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1、MODERN OPERATING SYSTEMS SECOND EDITIONPROBLEM SOLUTIONSANDREW S.TANENBAUM Vrije Universiteit Amsterdam,The NetherlandsPRENTICE HALLUPPER SADDLE RIVER,NJ 07458SOLUTIONS TO CHAPTER 1 PROBLEMS1.An operating system must provide the users with an extended(i.e.,virtual)machine,and it must manage the I/O
2、devices and other system resources.2.Multiprogramming is the rapid switching of the CPU between multiple processes in memory.It is commonly used to keep the CPU busy while one or more processes are doing I/O.3.Input spooling is the technique of reading in jobs,for example,from cards,onto the disk,so
3、 that when the currently executing processes are finished,there will be work waiting for the CPU.Output spooling consists of first copying printable files to disk before printing them,rather than printing directly as the output is generated.Input spooling on a personal computer is not very likely,bu
4、t output spooling is.4.The prime reason for multiprogramming is to give the CPU something to do while waiting for I/O to complete.If there is no DMA,the CPU is fully occupied doing I/O,so there is nothing to be gained(at least in terms of CPU utilization)by multiprogramming.No matter how much I/O a
5、program does,the CPU will be 100 percent busy.This of course assumes the major delay is the wait while data are copied.A CPU could do other work if the I/O were slow for other reasons(arriving on a serial line,for instance).5.Second generation computers did not have the necessary hardware to protect
6、 the operating system from malicious user programs.6.It is still alive.For example,Intel makes Pentium I,II,and III,and 4 CPUs with a variety of different properties including speed and power consumption.All of these machines are architecturally compatible.They differ only in price and performance,w
7、hich is the essence of the family idea.7.A 25 80 character monochrome text screen requires a 2000-byte buffer.The 1024 768 pixel 24-bit color bitmap requires 2,359,296 bytes.In 1980 these two options would have cost$10 and$11,520,respectively.For current prices,check on how much RAM currently costs,
8、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 single user.Mainframe systems nearly always emphasize batch or timesharing with many users.Protection is much more of an issue on mainframe systems,
9、as is efficient use of all resources.10.Every nanosecond one instruction emerges from the pipeline.This means the machine is executing 1 billion instructions per second.It does not matter at all how many stages the pipeline has.A 10-stage pipeline with 1 nsec per2 PROBLEM SOLUTIONS FOR CHAPTER 1stag
10、e would also execute 1 billion instructions per second.All that matters is how often a finished instructions pops out the end of the pipeline.11.The manuscript contains 80 50 700=2.8 million characters.This is,of course,impossible to fit into the registers of any currently available CPU and is too b
11、ig for a 1-MB cache,but if such hardware were available,the manuscript could be scanned in 2.8 msec from the registers or 5.8 msec from the cache.There are approximately 2700 1024-byte blocks of data,so scanning from the disk would require about 27 seconds,and from tape 2 minutes 7 seconds.Of course
12、,these times are just to read the data.Processing and rewriting the data would increase the time.12.Logically,it does not matter if the limit register uses a virtual address or a physical address.However,the performance of the former is better.If virtual addresses are used,the addition of the virtua
13、l address and the base register can start simultaneously with the comparison and then can run in parallel.If physical addresses are used,the comparison cannot start until the addition is complete,increasing the access time.13.Maybe.If the caller gets control back and immediately overwrites the data,
14、when the write finally occurs,the wrong data will be written.However,if the driver first copies the data to a private buffer before returning,then the caller can be allowed to continue immediately.Another possibility is to allow the caller to continue and give it a signal when the buffer may be reus
15、ed,but this is tricky and error prone.14.A trap is caused by the program and is synchronous with it.If the program is run again and again,the trap will always occur at exactly the same position in the instruction stream.An interrupt is caused by an external event and its timing is not reproducible.1
16、5.Base=40,000 and limit=10,000.An answer of limit=50,000 is incorrect for the way the system was described in this book.It could have been implemented that way,but doing so would have required waiting until the address+base calculation was completed before starting the limit check,thus slowing down
17、the computer.16.The process table is needed to store the state of a process that is currently suspended,either ready or blocked.It is not needed in a single process system because the single process is never suspended.17.Mounting a file system makes any files already in the mount point directory ina
18、ccessible,so mount points are normally empty.However,a system administrator might want to copy some of the most important files normally located in the mounted directory to the mount point so they could be found in their normal path in an emergency when the mounted device was being checked or repair
19、ed.PROBLEM SOLUTIONS FOR CHAPTER 1 318.Fork can fail if there are no free slots left in the process table(and possibly if there is no memory or swap space left).Exec can fail if the file name given does not exist or is not a valid executable file.Unlink can fail if the file to be unlinked does not e
20、xist or the calling process does not have the authority to unlink it.19.If the call fails,for example because fd is incorrect,it can return 1.It can also fail because the disk is full and it is not possible to write the number of bytes requested.On a correct termination,it always returns nbytes.20.I
21、t contains the bytes:1,5,9,2.21.Block special files consist of numbered blocks,each of which can be read or written independently of all the other ones.It is possible to seek to any block and start reading or writing.This is not possible with character special files.22.System calls do not really hav
22、e names,other than in a documentation sense.When the library procedure read traps to the kernel,it puts the number of the system call in a register or on the stack.This number is used to index into a table.There is really no name used anywhere.On the other hand,the name of the library procedure is v
23、ery important,since that is what appears in the program.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 a library procedure results in a system call.But if performance is an issue,if a task can be accompl
24、ished without a system call the program will run faster.Every system call involves overhead time in switching from the user context to the kernel context.Furthermore,on a multiuser system the operating system may schedule another process to run when a system call completes,further slowing the progre
25、ss in real time of a calling process.25.Several UNIX calls have no counterpart in the Win32 API:Link:a Win32 program cannot refer to a file by an alternate name or see it in more than one directory.Also,attempting to create a link is a convenient way to test for and create a lock on a file.Mount and
26、 umount:a Windows program cannot make assumptions about standard path names because on systems with multiple disk drives the drive name part of the path may be different.Chmod:Windows programmers have to assume that every user can access every file.Kill:Windows programmers cannot kill a misbehaving
27、program that is not cooperating.4 PROBLEM 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 240 bytes,which is 1,099,511,627,776 bytes.(d)It is 6 1024 kg.SOLUTIONS TO CHAPTER 2 PROBLEMS1.The transition from
28、 blocked to running is conceivable.Suppose that a process is blocked on I/O and the I/O finishes.If the CPU is otherwise idle,the process could go directly from blocked to running.The other missing transition,from ready to blocked,is impossible.A ready process cannot do I/O or anything else that mig
29、ht block it.Only a running process can block.2.You could have a register containing a pointer to the current process table entry.When I/O completed,the CPU would store the current machine state in the current process table entry.Then it would go to the interrupt vector for the interrupting device an
30、d fetch a pointer to another process table entry(the service procedure).This process would then be started up.3.Generally,high-level languages do not allow one the kind of access to CPU hardware that is required.For instance,an interrupt handler may be required to enable and disable the interrupt se
31、rvicing a particular device,or to manipulate data within a processstack area.Also,interrupt service routines must execute as rapidly as possible.4.There are several reasons for using a separate stack for the kernel.Two of them are as follows.First,you do not want the operating system to crash becaus
32、e a poorly written user program does not allow for enough stack space.Second,if the kernel leaves stack data in a user program s memory space upon return from a system call,a malicious user might be able to use this data to find out information about other processes.5.It would be difficult,if not im
33、possible,to keep the file system consistent.Suppose that a client process sends a request to server process 1 to update a file.This process updates the cache entry in its memory.Shortly thereafter,another client process sends a request to server 2 to read that file.Unfortunately,if the file is also
34、cached there,server 2,in its innocence,will return obsolete data.If the first process writes the file through to the disk after caching it,and server 2 checks the disk on every read to see if its cached copy is up-to-date,the system can be made to work,but it is precisely all these disk accesses tha
35、t the caching system is trying to avoid.PROBLEM SOLUTIONS FOR CHAPTER 2 56.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.Timesharing threads is no different than timesharing processes,so each thread needs it
36、s own 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.If user-level threads are being used,this action will block the entire process,destroying the value of multithreading.Thus it
37、 is essential that kernel threads are 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 yielding is needed for the good of the application,then a thread will yield.After all,it is usually the same programmer who
38、writes the code for all of them.10.User-level threads cannot be preempted by the clock uless the whole process quantum has been used up.Kernel-level threads can be preempted individually.In the latter case,if a thread runs too long,the clock will interrupt the current process and thus the current th
39、read.The kernel is free to pick a different 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 take 90 msec.The weighted average is 2/3 15+1/3 90.Thus the mean request takes 40 msec and the server can do 25 per second
40、.For a multithreaded server,all the waiting for the disk is overlapped,so every request takes 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 multiple threads.It just adds unnecessary complexity.As an example,consider a
41、 telephone directory assistance number(like 555-1212)for an area with 1 million people.If each(name,telephone number)record is,say,64 characters,the entire database takes 64 megabytes,and can easily be kept in the server s memory to provide fast lookup.13.The pointers are really necessary because th
42、e size of the global variable is unknown.It could be anything from a character to an array of floating-point numbers.If the value were stored,one would have to give the size to create3global,which is all right,but what type should the second parameter of set3globalbe,and what type should the value o
43、f read3globalbe?14.It could happen that the runtime system is precisely at the point of blocking or unblocking a thread,and is busy manipulating the scheduling queues.This would be a very inopportune moment for the clock interrupt handler to begin inspecting those queues to see if it was time to do
44、thread switching,since they might be in an inconsistent state.One solution is to set a flag when the runtime system is entered.The clock handler would see this and set its own flag,6 PROBLEM SOLUTIONS FOR CHAPTER 2then return.When the runtime system finished,it would check the clock flag,see that a
45、clock interrupt occurred,and now run the clock handler.15.Yes it is possible,but inefficient.A thread wanting to do a system call first sets an alarm timer,then does the call.If the call blocks,the timer returns control to the threads package.Of course,most of the time the call will not block,and th
46、e timer has to be cleared.Thus each system call that might block 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 a threads package.16.The priority inversion problem occurs when a low-priority process is in
47、 its critical region and suddenly a high-priority process becomes ready and is scheduled.If it uses busy waiting,it will run forever.With user-level threads,it cannot happen that a low-priority thread is suddenly preempted to allow a high-priority thread run.There is no preemption.With kernel-level
48、threads this problem can arise.17.Each thread calls procedures on its own,so it must have its own stack for the local variables,return addresses,and so on.This is equally true for user-level threads as for kernel-level threads.18.A race condition is a situation in which two(or more)processes are abo
49、ut to perform some action.Depending on the exact timing,one or the other goes first.If one of the processes goes first,everything works,but if another one goes first,a fatal error occurs.19.Yes.The simulated computer could be multiprogrammed.For example,while process A is running,it reads out some s
50、hared variable.Then a simulated clock tick happens and process B runs.It also reads out the same variable.Then it adds 1 to the variable.When process A runs,if it also adds one to the variable,we have a race condition.20.Yes,it still works,but it still is busy waiting,of course.21.It certainly works