstsets an alarm timer,then does the call.If the call blocks,the timer returnscontrol to the threads package.O

46、f course,most of the time the call will notblock,and the timer has to be cleared.Thus each system call that mightblock has to be executed as three system calls.If timers go off prematurely,all kinds of problems can develop.This is not an attractive way to build athreads package.16.The priority inver

47、sion problem occurs when a low-priority process is in itscritical region and suddenly a high-priority process becomes ready and isscheduled.If it uses busy waiting,it will run forever.With user-levelthreads,it cannot happen that a low-priority thread is suddenly preempted toallow a high-priority thr

48、ead run.There is no preemption.With kernel-levelthreads this problem can arise.17.Each thread calls procedures on its own,so it must have its own stack for thelocal variables,return addresses,and so on.This is equally true for user-levelthreads as for kernel-level threads.18.A race condition is a si

49、tuation in which two(or more)processes are about toperform some action.Depending on the exact timing,one or the other goesfirst.If one of the processes goes first,everything works,but if another onegoes first,a fatal error occurs.19.Yes.The simulated computer could be multiprogrammed.For example,whi

50、le process A is running,it reads out some shared variable.Then a simu-lated clock tick happens and process B runs.It also reads out the same vari-able.Then it adds 1 to the variable.When process A runs,if it also adds oneto the variable,we have a race condition.20.Yes,it still works,but it still is