计算机操作系统chapter1

计算机软件操作技巧大全集Chapter 1: 文件管理技巧计算机软件中，文件管理是日常操作中的重要部分。

掌握一些文件管理技巧可以提高日常工作和学习的效率。

1.1 创建和删除文件夹在文件管理中，合理的文件夹组织结构可以帮助我们更好地管理文件。

创建文件夹可以通过右键点击桌面或文件管理器中的空白处，选择“新建文件夹”来快速创建。

同样地，删除文件夹也可以通过右键点击所要删除的文件夹，选择“删除”来完成。

1.2 文件夹的重命名与复制当需要重命名文件夹时，选中该文件夹并点击右键，选择“重命名”，然后输入新的名字即可。

而复制文件夹可以通过右键点击所需复制的文件夹，选择“复制”，然后在想要复制的位置点击右键，选择“粘贴”。

1.3 快速搜索文件当文件数量过多时，可以通过搜索功能来快速定位所需文件。

在文件管理器的搜索框中键入关键词，即可显示含有该关键词的文件列表。

Chapter 2: 文本编辑技巧文本编辑是计算机操作中常用的技能。

掌握一些文本编辑技巧可以方便我们进行文本的修改和处理。

2.1 快速选择文本在文本中使用鼠标可以进行快速选择。

双击单词可选中整个单词，三次点击可选中整个段落。

按住Shift键加方向键，可以选择连续的文本。

2.2 剪切、复制和粘贴剪切文本可以通过选中所需文本后按Ctrl+X来实现，复制文本可以通过Ctrl+C来完成。

粘贴文本可以通过Ctrl+V或右键点击目标位置，选择“粘贴”。

2.3 查找和替换文本在文本编辑器中，按Ctrl+F可以打开查找窗口，输入所需查找的文本后，点击“下一个”按钮可以定位到下一个匹配文本所在位置。

而替换文本可以通过点击替换窗口，输入被替换的内容和替换的内容，然后点击“替换”或“替换全部”按钮来完成。

Chapter 3: 网络使用技巧互联网的普及使得网络使用成为我们生活中必不可少的一部分。

了解一些网络使用技巧可以帮助我们更好地利用网络资源。

3.1 使用搜索引擎搜索引擎是我们获取信息的主要工具，善于使用搜索引擎可以提高信息查找的效率。

MODERNOPERATINGSYSTEMSTHIRD EDITION PROBLEM SOLUTIONSANDREW S.TANENBAUMVrije UniversiteitAmsterdam,The NetherlandsPRENTICE HALLUPPER SADDLE RIVER,NJ07458Copyright Pearson Education,Inc.2008SOLUTIONS TO CHAPTER1PROBLEMS1.Multiprogramming is the rapid switching of the CPU between multiple proc-esses in memory.It is commonly used to keep the CPU busy while one or more processes are doing I/O.2.Input spooling is the technique of reading in jobs,for example,from cards,onto the disk,so that when the currently executing processes arefinished, there will be work waiting for the CPU.Output spooling consists offirst copying printablefiles to disk before printing them,rather than printing di-rectly as the output is generated.Input spooling on a personal computer is not very likely,but output spooling is.3.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 matter how much I/O a program does,the CPU will be100%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).4.It is still alive.For example,Intel makes Pentium I,II,and III,and4CPUswith a variety of different properties including speed and power consumption.All of these machines are architecturally compatible.They differ only in price and performance,which is the essence of the family idea.5.A25×80character monochrome text screen requires a2000-byte buffer.The1024×768pixel24-bit color bitmap requires2,359,296bytes.In1980these two options would have cost$10and$11,520,respectively.For current prices,check on how much RAM currently costs,probably less than$1/MB.6.Consider fairness and real time.Fairness requires that each process be allo-cated its resources in a fair way,with no process getting more than its fair share.On the other hand,real time requires that resources be allocated based on the times when different processes must complete their execution.A real-time process may get a disproportionate share of the resources.7.Choices(a),(c),and(d)should be restricted to kernel mode.8.It may take20,25or30msec to complete the execution of these programsdepending on how the operating system schedules them.If P0and P1are scheduled on the same CPU and P2is scheduled on the other CPU,it will take20mses.If P0and P2are scheduled on the same CPU and P1is scheduled on the other CPU,it will take25msec.If P1and P2are scheduled on the same CPU and P0is scheduled on the other CPU,it will take30msec.If all three are on the same CPU,it will take35msec.2PROBLEM SOLUTIONS FOR CHAPTER19.Every nanosecond one instruction emerges from the pipeline.This means themachine is executing1billion instructions per second.It does not matter at all how many stages the pipeline has.A10-stage pipeline with1nsec per stage would also execute1billion instructions per second.All that matters is how often afinished instruction pops out the end of the pipeline.10.Average access time=0.95×2nsec(word is cache)+0.05×0.99×10nsec(word is in RAM,but not in cache)+0.05×0.01×10,000,000nsec(word on disk only)=5002.395nsec=5.002395μsec11.The manuscript contains80×50×700=2.8million characters.This is,ofcourse,impossible tofit into the registers of any currently available CPU and is too big for a1-MB cache,but if such hardware were available,the manuscript could be scanned in2.8msec from the registers or5.8msec from the cache.There are approximately27001024-byte blocks of data,so scan-ning from the disk would require about27seconds,and from tape2minutes7 seconds.Of course,these times are just to read the data.Processing and rewriting the data would increase the time.12.Maybe.If the caller gets control back and immediately overwrites the data,when the writefinally occurs,the wrong data will be written.However,if the driverfirst 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 reused,but this is tricky and error prone.13.A trap instruction switches the execution mode of a CPU from the user modeto the kernel mode.This instruction allows a user program to invoke func-tions in the operating system kernel.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 in the instruction stream.An interrupt is caused by an external event and its timing is not reproducible.15.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.16.Mounting afile system makes anyfiles already in the mount point directoryinaccessible,so mount points are normally empty.However,a system admin-istrator might want to copy some of the most importantfiles 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 repaired.PROBLEM SOLUTIONS FOR CHAPTER13 17.A system call allows a user process to access and execute operating systemfunctions inside the er programs use system calls to invoke operat-ing system services.18.Fork can fail if there are no free slots left in the process table(and possibly ifthere is no memory or swap space left).Exec can fail if thefile name given does not exist or is not a valid executablefile.Unlink can fail if thefile to be unlinked does not exist 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 canalso 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.It contains the bytes:1,5,9,2.21.Time to retrieve thefile=1*50ms(Time to move the arm over track#50)+5ms(Time for thefirst sector to rotate under the head)+10/100*1000ms(Read10MB)=155ms22.Block specialfiles consist of numbered blocks,each of which can be read orwritten 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 specialfiles.23.System 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 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 very important,since that is what appears in the program.24.Yes it can,especially if the kernel is a message-passing system.25.As far as program logic is concerned it does not matter whether a call to a li-brary procedure results in a system call.But if performance is an issue,if a task can be accomplished 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 sys-tem may schedule another process to run when a system call completes, further slowing the progress in real time of a calling process.26.Several UNIX calls have no counterpart in the Win32API:Link:a Win32program cannot refer to afile by an alternative 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 afile.4PROBLEM SOLUTIONS FOR CHAPTER1Mount and 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 uses access control listsKill:Windows programmers cannot kill a misbehaving program that is not cooperating.27.Every system architecture has its own set of instructions that it can execute.Thus a Pentium cannot execute SPARC programs and a SPARC cannot exe-cute Pentium programs.Also,different architectures differ in bus architecture used(such as VME,ISA,PCI,MCA,SBus,...)as well as the word size of the CPU(usually32or64bit).Because of these differences in hardware,it is not feasible to build an operating system that is completely portable.A highly portable operating system will consist of two high-level layers---a machine-dependent layer and a machine independent layer.The machine-dependent layer addresses the specifics of the hardware,and must be implemented sepa-rately for every architecture.This layer provides a uniform interface on which the machine-independent layer is built.The machine-independent layer has to be implemented only once.To be highly portable,the size of the machine-dependent layer must be kept as small as possible.28.Separation of policy and mechanism allows OS designers to implement asmall number of basic primitives in the kernel.These primitives are sim-plified,because they are not dependent of any specific policy.They can then be used to implement more complex mechanisms and policies at the user level.29.The conversions are straightforward:(a)A micro year is10−6×365×24×3600=31.536sec.(b)1000meters or1km.(c)There are240bytes,which is1,099,511,627,776bytes.(d)It is6×1024kg.SOLUTIONS TO CHAPTER2PROBLEMS1.The transition from blocked to running is conceivable.Suppose that a processis blocked on I/O and the I/Ofinishes.If the CPU is otherwise idle,the proc-ess 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 any-thing else that might block it.Only a running process can block.PROBLEM SOLUTIONS FOR CHAPTER25 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 state in the current process table entry.Then it would go to the interrupt vector for the interrupting device and fetch a pointer to another process table entry(the ser-vice procedure).This process would then be started up.3.Generally,high-level languages do not allow the kind of access to CPU hard-ware that is required.For instance,an interrupt handler may be required to enable and disable the interrupt servicing a particular device,or to manipulate data within a process’stack area.Also,interrupt service routines must exe-cute 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 operating system to crash be-cause 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 tofind out information about other processes.5.If each job has50%I/O wait,then it will take20minutes to complete in theabsence of competition.If run sequentially,the second one willfinish40 minutes after thefirst one starts.With two jobs,the approximate CPU utiliza-tion is1−0.52.Thus each one gets0.375CPU minute per minute of real time.To accumulate10minutes of CPU time,a job must run for10/0.375 minutes,or about26.67minutes.Thus running sequentially the jobsfinish after40minutes,but running in parallel theyfinish after26.67minutes.6.It would be difficult,if not impossible,to keep thefile system consistent.Sup-pose that a client process sends a request to server process1to update afile.This process updates the cache entry in its memory.Shortly thereafter,anoth-er client process sends a request to server2to read thatfile.Unfortunately,if thefile is also cached there,server2,in its innocence,will return obsolete data.If thefirst process writes thefile through to the disk after caching it, and server2checks 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 ac-cesses that the caching system is trying to avoid.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, destroying the value of multithreading.Thus it is essential that kernel threads are used to permit some threads to block without affecting the others.9.Yes.If the server is entirely CPU bound,there is no need to have multiplethreads.It just adds unnecessary complexity.As an example,consider a tele-phone directory assistance number(like555-1212)for an area with1million6PROBLEM SOLUTIONS FOR CHAPTER2people.If each(name,telephone number)record is,say,64characters,the entire database takes64megabytes,and can easily be kept in the server’s memory to provide fast lookup.10.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.Multipro-gramming threads is no different than multiprogramming processes,so each thread needs its own register save area.11.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.After all,it is usually the same programmer who writes the code for all of them. er-level threads cannot be preempted by the clock unless the whole proc-ess’quantum has been used up.Kernel-level threads can be preempted indivi-dually.In the latter case,if a thread runs too long,the clock will interrupt the current 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.13.In the single-threaded case,the cache hits take15msec and cache misses take90msec.The weighted average is2/3×15+1/3×90.Thus the mean re-quest takes40msec and the server can do25per second.For a multithreaded server,all the waiting for the disk is overlapped,so every request takes15 msec,and the server can handle662/3requests per second.14.The biggest advantage is the efficiency.No traps to the kernel are needed toswitch threads.The biggest disadvantage is that if one thread blocks,the en-tire process blocks.15.Yes,it can be done.After each call to pthread create,the main programcould do a pthread join to wait until the thread just created has exited before creating the next thread.16.The pointers are really necessary because the size of the global variable isunknown.It could be anything from a character to an array offloating-point numbers.If the value were stored,one would have to give the size to create global,which is all right,but what type should the second parameter of set global be,and what type should the value of read global be?17.It could happen that the runtime system is precisely at the point of blocking orunblocking 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 thread switching,since they might be in an inconsistent state.One solution is to set aflag when the run-time system is entered.The clock handler would see this and set its ownflag, then return.When the runtime systemfinished,it would check the clockflag, see that a clock interrupt occurred,and now run the clock handler.PROBLEM SOLUTIONS FOR CHAPTER27 18.Yes it is possible,but inefficient.A thread wanting to do a system callfirstsets 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 the 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.19.The priority inversion problem occurs when a low-priority process is in itscritical 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 threads this problem can arise.20.With round-robin scheduling it works.Sooner or later L will run,and eventu-ally it will leave its critical region.The point is,with priority scheduling,L never gets to run at all;with round robin,it gets a normal time slice periodi-cally,so it has the chance to leave its critical region.21.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-level threads as for kernel-level threads.22.Yes.The simulated computer could be multiprogrammed.For example,while process A is running,it reads out some shared variable.Then a simula-ted clock tick happens and process B runs.It also reads out the same vari-able.Then it adds1to the variable.When process A runs,if it also adds one to the variable,we have a race condition.23.Yes,it still works,but it still is busy waiting,of course.24.It certainly works with preemptive scheduling.In fact,it was designed forthat case.When scheduling is nonpreemptive,it might fail.Consider the case in which turn is initially0but process1runsfirst.It will just loop forever and never release the CPU.25.To do a semaphore operation,the operating systemfirst disables interrupts.Then it reads the value of the semaphore.If it is doing a down and the sema-phore is equal to zero,it puts the calling process on a list of blocked processes associated with the semaphore.If it is doing an up,it must check to see if any processes are blocked on the semaphore.If one or more processes are block-ed,one of them is removed from the list of blocked processes and made run-nable.When all these operations have been completed,interrupts can be enabled again.8PROBLEM SOLUTIONS FOR CHAPTER226.Associated with each counting semaphore are two binary semaphores,M,used for mutual exclusion,and B,used for blocking.Also associated with each counting semaphore is a counter that holds the number of up s minus the number of down s,and a list of processes blocked on that semaphore.To im-plement down,a processfirst gains exclusive access to the semaphores, counter,and list by doing a down on M.It then decrements the counter.If it is zero or more,it just does an up on M and exits.If M is negative,the proc-ess is put on the list of blocked processes.Then an up is done on M and a down is done on B to block the process.To implement up,first M is down ed to get mutual exclusion,and then the counter is incremented.If it is more than zero,no one was blocked,so all that needs to be done is to up M.If, however,the counter is now negative or zero,some process must be removed from the list.Finally,an up is done on B and M in that order.27.If the program operates in phases and neither process may enter the nextphase until both arefinished with the current phase,it makes perfect sense to use a barrier.28.With kernel threads,a thread can block on a semaphore and the kernel canrun some other thread in the same process.Consequently,there is no problem using semaphores.With user-level threads,when one thread blocks on a semaphore,the kernel thinks the entire process is blocked and does not run it ever again.Consequently,the process fails.29.It is very expensive to implement.Each time any variable that appears in apredicate on which some process is waiting changes,the run-time system must re-evaluate the predicate to see if the process can be unblocked.With the Hoare and Brinch Hansen monitors,processes can only be awakened on a signal primitive.30.The employees communicate by passing messages:orders,food,and bags inthis case.In UNIX terms,the four processes are connected by pipes.31.It does not lead to race conditions(nothing is ever lost),but it is effectivelybusy waiting.32.It will take nT sec.33.In simple cases it may be possible to determine whether I/O will be limitingby looking at source code.For instance a program that reads all its inputfiles into buffers at the start will probably not be I/O bound,but a problem that reads and writes incrementally to a number of differentfiles(such as a compi-ler)is likely to be I/O bound.If the operating system provides a facility such as the UNIX ps command that can tell you the amount of CPU time used by a program,you can compare this with the total time to complete execution of the program.This is,of course,most meaningful on a system where you are the only user.34.For multiple processes in a pipeline,the common parent could pass to the op-erating system information about the flow of data.With this information the OS could,for instance,determine which process could supply output to a process blocking on a call for input.35.The CPU efficiency is the useful CPU time divided by the total CPU time.When Q ≥T ,the basic cycle is for the process to run for T and undergo a process switch for S .Thus (a)and (b)have an efficiency of T /(S +T ).When the quantum is shorter than T ,each run of T will require T /Q process switches,wasting a time ST /Q .The efficiency here is thenT +ST /QT which reduces to Q /(Q +S ),which is the answer to (c).For (d),we just sub-stitute Q for S and find that the efficiency is 50%.Finally,for (e),as Q →0the efficiency goes to 0.36.Shortest job first is the way to minimize average response time.0<X ≤3:X ,3,5,6,9.3<X ≤5:3,X ,5,6,9.5<X ≤6:3,5,X ,6,9.6<X ≤9:3,5,6,X ,9.X >9:3,5,6,9,X.37.For round robin,during the first 10minutes each job gets 1/5of the CPU.Atthe end of 10minutes,C finishes.During the next 8minutes,each job gets 1/4of the CPU,after which time D finishes.Then each of the three remaining jobs gets 1/3of the CPU for 6minutes,until B finishes,and so on.The fin-ishing times for the five jobs are 10,18,24,28,and 30,for an average of 22minutes.For priority scheduling,B is run first.After 6minutes it is finished.The other jobs finish at 14,24,26,and 30,for an average of 18.8minutes.If the jobs run in the order A through E ,they finish at 10,16,18,22,and 30,for an average of 19.2minutes.Finally,shortest job first yields finishing times of 2,6,12,20,and 30,for an average of 14minutes.38.The first time it gets 1quantum.On succeeding runs it gets 2,4,8,and 15,soit must be swapped in 5times.39.A check could be made to see if the program was expecting input and didanything with it.A program that was not expecting input and did not process it would not get any special priority boost.40.The sequence of predictions is 40,30,35,and now 25.41.The fraction of the CPU used is35/50+20/100+10/200+x/250.To beschedulable,this must be less than1.Thus x must be less than12.5msec. 42.Two-level scheduling is needed when memory is too small to hold all theready processes.Some set of them is put into memory,and a choice is made from that set.From time to time,the set of in-core processes is adjusted.This algorithm is easy to implement and reasonably efficient,certainly a lot better than,say,round robin without regard to whether a process was in memory or not.43.Each voice call runs200times/second and uses up1msec per burst,so eachvoice call needs200msec per second or400msec for the two of them.The video runs25times a second and uses up20msec each time,for a total of 500msec per second.Together they consume900msec per second,so there is time left over and the system is schedulable.44.The kernel could schedule processes by any means it wishes,but within eachprocess it runs threads strictly in priority order.By letting the user process set the priority of its own threads,the user controls the policy but the kernel handles the mechanism.45.The change would mean that after a philosopher stopped eating,neither of hisneighbors could be chosen next.In fact,they would never be chosen.Sup-pose that philosopher2finished eating.He would run test for philosophers1 and3,and neither would be started,even though both were hungry and both forks were available.Similarly,if philosopher4finished eating,philosopher3 would not be started.Nothing would start him.46.If a philosopher blocks,neighbors can later see that she is hungry by checkinghis state,in test,so he can be awakened when the forks are available.47.Variation1:readers have priority.No writer may start when a reader is ac-tive.When a new reader appears,it may start immediately unless a writer is currently active.When a writerfinishes,if readers are waiting,they are all started,regardless of the presence of waiting writers.Variation2:Writers have priority.No reader may start when a writer is waiting.When the last ac-tive processfinishes,a writer is started,if there is one;otherwise,all the readers(if any)are started.Variation3:symmetric version.When a reader is active,new readers may start immediately.When a writerfinishes,a new writer has priority,if one is waiting.In other words,once we have started reading,we keep reading until there are no readers left.Similarly,once we have started writing,all pending writers are allowed to run.48.A possible shell script might beif[!–f numbers];then echo0>numbers;ficount=0while(test$count!=200)docount=‘expr$count+1‘n=‘tail–1numbers‘expr$n+1>>numbersdoneRun the script twice simultaneously,by starting it once in the background (using&)and again in the foreground.Then examine thefile numbers.It will probably start out looking like an orderly list of numbers,but at some point it will lose its orderliness,due to the race condition created by running two cop-ies of the script.The race can be avoided by having each copy of the script test for and set a lock on thefile before entering the critical area,and unlock-ing it upon leaving the critical area.This can be done like this:if ln numbers numbers.lockthenn=‘tail–1numbers‘expr$n+1>>numbersrm numbers.lockfiThis version will just skip a turn when thefile is inaccessible,variant solu-tions could put the process to sleep,do busy waiting,or count only loops in which the operation is successful.SOLUTIONS TO CHAPTER3PROBLEMS1.It is an accident.The base register is16,384because the program happened tobe loaded at address16,384.It could have been loaded anywhere.The limit register is16,384because the program contains16,384bytes.It could have been any length.That the load address happens to exactly match the program length is pure coincidence.2.Almost the entire memory has to be copied,which requires each word to beread and then rewritten at a different location.Reading4bytes takes10nsec, so reading1byte takes2.5nsec and writing it takes another2.5nsec,for a total of5nsec per byte compacted.This is a rate of200,000,000bytes/sec.To copy128MB(227bytes,which is about1.34×108bytes),the computer needs227/200,000,000sec,which is about671msec.This number is slightly pessimistic because if the initial hole at the bottom of memory is k bytes, those k bytes do not need to be copied.However,if there are many holes andmany data segments,the holes will be small,so k will be small and the error in the calculation will also be small.3.The bitmap needs1bit per allocation unit.With227/n allocation units,this is224/n bytes.The linked list has227/216or211nodes,each of8bytes,for a total of214bytes.For small n,the linked list is better.For large n,the bitmap is better.The crossover point can be calculated by equating these two formu-las and solving for n.The result is1KB.For n smaller than1KB,a linked list is better.For n larger than1KB,a bitmap is better.Of course,the assumption of segments and holes alternating every64KB is very unrealistic.Also,we need n<=64KB if the segments and holes are64KB.4.Firstfit takes20KB,10KB,18KB.Bestfit takes12KB,10KB,and9KB.Worstfit takes20KB,18KB,and15KB.Nextfit takes20KB,18KB,and9 KB.5.For a4-KB page size the(page,offset)pairs are(4,3616),(8,0),and(14,2656).For an8-KB page size they are(2,3616),(4,0),and(7,2656).6.They built an MMU and inserted it between the8086and the bus.Thus all8086physical addresses went into the MMU as virtual addresses.The MMU then mapped them onto physical addresses,which went to the bus.7.(a)M has to be at least4,096to ensure a TLB miss for every access to an ele-ment of X.Since N only affects how many times X is accessed,any value of N will do.(b)M should still be atleast4,096to ensure a TLB miss for every access to anelement of X.But now N should be greater than64K to thrash the TLB, that is,X should exceed256KB.8.The total virtual address space for all the processes combined is nv,so thismuch storage is needed for pages.However,an amount r can be in RAM,so the amount of disk storage required is only nv−r.This amount is far more than is ever needed in practice because rarely will there be n processes ac-tually running and even more rarely will all of them need the maximum al-lowed virtual memory.9.The page table contains232/213entries,which is524,288.Loading the pagetable takes52msec.If a process gets100msec,this consists of52msec for loading the page table and48msec for running.Thus52%of the time is spent loading page tables.10.(a)We need one entry for each page,or224=16×1024×1024entries,sincethere are36=48−12bits in the page numberfield.。

1.计算机存储数据的基本单位是（）A.bitB.ByteC.字D.字符2.多年来，人们习惯于以计算机主机所使用的主要元器件的发展进行分代，所谓第四代计算机使用的主要元器件是（）A.电子管B.晶体管C.中小规模集成电路D.大规模和超大规模集成电路3.在计算机的不同发展阶段，操作系统最先出现在（）A.第一代计算机B.第二代计算机C.第三代计算机D.第四代计算机4.运算器的主要功能是进行（）A．只做加法B．逻辑运算C．算术运算和逻辑运算D．算术运算5.计算机硬件的五大基本构件包括运算器、存储器、输入设备、输出设备和（）A.显示器B.控制器C.磁盘驱动器D.鼠标器6.关于冯.诺依曼计算机，下列说法正确的是（）A.冯.诺依曼计算机的程序和数据是靠输入设备送入计算机的寄存器保存的B.冯.诺依曼计算机工作时是由数据流驱动控制流工作的C.冯.诺依曼计算机的基本特点可以用“存储程序”和“程序控制”高度概括D.随着计算机技术的发展，冯.诺依曼计算机目前已经被淘汰7.冯.诺依曼计算机的核心思想是（），冯.诺依曼计算机的工作特点是（）（1） A.采用二进制 B.存储程序 C.并行计算 D.指令系统（2）A.堆栈操作 B.存储器按内容访问C.按地址访问并顺序执行指令D.多指令流单数据流8.一个完整的计算机系统包括（）A.主机、键盘、显示器B.主机及外围设备C.系统软件与应用软件D.硬件系统与软件系统9.下列软件中，不属于系统软件的是（）A．编译软件B．操作系统C．数据库管理系统D．C语言程序解析：计算机的软件分为系统软件和应用软件。

系统软件是为了计算机能正常、高效工件所配备的各种管理、监控和维护系统的程序及其有关资料。

系统软件主要包括如下几个方面：（1）操作系统软件，这是软件的核心（2）各种语言的解释程序和编译程序（如BASIC语言解释程序等）（3）各种服务性程序（如机器的调试、故障检查和诊断程序等）（4）各种数据库管理系统（FoxPro等）10.某单位的人事档案管理程序属于（）A.工具软件B.应用软件C.系统软件D.字表处理软件11.下列选项中，描述浮点数操作速度的指标是（）A.MIPSB.CPIC.IPCD.MFLOP12.半个世纪以来，对计算机发展的阶段有过多种描述。

第一章计算机系统概论1.操作系统类似于计算机硬件和人类用户之间的接口。

答案：T。

2.处理器的一个主要功能是与内存交换数据。

答案：T。

3.一般用户对系统程序无障碍，对应用程序有障碍。

答案：F4.数据寄存器一般是通用的，但可能局限于像浮点数运算这样的特定任务。

T5.程序状态字（PSW）通常包含条件码等状态信息。

条件码是由程序员为操作结果设置的位。

答案：F6.一个单一的指令需要的处理称为执行周期。

答案：F（称为指令周期）7.取到的指令通常被存放在指令寄存器中（IR）。

答案：T8.中断是系统模块暂停处理器正常处理过程所采用的一种机制。

答案：T9.为适应中断产生的情况，必须在指令周期中增加一个额外的读取阶段。

F10.在处理器控制控制例行的中断处理器之前，需要储存的最少信息有程序状态字和当前指令地址。

答案：F11.多中断的一个处理方法是在处理一个中断时禁止再发生中断。

答案：T12.多道程序设计允许处理器使用长时间等待的中断处理的空闲时间。

答案：T13.在两级存取优先级中，命中率定义为对较慢存储器的访问次数与对所有存储器访问次数的比值。

答案：F14.高速缓冲存储器的开发利用了局部性原理，即在处理器与主存储器之间提供一个容量小而快速的存储器。

T15.在高速缓冲存储器的设计中，块大小与高速缓冲存储器和主存储器间的数据交换单位有关。

答案：T16.可编程I/O的一个主要问题是，处理器必须等到I/O模块准备完毕，并且在等待的过程中必须反复不停的检查I/O模块的状态。

答案：T第二章操作系统概述1．操作系统是控制应用程序执行的程序，并充当应用程序和计算机硬件之间的接口。

（对）2．在多用户系统中，操作系统管理那些用作重要目的的资源。

（对）3．操作系统通常在它的专用O/S处理器上并行应用程序。

（错）4．操作系统演化的动力之一就是基本硬件技术的进步。

（对）5. 早期的计算机中没有操作系统，用户直接与硬件打交道。

（对）6 在一个批处理系统，“control is passed to a job”意味着处理器正在取指令和执行用户程序。

Linux常见命令使用方法Chapter 1 介绍Linux常见命令是指Linux操作系统中常用的一些命令，它们可以帮助用户在Linux系统上进行各种操作。

这些命令具有很强的专业性，对于Linux系统的管理者和开发人员来说是必不可少的工具。

本文将介绍一些常见的Linux命令及其使用方法，包括文件和目录操作、系统管理、软件安装和网络配置等方面的内容。

Chapter 2 文件和目录操作2.1 cd命令cd命令是Linux中进入目录的命令。

在Linux中，所有的文件和目录都是以根目录“/”为开始的。

如果想要进入某个目录，可以使用cd命令。

例如，如果想要进入主目录，可以使用以下命令：cd ~如果想要进入某个子目录，可以使用以下命令：cd 目录路径2.2 ls命令ls命令可以列出指定目录中的所有文件和子目录。

例如，如果要列出当前目录中的所有文件和子目录，可以使用以下命令：ls如果想要列出指定目录中的所有文件和子目录，可以使用以下命令：ls 目录路径2.3 mkdir命令mkdir命令可以创建新目录。

例如，如果想要在当前目录下创建一个名为“test”的目录，可以使用以下命令：mkdir test2.4 rm命令rm命令可以删除指定的文件或目录。

例如，如果要删除一个名为“example.txt”的文件，可以使用以下命令：rm example.txt如果想要删除整个目录及其子目录，可以使用以下命令：rm -rf 目录路径Chapter 3 系统管理3.1 su命令su命令可以用于切换用户，例如从普通用户切换到超级用户。

例如，如果要切换到超级用户，可以使用以下命令：su在输入密码后就可以切换到超级用户了。

3.2 sudo命令sudo命令可以用于在不切换用户的情况下执行超级用户身份的操作。

例如，如果要以超级用户身份执行apt-get install命令来安装软件，可以使用以下命令：sudo apt-get install 软件包名3.3 ps命令ps命令可以显示当前系统中正在运行的进程。