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Level 4 computer network engineer - Summary of notes on some topics of operating system [11 ~ 20 questions]

Computer level 4 notes

The operating system part :(11~20 topic )

Because the length is too long , To ensure the quality of learning , Then it is divided into four parts ( Four blogs )
Every time 10 Entitled an article , Other topics are in my Level 4 computer network engineer Columns can be found in

The first 11 topic :

  1. Which activity exists only Synchronous relationship ( Complete the same task ): Each process on the automobile assembly line
  2. When a process is waiting for another process to send a message to it , The relationship between them is : Synchronous relationship
  3. You can't use P、V Operation to achieve : Process sharing
  4. When a process executes concurrently , If two concurrent processes contain The same shared variable , Then they exist : Synchronous relationship
  5. When using P、V When the operation protects the use of shared resources , The code segment that reads and writes to the shared resource is called : A critical region
  6. The process accesses the critical area Follow the relevant guidelines : When you are free, go into , Busy, wait , Let the right waiting , Limited waiting
  7. Mutual exclusion ( Compete for the same resource ) ,eg: booking
  8. When a system prints data , Data reading process 、 The relationship between the process of processing data and the process of printing results is : Synchronous relationship
  9. When a system prints data , Data reading process 、 The process of processing data and printing results Through which object is associated : buffer
  10. The two of them are Don't feel ( Have no idea if other processes exist ) The process of , The potential control problem is : The two processes compete with each other , May cause mutual exclusion 、 Deadlock or hunger , They are Runtime correlation Described as : The result of one process has no effect on the result of another process .
  11. The two of them are Indirect perception ( Interact with third parties , For example, using the same resource ) The process of , The potential control problem is : Two processes share and collaborate , But it may cause mutual exclusion 、 Deadlock or hunger
  12. The two of them are Direct perception ( Direct interaction between the two sides , For example, sending messages to each other ) The process of , The potential control problem is : The two processes cooperate through communication , But it may cause mutual exclusion 、 Deadlock or hunger
  13. There are both synchronization and mutual exclusion : Different users play video games playing football on the same game console ( Because playing in the same team is synchronous )
  14. There is neither process synchronization nor process mutual exclusion : Different users compile programs simultaneously on their computers

The first 12 topic :

  1. The value of the semaphore is : Initial value of semaphore subtract Total semaphores ( Initial value of semaphore mutex It's usually 1, The total number of semaphores is mutex add Wait for the value in the queue .)
  2. P、V operation :P yes Minus one ,V yes Add one
  3. Full mail slot The sending process can no longer apply for mutex .
  4. Most suitable for transmitting a large amount of information : Shared memory ( Not suitable for P、V operation )
  5. In process synchronization , What cannot operate on semaphores is : Add and subtract (P、V operation )
  6. The following semaphores S In the numerical range of , Which one is right : Because there is 4 Printers , So the first number is 4 , Because the maximum number of processes applying for printers in the system is 20 individual , So the second number is -16(4-20) individual , So the final answer is :[4,-16]
  7. In order to ensure the correct use of critical resources , When a process accesses a critical resource , The order of calling code is : Enter the area P—— A critical region —— Exit zone ( Leave )V—— The remaining area
  8. Process before accessing critical resources , Which area of code must be called first : Enter the area
  9. When solving the problem of process synchronization and mutual exclusion , On the semaphore P Primitive operations Which of the following types of code is used to complete : Enter the area
  10. The so-called... In the semaphore mechanism “ A critical region ” Refer to : Code that accesses critical resources .
  11. [ Failed to transfer the external chain picture , The origin station may have anti-theft chain mechanism , It is suggested to save the pictures and upload them directly (img-BJxuekkq-1652247327952)(C:\Users\86152\AppData\Roaming\Typora\typora-user-images\image-20220118155041055.png)] Insert picture description here

" c = …" It's a split line , Itself and the following parts are : A critical region , The above part is the access area ( or regard as while For the split line ,while And above are access areas , The following is the critical zone .)

“ FALSE ” It's a split line , As such : Exit zone , The following part is : The remaining area

The first 13 topic :

  1. Whether Xiao Wang's program can meet the requirements of the selected course A, Drop out of the course again B Purpose : There may be

    analysis :P(count.A); If you can't take a course A, Then wait again select_course(A) ; If you choose a course A; relese_rourse(B); V(count.B) Then put B Refund . Although there is no problem with the procedure , But the key still depends on whether there are courses A, There is no explanation in the title , So the result is : There may be .

  2. First come, first served Is not preemptive

  3. Producers should use... Before putting products into the buffer P The operation ensures that the buffer has free slots

  4. Which method of interprocess communication will not produce multiple identical copies in the system : Shared memory

  5. In direct communication , The system provides Send primitive yes :send(receiver,message) ( Brief notes : The first letter SRM)

  6. In direct communication , The system provides Receive the original yes :receive(sender,message) ( Brief notes : The first letter RSM)

  7. Between processes that communicate with each other Set a common memory area , A group of processes to the common memory Write , Another set of processes reads from this common memory , In this way, the information exchange between two groups of processes is called : Shared memory

  8. utilize Several common buffers in memory are organized into queues , The communication mode to realize the information exchange between processes is called : Message mechanism

  9. By connecting two processes An open shared file , Data communication between processes can be realized , This mode of communication is called : Pipeline communication

  10. Pipeline communication yes synchronous ( One party sends , The other party receives... At the same time )

  11. The mutual exclusion of the tube side is caused by Oneself Realized , No By semaphores and PV The operation realizes .

  12. use Shared memory When process communication is carried out in the mode of , There are two problems :1) Who provides shared memory **( Provided by the operating system )**2) Reading and writing are mutually exclusive ( The programmer solves this problem )

  13. use Message buffering method Can complete inter process communication , Which of the following contents does the communication mechanism include : news buffer District 、 news queue Sync Mutually exclusive semaphores and Send and receive Message primitives

  14. use Pipeline mode Can complete inter process communication , The disadvantage of this method is : The communication speed is slow ; Advantage is : Large communication capacity

  15. use Mailbox communication mode Can complete inter process communication , The advantage of this method is : The sender and receiver can send and receive messages asynchronously , No time limit .

  16. “ Shared memory ” communication mode : stay Set up a common area between processes communicating with each other , A group of processes to the public area Write , Another set of processes reads from this common area ( remember : key word

  17. “ Message buffering ” communication mode : Open up several areas in memory , The sending process requests an area , And send the information to , After the software, insert it into the corresponding queue of the receiving process , Then notify the receiving process .( remember : key word

  18. ” mail “ communication mode : The sending process and the receiving process do not establish direct contact , The receiving process can take information from the mechanism at any time ( remember : key word

  19. ” The Conduit “ communication mode : Send the process creation number to connect the two processes An open shared file , Then write the data stream ; The receiving process can read data from the shared file when it needs a pipeline , The length of written and read data is variable ( remember : key word

  20. Inferior Communication method : Semaphore (P、V operation )

The first 14 topic :

  1. Storage management solution , allow Dynamically expand memory capacity The plan is : Virtual page ( Introduce external memory to expand memory )
  2. When a program is loaded into memory by dynamic address mapping , Its address translation When was the work finished : The moment each instruction is executed
  3. Concentrate the fragmented free area into a large free area : Memory crunch
  4. Both It can meet the requirements of multi-channel programming And The design is the simplest Variable partition
  5. Each process gets the processor running before , Which of the following must be done first : Partially loaded into memory
  6. The process of assembling multiple object programs into runnable programs is called : link
  7. Operating system , take Logical address Convert to Physical memory address The process is called : relocation
  8. use Mobile technology solves fragmentation The problem is : Variable partition
  9. When loading a program , First direct Put the program Load into In the allocated memory area , Then in the program In the process of execution , whenever Execute an order Then, the hardware address conversion mechanism converts the... In the instruction Convert logical address to physical address , This process is called : Dynamic relocation
  10. When loading a program , Put the instruction address and data address in the program After calculation , Convert all to physical addresses after , Again Load physical memory , This process is called : Static address assignment
  11. Each process has its own relatively independent process address space , If the address generated by the process at run time Out of its address space , It is said that : The address is out of bounds
  12. Operating system , hold Logical address convert to Absolute address The work of is called : relocation
  13. What can't solve the debris problem with mobile technology is : Page
  14. If you want to ** Ensure that a program can be executed correctly even after the storage area is changed ,** Which technology can be used : Dynamic relocation ( Advantages of dynamic relocation )
  15. When loading a program , Put in the program Instruction address and Data address All converted to Absolute address , And in the process No more address translation in execution Work . This method of address transfer is called : Static repositioning
  16. [ Failed to transfer the external chain picture , The origin station may have anti-theft chain mechanism , It is suggested to save the pictures and upload them directly (img-PdX2hhec-1652247327953)(C:\Users\86152\AppData\Roaming\Typora\typora-user-images\image-20220119114948732.png)] Insert picture description here

​ The processes in the table P2 The flag of the line is set to NULL, And will Upper neighbor The free area is combined with it to form a free area , Modify the free zone table .

Method : See which process exits , The free area adjacent to it (NULL) To merge with it , Cannot merge across addresses

The first 15 topic :

  1. If the partition Start address plus length Equal to the starting address of the free area represented by a registration item in the free area table , said : The next partition of the recycle partition is idle

  2. by Speed up memory allocation , When using the best adaptation algorithm , The organization of the free zone should be : Arrange in ascending order of free area size

  3. Which method makes the memory High utilization and simple management Page allocation

  4. [ Failed to transfer the external chain picture , The origin station may have anti-theft chain mechanism , It is suggested to save the pictures and upload them directly (img-j8CpDA8j-1652247327954)(C:\Users\86152\AppData\Roaming\Typora\typora-user-images\image-20220119143213414.png)] Insert picture description here

    analysis : First , First draw a total of memory 1024 The size of the container , First the 128 Load , And then 320 Load ,224 Load 、288 Load , then 224 sign out , Leave a free space , And then 120 Load , Because there are 224 The free area of , Therefore, priority is given to loading into the free area , The remaining free space is 104; then 320 sign out ,224 Then in , The remaining free space is 96; Because the container has a total of 1024 size , So the last remaining area is 64. Select the maximum value from the remaining areas . So it's 104.

  5. In variable partition memory management , Algorithms that prefer to use low address free areas first are : First adaptation algorithm

  6. In the storage management system using switching and coverage technology , Process switching refers to : Processes that will not be used for the time being Code 、 Data and partial process control blocks Switch to disk

  7. In a virtual page system , When page replacement is needed , The selection will reside in memory The longest time The one page call-up strategy is : First in first out page replacement algorithm (FIFO)

  8. When performing page replacement in a virtual page system , First replacement ** The longest unused page ,** This strategy is called : Most recently used page replacement algorithm (LRU)

  9. When performing page replacement in a virtual page system , According to a period of time Choose how many times the page is used Pages that can be called up , This strategy is called : Recently, page replacement algorithm is the least frequently used (LFU)

  10. When performing page replacement in a virtual page system , After replacement Pages that are no longer needed or will not be used for the longest time , This strategy is called : Ideal page replacement algorithm (OPT)

  11. When performing page replacement in a virtual page system , Check the page that has been in memory for the longest time R position , If it is 0, Then replace the page ; If it is 1, will R Positional clarity 0, And put the page at the end of the list , Modify its entry time , And then continue to search , This strategy is called : Second chance page replacement algorithm

  12. When performing page replacement in a virtual page system , Call out the first page loaded into memory The strategy is : First in first out page replacement algorithm (FIFO)

  13. When performing page replacement in a virtual page system , Always choose I haven't been visited for the longest time Call up the page of , This strategy is called : Least recently used page replacement algorithm (LRU)

  14. When performing page replacement in a virtual page system , In the nearest The clock is ticking Try to replace one Not interviewed And unmodified pages , This strategy is called : Page replacement algorithm not used recently (NRU)

  15. When performing page replacement in a virtual page system , The operating system is based on the cycle T Inside , choice The least used page Call out , This strategy is called : The least commonly used page replacement algorithm recently (LFU)

  16. [ Failed to transfer the external chain picture , The origin station may have anti-theft chain mechanism , It is suggested to save the pictures and upload them directly (img-9dkJY0LU-1652247327954)(C:\Users\86152\AppData\Roaming\Typora\typora-user-images\image-20220119152343735.png)] Insert picture description here

​ analysis : The least used is Access to a Zero ( That's it 15 and 34 了 ), lately , namely : The longest time from now ,( That's it 15 了 , Because it has the smallest page number , That means it first appeared ( The longest time since now )).

  1. [ Failed to transfer the external chain picture , The origin station may have anti-theft chain mechanism , It is suggested to save the pictures and upload them directly (img-UG3I05Gj-1652247327955)(C:\Users\86152\AppData\Roaming\Typora\typora-user-images\image-20220119164723353.png)] Insert picture description here

    analysis : fifo , Who came first , Choose who .11 At the top , So choose A.

  2. [ Failed to transfer the external chain picture , The origin station may have anti-theft chain mechanism , It is suggested to save the pictures and upload them directly (img-xATtasq9-1652247327956)(C:\Users\86152\AppData\Roaming\Typora\typora-user-images\image-20220119164947679.png)] Insert picture description here

analysis : choose T The least visited in time . namely ,22 Number , choose B

  1. [ Failed to transfer the external chain picture , The origin station may have anti-theft chain mechanism , It is suggested to save the pictures and upload them directly (img-nfCphMUl-1652247327957)(C:\Users\86152\AppData\Roaming\Typora\typora-user-images\image-20220119165148159.png)] Insert picture description here

analysis : choose Access bit and modify bit All zero , Should choose B

  1. [ Failed to transfer the external chain picture , The origin station may have anti-theft chain mechanism , It is suggested to save the pictures and upload them directly (img-01b0WlSs-1652247327958)(C:\Users\86152\AppData\Roaming\Typora\typora-user-images\image-20220119165517248.png)] Insert picture description here

analysis :56 The number is on the top , They came first , So choose A.

The first 16 topic :

  1. [ Failed to transfer the external chain picture , The origin station may have anti-theft chain mechanism , It is suggested to save the pictures and upload them directly (img-qlmvCFK7-1652247327958)(C:\Users\86152\AppData\Roaming\Typora\typora-user-images\image-20220119172644510.png)] Insert picture description here
    analysis :1) Find page table items :2G/4K;2G = 2 x 1024 x 1024 = 2097152 K, And then calculate 2G/4K = 524288.

    ​ 2) Use page table entries x 4 ( byte ), Again /4(KB), And then again ÷ 1024.

  2. In a system with a page based storage management scheme ,** Logical address 32 Who said ,** The memory block size is 210**. How many pages can the user program be divided into :**232 - 2^10 = 2^22.( Logical address - Memory block size )

  3. [ Failed to transfer the external chain picture , The origin station may have anti-theft chain mechanism , It is suggested to save the pictures and upload them directly (img-8PnYgzSd-1652247327959)(C:\Users\86152\AppData\Roaming\Typora\typora-user-images\image-20220120093836470.png)] Insert picture description here

​ analysis : Because the hit rate is not enough 100%, So there will be delays , Then the effective access time must be longer than 0.4 longer ; Hit rate is 90%, The miss rate is 10%,0.4 Of 10% by 0.04, Therefore, the effective access time is :0.4+0.04 = 0.44.

  1. A virtual page storage management system adopts Second level page table Address translation , If cache and fast table are not considered , Then the process Each instruction needs to be accessed at least several times :3 Time .( The first time is to access the index , Second access to the secondary page table , The third time is to access memory )

  2. At some point CPU The utilization rate of is 50%, The disk's busy rate is 3%. In this case , Which of the following is reasonable : Appropriately increase the number of processes to improve system efficiency ( Too idle increases )

  3. At some point CPU The utilization rate of is 3%, The disk's busy rate is 97%. In this case , Which of the following is reasonable : Properly reduce the number of processes to improve system efficiency ( Too busy reduces )

  4. [ Failed to transfer the external chain picture , The origin station may have anti-theft chain mechanism , It is suggested to save the pictures and upload them directly (img-TJbcZyD3-1652247327960)(C:\Users\86152\AppData\Roaming\Typora\typora-user-images\image-20220120095436827.png)] Insert picture description here

  5. [ Failed to transfer the external chain picture , The origin station may have anti-theft chain mechanism , It is suggested to save the pictures and upload them directly (img-BYanl0SU-1652247327961)(C:\Users\86152\AppData\Roaming\Typora\typora-user-images\image-20220120095620037.png)] Insert picture description here

  6. [ Failed to transfer the external chain picture , The origin station may have anti-theft chain mechanism , It is suggested to save the pictures and upload them directly (img-IacCRGLo-1652247327961)(C:\Users\86152\AppData\Roaming\Typora\typora-user-images\image-20220120095808511.png)] Insert picture description here

    analysis : Note that the unit of address length is byte ,1 Bytes equal to 8 Binary bits , It needs to be changed here

  7. [ Failed to transfer the external chain picture , The origin station may have anti-theft chain mechanism , It is suggested to save the pictures and upload them directly (img-BMXZLNKk-1652247327962)(C:\Users\86152\AppData\Roaming\Typora\typora-user-images\image-20220120100026980.png)] Insert picture description here

analysis : ask while knowing the answer , How many bytes does a process have , Just ask How big is the total space of the process , That is to ask How big is the virtual address used by the system , namely : 2^20

  1. [ Failed to transfer the external chain picture , The origin station may have anti-theft chain mechanism , It is suggested to save the pictures and upload them directly (img-xQQXaTms-1652247327962)(C:\Users\86152\AppData\Roaming\Typora\typora-user-images\image-20220120100906093.png)] Insert picture description here

analysis : This question feels a little strange ,,, No matter the , Just remember !! Should be and The memory in the stem is addressed in bytes of .220-212 = 2^8

  1. [ Failed to transfer the external chain picture , The origin station may have anti-theft chain mechanism , It is suggested to save the pictures and upload them directly (img-sNyPe32Z-1652247327963)(C:\Users\86152\AppData\Roaming\Typora\typora-user-images\image-20220120101227147.png)] Insert picture description here

analysis : Pay attention to and 10 topic distinguish , In this question 20 A binary bit represents a virtual Page number , and 10 In question It means virtual Address , So the calculation method of this problem is :2^20 + 2^12 = 2^32.

The first 17 topic :

  1. Which permutation algorithm is likely to produce Belady Abnormal phenomenon :FIFO
  2. Virtual storage space The size of is limited by what factor : Computer address bit width ( Computer processor word length )
  3. [ Failed to transfer the external chain picture , The origin station may have anti-theft chain mechanism , It is suggested to save the pictures and upload them directly (img-gdRGHhcD-1652247327964)(C:\Users\86152\AppData\Roaming\Typora\typora-user-images\image-20220120102320703.png)] Insert picture description here

​ analysis : Just look at WRITE[3,100] Just go , The meaning of the sentence is : stay Page No 3 In my line of work Write operations (W), By observing the data in the table, we can know , Page No 3 Of course , Access control is R, It can only be carried out Read operations , So it happens : Write protect interrupt .

  1. [ Failed to transfer the external chain picture , The origin station may have anti-theft chain mechanism , It is suggested to save the pictures and upload them directly (img-O7iouyzL-1652247327964)(C:\Users\86152\AppData\Roaming\Typora\typora-user-images\image-20220120103937544.png)] Insert picture description here

analysis : This question is terrible ( The calculation is very complicated ), Just remember the answer .qwq

  1. [ Failed to transfer the external chain picture , The origin station may have anti-theft chain mechanism , It is suggested to save the pictures and upload them directly (img-oMeVu7Uj-1652247327965)(C:\Users\86152\AppData\Roaming\Typora\typora-user-images\image-20220120104342918.png)] Insert picture description here

analysis : Divide the page size by logical address , Take the first integer of their remainder , Then look at the table below , Whose significant bit is 0 , Then a page missing interrupt occurs .

  1. [ Failed to transfer the external chain picture , The origin station may have anti-theft chain mechanism , It is suggested to save the pictures and upload them directly (img-rjZP4BSw-1652247327967)(C:\Users\86152\AppData\Roaming\Typora\typora-user-images\image-20220120105818725.png)] Insert picture description here

  2. Address translation is performed by operating system Hardware Accomplished

  3. The address on the page is the of the address The high part The lower part

  4. Watch it Store in register Cache memory

  5. The paging daemon is The front desk backstage perform

  6. The page table length register holds the page table length of the running process

  7. The starting address of the page table belongs to the field information of the process , Stored in the process control block of the process

  8. Logical addresses are addressed from zero

  9. Logical addresses are contiguous , Physical pages can be non contiguous

  10. Paging is a consideration for the operating system

  11. The calculation formula of physical address is : Memory block number x Block length + Page address

  12. [ Failed to transfer the external chain picture , The origin station may have anti-theft chain mechanism , It is suggested to save the pictures and upload them directly (img-hgIz0Kko-1652247327968)(C:\Users\86152\AppData\Roaming\Typora\typora-user-images\image-20220120142341633.png)] Insert picture description here

analysis **: Page table length = Process address space ÷ Page size ;**512MB= 524288KB ,524288 ÷ 1 = 2^19. Regardless of the physical memory size

( Such questions , It is recommended to directly remember , Calculation is too cumbersome )

  1. [ Failed to transfer the external chain picture , The origin station may have anti-theft chain mechanism , It is suggested to save the pictures and upload them directly (img-U0qI9KS0-1652247327969)(C:\Users\86152\AppData\Roaming\Typora\typora-user-images\image-20220120143231396.png)] Insert picture description here

  2. [ Failed to transfer the external chain picture , The origin station may have anti-theft chain mechanism , It is suggested to save the pictures and upload them directly (img-Pp5T8wUm-1652247327969)(C:\Users\86152\AppData\Roaming\Typora\typora-user-images\image-20220120143310184.png)] Insert picture description here

  3. [ Failed to transfer the external chain picture , The origin station may have anti-theft chain mechanism , It is suggested to save the pictures and upload them directly (img-q1DOjIqR-1652247327970)(C:\Users\86152\AppData\Roaming\Typora\typora-user-images\image-20220120143346862.png)] Insert picture description here

  4. [ Failed to transfer the external chain picture , The origin station may have anti-theft chain mechanism , It is suggested to save the pictures and upload them directly (img-DoOH72o0-1652247327971)(C:\Users\86152\AppData\Roaming\Typora\typora-user-images\image-20220120143638624.png)] Insert picture description here

  5. [ Failed to transfer the external chain picture , The origin station may have anti-theft chain mechanism , It is suggested to save the pictures and upload them directly (img-0kCYbydC-1652247327973)(C:\Users\86152\AppData\Roaming\Typora\typora-user-images\image-20220120143721776.png)] Insert picture description here

  6. [ Failed to transfer the external chain picture , The origin station may have anti-theft chain mechanism , It is suggested to save the pictures and upload them directly (img-un8ThJSD-1652247327974)(C:\Users\86152\AppData\Roaming\Typora\typora-user-images\image-20220120143856847.png)] Insert picture description here

The first 18 topic :

  1. happen ” shake “ The phenomenon is caused by : Page replacement algorithm is unreasonable Caused by the
  2. LRU Algorithm : First of all, replace the pages that haven't been visited for the longest time in the near future
  3. decision Maximum capacity of virtual storage space The key elements are : Computer system Address bit width
  4. In order to improve memory utilization and Reduce internal debris , Page Division : And the number of page tables relevant , Can find Balance point
  5. Realization Virtual page storage management The hardware foundation of is : Page missing interrupt mechanism
  6. To improve memory utilization , You can use a variety of pages of different sizes as needed
  7. Page and page frame size in page storage management Can only Adopt one size
  8. User processes can change Page size
  9. Operating system in Every time it starts Determine the page size of this run according to the physical memory size
  10. [ Failed to transfer the external chain picture , The origin station may have anti-theft chain mechanism , It is suggested to save the pictures and upload them directly (img-8ukbUI15-1652247327975)(C:\Users\86152\AppData\Roaming\Typora\typora-user-images\image-20220120145331352.png)] Insert picture description here

analysis :for i,for j Type of question , Method :(i Value 150 x j Value 100**) ÷ Integer variable **( In this question is 150)

  1. [ Failed to transfer the external chain picture , The origin station may have anti-theft chain mechanism , It is suggested to save the pictures and upload them directly (img-leqn7ht6-1652247327975)(C:\Users\86152\AppData\Roaming\Typora\typora-user-images\image-20220120145946906.png)] Insert picture description here

analysis :for j,for i Type of question , Method : Such questions , The correct answer must be :“ So-and-so x So-and-so ” In the form of , Exclude first A、B term ;

Be careful :“ So-and-so x 1” This option has to be ruled out The question teacher is too dog

​ secondly , If ** Integer variables and i、j Different values of ,** The answer is :j Value x Different values ( Sometimes it's a value that doesn't appear in the question stem );

​ If ** Integer variables and i、j When the values of are the same ( The three values are equal ),** The answer is :j Value x The same value ;

  1. [ Failed to transfer the external chain picture , The origin station may have anti-theft chain mechanism , It is suggested to save the pictures and upload them directly (img-LYpDWl7y-1652247327976)(C:\Users\86152\AppData\Roaming\Typora\typora-user-images\image-20220120151216847.png)] Insert picture description here

13.[ Failed to transfer the external chain picture , The origin station may have anti-theft chain mechanism , It is suggested to save the pictures and upload them directly (img-LLYbl5t2-1652247327977)(C:\Users\86152\AppData\Roaming\Typora\typora-user-images\image-20220120151323518.png)] Insert picture description here

The first 19 topic :

  1. [ Failed to transfer the external chain picture , The origin station may have anti-theft chain mechanism , It is suggested to save the pictures and upload them directly (img-wLb9OP09-1652247327977)(C:\Users\86152\AppData\Roaming\Typora\typora-user-images\image-20220120152345224.png)] Insert picture description here

analysis : The best page replacement algorithm , Also known as residual value algorithm , The condition for deciding whether to replace the current value is , Whether this value exists in the future access sequence .

  1. If page transfer request is adopted , When a user needs to load a new page , The page it calls in comes from : Disk file area

  2. In order to prevent the occurrence of ** shake ( Bumpy )** The phenomenon , Which of the following methods can be used : Using working set algorithm

  3. The principle of program locality is divided into space locality and time locality , Spatial locality Refer to : Of the program code Sequence , Temporal locality : Program exists A lot of cycles .

  4. [ Failed to transfer the external chain picture , The origin station may have anti-theft chain mechanism , It is suggested to save the pictures and upload them directly (img-IiU4PNCB-1652247327978)(C:\Users\86152\AppData\Roaming\Typora\typora-user-images\image-20220120155008203.png)] Insert picture description here

  5. [ Failed to transfer the external chain picture , The origin station may have anti-theft chain mechanism , It is suggested to save the pictures and upload them directly (img-WqLwifTP-1652247327979)(C:\Users\86152\AppData\Roaming\Typora\typora-user-images\image-20220120155300001.png)] Insert picture description here

  6. [ Failed to transfer the external chain picture , The origin station may have anti-theft chain mechanism , It is suggested to save the pictures and upload them directly (img-QSGboO4Q-1652247327980)(C:\Users\86152\AppData\Roaming\Typora\typora-user-images\image-20220120155459481.png)] Insert picture description here

​ analysis : take 30 Divided into 10 and 20 , yes Best fit ;30 It is also the maximum capacity value , So it also belongs to First fit , But cause The priority of the first adaptation is higher than Best fit , So choose First fit

  1. [ Failed to transfer the external chain picture , The origin station may have anti-theft chain mechanism , It is suggested to save the pictures and upload them directly (img-U3J914Wd-1652247327981)(C:\Users\86152\AppData\Roaming\Typora\typora-user-images\image-20220120160039575.png)] Insert picture description here

  2. [ Failed to transfer the external chain picture , The origin station may have anti-theft chain mechanism , It is suggested to save the pictures and upload them directly (img-eabxUAVF-1652247327981)(C:\Users\86152\AppData\Roaming\Typora\typora-user-images\image-20220120160537322.png)] Insert picture description here

  3. [ Failed to transfer the external chain picture , The origin station may have anti-theft chain mechanism , It is suggested to save the pictures and upload them directly (img-dwh0VuP7-1652247327982)(C:\Users\86152\AppData\Roaming\Typora\typora-user-images\image-20220120161610893.png)] Insert picture description here

​ analysis : Regardless of size , Come up and divide —— First adapt to the algorithm

The first 20 topic :

  1. The way files are accessed depends on : The physical structure of the file And the physical characteristics of the device where the files are stored
  2. In the file system , The logical block of the file is stored with the physical block on the storage medium In the same order The physical structure of is : Sequential structure
  3. In the file system , The allocation unit of file storage space is usually : Data blocks
  4. From the perspective of users , File control block (FCB) The most important field in is : file name
  5. From the perspective of users , The main goal of establishing a file system is : Realize file name access
  6. From the perspective of users , Create a multi-level tree directory The main objectives of : Solve the problem of file name duplication
  7. [ Failed to transfer the external chain picture , The origin station may have anti-theft chain mechanism , It is suggested to save the pictures and upload them directly (img-F77cqPvS-1652247327985)(C:\Users\86152\AppData\Roaming\Typora\typora-user-images\image-20220120163306997.png Insert picture description here
    )]

[ Failed to transfer the external chain picture , The origin station may have anti-theft chain mechanism , It is suggested to save the pictures and upload them directly (img-HVGVl1I7-1652247327986)(C:\Users\86152\AppData\Roaming\Typora\typora-user-images\image-20220120163344635.png)] Insert picture description here

analysis : Yes 10 individual Point directly to the disk block number , then 32 position Divide 8 be equal to 4 byte , Reuse 1kB = 1024B,1024 Divide 4 be equal to 256, Through the theme , So choose C

  1. File system , If the logically continuous file information is scattered and stored in several discontinuous disk blocks , Set a pointer to the next disk block in each disk block , This structure is called : Link structure
  2. File system , If the logically continuous file information is scattered and stored in several discontinuous disk blocks , And the addresses of all disk blocks ** Centralized storage in one table ,** This structure is called : Index structure
  3. With Keep creating and deleting files , Which leads to Generate disk fragmentation The physical structure of the file is : Sequential structure
  4. What kind of file's physical structure The retrieval speed is slow , And not suitable for random access files Link structure
  5. The basic unit constituting the content of the document is called : Information item
  6. The content of the document No, Order relations
  7. The contents of the document are made up of operating system user Explain and use
  8. user Need to pay attention to Where files are stored on disk
  9. Too absolute not to choose
  10. All files in the file system once created , Unless deleted or beyond the shelf life , Otherwise there will always be .
  11. File names in all file systems There are Suffix and uniformly specify the meaning of suffix ( This is not necessarily )
  12. Special files are usually closely associated with device drivers
  13. For system files , Only users are allowed to access them through system calls
  14. linux Of EXT2 file system Case insensitive for file names Capital must be written.
  15. Directory files are system files
  16. UNIX In the operating system I/O The equipment is regarded as a special document
  17. from Use Focusing on the organization form of documents is called the organization form of documents Logical structure
  18. from lookup From the perspective of documents, the organization of documents is called Logical structure
  19. from access From the perspective of documents, the organization of documents is called Physical structure
  20. Files kept on permanent storage media for verification and recovery are called archive files
  21. File system , Of documents Logical structure Refer to : The operating system is provided to users Of Organization of documents ( Presentation
  22. File system , Of documents Physical structure Refer to : The organization of files on disk ( File organization form of operating system management )
  23. When users create and use files , It has to be given : file name

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