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values | SubDomain stringclasses 24
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value | Question stringlengths 15 717 | A stringlengths 1 292 | B stringlengths 1 232 | C stringlengths 1 217 | D stringlengths 1 192 | Answer stringclasses 4
values | Explanation stringlengths 21 1.43k ⌀ |
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1,820 | Test | Operating System | Memory Management | Multiple-choice | Knowledge | English | The correct statement about storage management is (). | The purpose of storage protection is to restrict the allocation of memory. | In a time-sharing system with memory M and N users, each user occupies M/N of the memory space. | In a virtual memory system, as long as the disk space is unlimited, a job can have an arbitrarily large addressing space. | To implement virtual memory management, corresponding hardware support is required. | D | null |
1,821 | Test | Operating System | Memory Management | Multiple-choice | Knowledge | English | Among the following memory management schemes, the () method can adopt static relocation. | Fixed Partition | Variable Partitioning | Paging | Segmental | A | null |
1,822 | Test | Operating System | Memory Management | Multiple-choice | Knowledge | English | A storage management that does not produce internal fragmentation is (). | Paged Memory Management | Segmented Memory Management | Fixed Partition Storage Management | Segmented Paging Memory Management | B | null |
1,823 | Test | Operating System | Memory Management | Multiple-choice | Knowledge | English | For the relocated storage management method, one should (). | Set up a relocation register throughout the system. | Set a relocation register for each program. | Set two relocation registers for each program. | Set a relocation register for each program and data. | A | null |
1,824 | Test | Operating System | Memory Management | Multiple-choice | Knowledge | English | Converting the logical addresses used in the working space to physical addresses in memory is called (). | Loading | Relocation | Physical Chemistry | Formalization | B | null |
1,825 | Test | Operating System | Memory Management | Multiple-choice | Knowledge | English | In segmented distribution, the CPU needs () memory accesses each time it fetches data from memory. | 1 | 3 | 2 | 4 | C | null |
1,826 | Test | Operating System | Memory Management | Multiple-choice | Knowledge | English | The operating system adopts a paging storage management method, which requires (). | Each process has its own page table, and the page table of the process resides in memory. | Each process has a page table, but only the page table of the executing process resides in memory. | All processes share a single page table to conserve limited memory space, but the page table must reside in memory. | All processes share a single page table, and only the pages currently in use in the page table must reside in memory to maximize the conservation of limited memory space. | A | null |
1,827 | Test | Operating System | Memory Management | Multiple-choice | Knowledge | English | Virtual storage technology is (). | Techniques for Supplementing Physical Memory Space | Techniques for Supplementing Logical Memory Space | Techniques for Supplementing External Storage Space | Techniques for Expanding Input/Output Buffers | B | null |
1,828 | Test | Operating System | Memory Management | Multiple-choice | Knowledge | English | To ensure that the virtual memory system functions effectively as intended, the characteristic that the running program should possess is (). | The program should not contain excessive I/O operations. | The size of the program should not exceed the actual memory capacity. | The program should have good locality. | The instruction relevance of the program should not be excessive. | C | null |
1,829 | Test | Operating System | Memory Management | Multiple-choice | Knowledge | English | The main characteristic of demand paging storage management is (). | Eliminated the fractional part within the page. | Expanded the memory | Facilitates dynamic linking | Facilitates information sharing | B | null |
1,830 | Test | Operating System | Memory Management | Multiple-choice | Knowledge | English | The storage management methods that provide virtual storage technology include (). | Dynamic Partition Storage Management | Paged Memory Management | Segmented Memory Management Request | Storage Overwriting Technology | C | null |
1,831 | Test | Operating System | Memory Management | Multiple-choice | Knowledge | English | When using swapping technology, if a process is in the middle of a critical section (), it cannot be swapped out of main memory. | Create | I/O operations | In the critical section | Deadlock | B | null |
1,832 | Test | Operating System | Memory Management | Multiple-choice | Knowledge | English | In storage management, the purpose of employing overlay and swapping techniques is (). | Save main memory space | Physically expanding the main memory capacity | Improve CPU Efficiency | Implementing main memory sharing | A | null |
1,833 | Test | Operating System | Memory Management | Multiple-choice | Knowledge | English | Memory protection needs to be implemented by the () to ensure that the process space is not accessed illegally. | Operating System | Hardware Organization | The operating system collaborates with the hardware architecture. | The operating system or hardware architecture operates independently. | C | null |
1,834 | Test | Operating System | Memory Management | Multiple-choice | Knowledge | English | In variable partition management, the purpose of using compaction technology is (). | Merge free space | Merge Partition | Increase main memory capacity | Facilitate address translation | A | null |
1,835 | Test | Operating System | Memory Management | Multiple-choice | Knowledge | English | When using the best-fit allocation algorithm in partition management, the free areas are registered in the free area table in () order. | Increasing length | Decreasing length | Incremental Addressing | Address Decrement | A | null |
1,836 | Test | Operating System | Memory Management | Multiple-choice | Knowledge | English | The pages in a paging system are for (). | user-perceived | The operating system perceives | perceived by the compilation system | The perceived connection assembly program | B | null |
1,837 | Test | Operating System | Memory Management | Multiple-choice | Knowledge | English | In page-based memory management, the starting address of the page table is stored in (). | memory | Page Table Storage | Translation Memory | Register | D | null |
1,838 | Test | Operating System | Memory Management | Multiple-choice | Knowledge | English | When using segmented storage management, how a program is divided into segments is determined at (). | Allocate Main Memory | User Programming | Assignment Submission | Program Execution | B | null |
1,839 | Test | Operating System | Memory Management | Multiple-choice | Knowledge | English | The following () method is conducive to the dynamic linking of the program. | Segmented Storage Management | Paging storage management | Variable Formula Partition Management | Fixed Partition Management | A | null |
1,840 | Test | Operating System | Memory Management | Multiple-choice | Knowledge | English | Reentrant programs improve system performance through the () method. | Change the length of the time slice | Change in the number of users | Increase the exchange rate | Reduce the number of permutations | D | null |
1,841 | Test | Operating System | Memory Management | Multiple-choice | Knowledge | English | Dynamic partitioning, also known as variable partitioning, is established dynamically during the system's runtime (). | During job loading | At the time of assignment creation | Upon completion of the assignment | When the assignment is not loaded | A | null |
1,842 | Test | Operating System | Memory Management | Multiple-choice | Knowledge | English | The purpose of storage management is (). | Convenient for users | Improve memory utilization | Facilitate users and enhance memory utilization. | Increase the actual capacity of memory | C | null |
1,843 | Test | Operating System | Memory Management | Multiple-choice | Knowledge | English | Access to the main memory, (). | in blocks (i.e., pages) or segments | In bytes or words | Varies with different memory management schemes | Based on the unit of user's logical records | B | null |
1,844 | Test | Operating System | Memory Management | Multiple-choice | Knowledge | English | In paging storage management, the allocation of main memory () | Perform operations in units of physical blocks. | Based on the size of the task | Addressing by physical segments | Based on the size of logical records | A | null |
1,845 | Test | Operating System | Memory Management | Multiple-choice | Knowledge | English | Which of the following is not a characteristic of virtual memory? (). | Disposable | Multiple times | Commutativity | Discreteness | A | null |
1,846 | Test | Operating System | Memory Management | Multiple-choice | Knowledge | English | The difference between demand paging and basic paging storage management is () | Address Redirection | There is no need to load the entire assignment into memory. | Using fast table technology | There is no need to load the assignment into a contiguous area. | B | null |
1,847 | Test | Operating System | Memory Management | Multiple-choice | Knowledge | English | Maximum capacity of virtual memory (). | For the sum of internal and external memory capacity | Determined by the computer's addressing structure | is arbitrary | Determined by the address space of the task. | B | null |
1,848 | Test | Operating System | Memory Management | Multiple-choice | Knowledge | English | The foundation of the virtual storage management system is the theory of (). | Dynamic nature | Virtuality | Locality | globality | C | null |
1,849 | Test | Operating System | Memory Management | Multiple-choice | Knowledge | English | The () method can be used to implement virtual storage. | Partition Merge | Coverage, Exchange | Translation Memory | Segment Merging | B | null |
1,850 | Test | Operating System | Memory Management | Multiple-choice | Knowledge | English | The main characteristic of paged virtual memory management is (). | Do not require that the job be loaded into a contiguous area of main memory. | Do not require loading all tasks into a contiguous area of the main memory at the same time. | Do not require handling of page faults. | No page replacement is required. | B | null |
1,851 | Test | Operating System | Memory Management | Multiple-choice | Reasoning | English | In segmented paging storage management, the address mapping table is (). | Each process has one segment table and two page tables. | Each process has a segment table for each of its segments, and a page table. | Each process has a segment table, and each segment has a page table. | Each process has one page table, and each segment has one segment table. | C | In a segmented paging system, a process is first divided into segments, and each segment is further divided into pages. |
1,852 | Test | Operating System | Memory Management | Multiple-choice | Reasoning | English | In the variable partition allocation scheme, after a process is completed, the system reclaims its main memory space and merges it with adjacent free areas. This requires modifying the free area table, resulting in a decrease in the number of free areas by 1 when (). | No upper adjacent free space and no lower adjacent free space. | There is a free area above but no free area below. | There is a free block below but no free block above. | There is a free area above and a free area below. | D | Merge the upper adjacent free block, the lower adjacent free block, and the reclaimed block into one free block, thus reducing the number of free blocks by one. However, when there is only an upper adjacent free block or a lower adjacent free block, the number of free blocks does not decrease. |
1,853 | Test | Operating System | Memory Management | Multiple-choice | Reasoning | English | First Fit Algorithm's free partition (). | In descending order of size, connected together. | In increasing order of size, connected together. | Sorted in ascending order by address | Sort by address in descending order | C | The free partitions in the first-fit algorithm are arranged in ascending order of addresses. |
1,854 | Test | Operating System | Memory Management | Multiple-choice | Reasoning | English | The following statement about virtual memory is correct (). | Before execution, the job must be fully loaded into memory and must remain in memory throughout its execution. | Before execution, the job does not need to be fully loaded into memory, and it does not need to remain in memory throughout its execution. | Before execution, the job does not need to be fully loaded into memory, but it must remain in memory throughout the execution process. | Before execution, the job must be fully loaded into memory, but it does not need to remain in memory throughout the execution process. | B | In non-virtual memory systems, a job must be entirely loaded into memory and remain resident in memory during execution; in virtual memory systems, a job does not need to be entirely loaded into memory nor does it need to remain resident in memory throughout its execution. This is one of the main differences between vi... |
1,855 | Test | Operating System | Memory Management | Multiple-choice | Reasoning | English | In page-based storage management, the selection of page size should consider the following () factors. Ⅰ. The advantage of larger pages is that there are fewer page tables. Ⅱ. The advantage of smaller pages is that they can reduce memory waste caused by internal fragmentation. Ⅲ. The main factor affecting disk access t... | Quadrants I and III | II and III | Type I and Type II | I, II, and III | C | Large pages result in fewer page tables for managing the pages, but lead to more internal fragmentation; small pages result in larger page tables, but less internal fragmentation. Appropriate calculations can yield an optimal page size and minimize system overhead. |
1,856 | Test | Operating System | Memory Management | Multiple-choice | Reasoning | English | An operating system manages memory using a paging storage management method, with a page size of (). | To determine based on the size of memory | Must be identical | To determine based on the CPU's address structure | Determine based on the size of external storage and internal memory. | B | An important issue in paging management is how to determine the page size. There are many factors to consider when determining page size, such as the average size of processes, the length occupied by the page table, etc. Once determined, all pages are of equal length (generally an integer power of 2) to facilitate syst... |
1,857 | Test | Operating System | Memory Management | Multiple-choice | Reasoning | English | In segmented paging, the CPU needs () memory accesses to fetch data from memory each time. | 1 | 3 | 2 | 4 | B | In segmented paging, when fetching data, the segment table is first looked up in memory, then the corresponding page table is accessed in memory, and finally, after forming the physical address, memory is accessed, requiring a total of 3 memory accesses. |
1,858 | Test | Operating System | Memory Management | Multiple-choice | Reasoning | English | Among the following statements about paging storage, the correct one is (). Ⅰ. In paging storage management, if the TLB is turned off, the memory must be accessed twice whenever an instruction is accessed or an operand is fetched. Ⅱ. Paging storage management does not produce internal fragmentation. Ⅲ. Pages in paging ... | I, II, IV | I, IV | Only Ⅰ | All correct | C | Ⅰ Correct: After disabling the TLB, each time an instruction is accessed or an operand is fetched, the page table (in memory) must be accessed first to obtain the physical address, followed by another memory access to perform the corresponding operation.
Ⅱ Incorrect: Remember, any fixed partitioning will result in int... |
1,859 | Test | Operating System | Memory Management | Multiple-choice | Reasoning | English | In demand paging storage management, if the page size is doubled while the maximum number of pages that can be accommodated remains unchanged, the number of page faults during sequential program execution will (). | Increase | reduce | Invariant | May increase or decrease | B | In a demand paging system, as the page size increases, the number of page frames required to store a program decreases, thus reducing the frequency of page faults. |
1,860 | Test | Operating System | Memory Management | Multiple-choice | Reasoning | English | When a process encounters a page fault interrupt during execution, the operating system should handle it and then allow it to execute the () instruction. | The previous one that was interrupted | The interrupted one | The subsequent one that was interrupted | The first line upon startup | B | A page fault interrupt is caused by a memory access instruction, indicating that the page to be accessed is not in memory. After handling the page fault interrupt and loading the required page, the memory access instruction should clearly be re-executed. |
1,861 | Test | Operating System | Memory Management | Multiple-choice | Reasoning | English | The following statement about the page scheduling algorithm in demand paging systems is incorrect (). | A good page scheduling algorithm should reduce and avoid thrashing. | The FIFO algorithm is simple to implement; it selects the page that entered the main memory first for eviction. | The LRU algorithm is based on the principle of locality and first evicts the page that has not been accessed for the longest time in the recent period. | The CLOCK algorithm first evicts pages that have been frequently accessed over a period of time. | D | The CLOCK algorithm selects the least recently used page for replacement, hence it is also known as the NRU (Not Recently Used) algorithm. |
1,862 | Test | Operating System | Memory Management | Multiple-choice | Reasoning | English | Consider the page replacement algorithm, the system has m physical blocks available for scheduling, all initially empty, the length of the page reference string is p, containing n different page numbers, regardless of the algorithm used, the number of page faults will not be less than (). | m | p | n | min(m,n) | C | Regardless of the page replacement algorithm used, each page cannot be in memory upon its first access, inevitably resulting in a page fault. Therefore, the number of page faults is greater than or equal to n. |
1,863 | Test | Operating System | Memory Management | Multiple-choice | Reasoning | English | Assuming the main memory capacity is 1MB and the external memory capacity is 400MB, with the computer system's address register being 32 bits, then the maximum capacity of the virtual memory is (). | 1MB | 401MB | 1MB + 2^32 MB | 2^32B | D | The maximum capacity of virtual memory is determined by the computer's addressing structure and is not necessarily related to the capacity of the main memory or external storage. Its virtual address space is 2^32B. |
1,864 | Test | Operating System | Memory Management | Multiple-choice | Reasoning | English | The reason why the implementation of the LRU algorithm is costly is (). | Requires special support from hardware. | Requires a special interrupt handler. | Special page types need to be marked in the page table. | All pages need to be sorted. | D | The LRU algorithm requires recording the time since the last access for all pages and replaces the one that has not been accessed for the longest time, which involves sorting and is too costly for a replacement algorithm. Therefore, it is necessary to add LRU bits to the page table entries. Option A can be considered a... |
1,865 | Test | Operating System | Memory Management | Multiple-choice | Reasoning | English | In the page table entries of a virtual memory system, what determines whether a page fault will occur is (). | Legal Position | Modify Bit | Page Type | Protection Code | A | The valid bit information in a page table entry indicates whether the page is in memory, thus determining whether a page fault will occur. |
1,866 | Test | Operating System | Memory Management | Multiple-choice | Reasoning | English | In page replacement strategies, the () strategy may cause thrashing. | FIFO | LRU | There is no single | All | D | Thrashing is the behavior of frequent page scheduling (page faults) during the page replacement process of a program, and no page scheduling strategy can completely avoid thrashing. |
1,867 | Test | Operating System | Memory Management | Multiple-choice | Reasoning | English | The main cause of memory churn is (). | Insufficient memory space | The CPU is running too slowly. | The CPU scheduling algorithm is unreasonable. | The page replacement algorithm is unreasonable. | D | Memory thrashing refers to the phenomenon where pages in main memory are frequently evicted and then immediately loaded back in, only to be quickly evicted again. This is caused by an inefficient page replacement algorithm and is something that page replacement algorithms should strive to avoid. |
1,868 | Test | Operating System | Memory Management | Multiple-choice | Reasoning | English | In page replacement algorithms, the algorithm that exhibits Belady's anomaly is (). | Optimal Page Replacement Algorithm (OPT) | First-In, First-Out replacement algorithm (FIFO) | Least Recently Used algorithm (LRU) | Recently Not Used algorithm (NRU) | B | FIFO is a queue-based algorithm that exhibits Belady's anomaly; options C and D are both stack-based algorithms, which can be theoretically proven not to exhibit Belady's anomaly. |
1,869 | Test | Operating System | Memory Management | Multiple-choice | Reasoning | English | In a virtual paging storage management system, if the page accessed by a process is not in the main memory and there are no free frames available in the main memory, the correct sequence of system handling is (). | Page Replacement → Page Eviction → Page Fault Interrupt → Page Fetch | Page Replacement → Page Fault → Page In → Page Out | Page Fault → Determine Page to Evict → Page Out → Page In | Page Fault → Page Replacement Decision → Page In → Page Out | C | According to the process of handling page faults, after a page fault occurs, the first step is to look for a free physical block in memory. If there are no free physical blocks available, a page replacement algorithm is used to determine which page to evict. Then, the evicted page is swapped out, followed by the swappi... |
1,870 | Test | Operating System | Memory Management | Multiple-choice | Reasoning | English | Given that the system uses a 32-bit physical address and a 48-bit virtual address, with a page size of 4KB and a page table entry size of 8B. Assuming the system employs pure paging, it will use a ()-level page table, and the page offset is () bits. | 3, 12 | 3.14 | 4, 12 | 4, 14 | C | The page size is 4KB, so the offset within a page is 12 bits. The system uses a 48-bit virtual address, so the virtual page number is 48-12=36 bits. When using a multi-level page table, the highest-level page table entries must not exceed one page size: the number of page table entries that can be accommodated per page... |
1,871 | Test | Operating System | Memory Management | Multiple-choice | Reasoning | English | Among the following statements, the correct one(s) is (are) (). Ⅰ. The First-In-First-Out (FIFO) page replacement algorithm can lead to Belady's anomaly. Ⅱ. The Least Recently Used (LRU) page replacement algorithm can lead to Belady's anomaly. Ⅲ. If all the pages of a process's working set are in virtual memory during ... | I, III | I, IV | II, III | II, IV | B | The FIFO algorithm may lead to Belady's anomaly. For example, with the page reference string 1,2,3,4,1,2,5,1,2,3,4,5, when 3 frames are allocated, there are 9 page faults, and when 4 frames are allocated, there are 10 page faults, so statement Ⅰ is correct. The Least Recently Used (LRU) algorithm does not produce Belad... |
1,872 | Test | Operating System | Memory Management | Multiple-choice | Reasoning | English | The measured partial state data of a computer system using a demand paging strategy are: CPU utilization is 20%, disk utilization for swap space is 97.7%, and the utilization of other devices is 5%. It is judged that the system is experiencing anomalies, and under these circumstances, () can improve system performance. | Install a faster hard drive | Increase swap space by expanding hard disk capacity. | Increase the number of running processes | Add RAM sticks to increase physical memory capacity. | D | The disk utilization rate for the swap space has reached 97.7%, while the utilization rate for other devices is at 5%, and the CPU utilization is at 20%. This indicates that, despite a low number of tasks, swapping operations are very frequent, leading to the conclusion that there is a severe shortage of physical memor... |
1,873 | Test | Operating System | Memory Management | Multiple-choice | Reasoning | English | Assuming a demand paging storage management system, the utilization rates of various related devices are measured as follows: CPU utilization is 10%, disk swap space utilization is 99.7%, and other I/O device utilization is 5%. The following measures () may improve CPU utilization: Ⅰ. Increase the capacity of memory Ⅱ.... | I, II, III, IV | I, III | II, III, V | II, IV | B | I Correct: Increase the capacity of memory. Increasing memory allows each program to have more page frames, which can reduce the page fault rate, thereby reducing the swapping in/out process and improving CPU utilization.
II Incorrect: Increase the capacity of the disk swap area. Since the system is already in a frequ... |
1,874 | Test | Operating System | File Management | Multiple-choice | Knowledge | English | The information stored in the directory file is (). | The data information stored in a certain file | The file directory of a certain document | All data file directories in this catalog | The directory containing all subdirectory files and data files within this directory. | D | null |
1,875 | Test | Operating System | File Management | Multiple-choice | Knowledge | English | If multiple processes share the same file F, then among the following statements, the correct one is (). | Each process can only open file F in "read" mode. | There is only one entry in the system open file table that contains the attributes of F. | The entries for F in the user-level open file tables of each process are identical. | When a process closes file F, the system removes the entry for F from the system-wide open file table. | A | null |
1,876 | Test | Operating System | File Management | Multiple-choice | Knowledge | English | Access to a file is often restricted by (). | User access rights and file attributes | User access permissions and user priority | Priority and File Attributes | File Attributes and Password | A | null |
1,877 | Test | Operating System | File Management | Multiple-choice | Knowledge | English | The logical structure of a file is designed for the convenience of (). | Storage Medium Characteristics | Operating System Management Methods | Main memory capacity | user | D | null |
1,878 | Test | Operating System | File Management | Multiple-choice | Knowledge | English | When a file exists on a tape, it is usually accessed using (). | Temporary storage method | Continuous and discrete storage methods | Discrete Storage Method | Sequential storage method | D | null |
1,879 | Test | Operating System | File Management | Multiple-choice | Knowledge | English | What is the function of a directory in achieving which of the following? | File's random access | Content-based file access | File's Named Access | Accessing files by size | C | null |
1,880 | Test | Operating System | File Management | Multiple-choice | Knowledge | English | The file structure includes logical structure and physical structure. Logical structure refers to: | The structure in which users organize data is derived from requirements. | The organizational structure of physical data blocks in an operating system. | The organization of a file's name, extension, and path. | The organizational form of a file's physical location and storage method on the storage medium. | A | null |
1,881 | Test | Operating System | File Management | Multiple-choice | Knowledge | English | The organization of physical files is determined by (). | Application Program | Main memory capacity | External storage capacity | Operating System | D | null |
1,882 | Test | Operating System | File Management | Multiple-choice | Knowledge | English | If there are two files with the same name in the file system, () should not be used. | Single-level directory structure | Two-level directory structure | Tree directory structure | Multi-level directory structure | A | null |
1,883 | Test | Operating System | File Management | Multiple-choice | Knowledge | English | In operating systems, input-output devices are commonly referred to as: | Peripheral Device | Special File | Virtual Device | Basic Components | A | null |
1,884 | Test | Operating System | File Management | Multiple-choice | Knowledge | English | From the user's perspective, the purpose of introducing a file system in an operating system is (). | Protecting User Data | Implement file access by name. | Implement virtual storage | Preserve user and system documents and data | B | null |
1,885 | Test | Operating System | File Management | Multiple-choice | Knowledge | English | In the UNIX operating system, input/output devices are treated as (). | regular file | Directory file | Index file | Special file | D | null |
1,886 | Test | Operating System | File Management | Multiple-choice | Knowledge | English | The main task of opening a file operation is (). | Copy the directory of the specified file to the designated area in memory. | Copy the specified file to the designated area in memory. | Locate the directory of the specified file on the designated storage medium. | Search for the specified file in memory. | A | null |
1,887 | Test | Operating System | File Management | Multiple-choice | Knowledge | English | The file directory entry of FAT32 does not include (). | File Name | File Access Permissions Description | Physical location of the File Control Block (FCB) | The physical location of the file | C | null |
1,888 | Test | Operating System | File Management | Multiple-choice | Knowledge | English | Some operating systems separate file descriptor information from directory entries, the benefit of doing this is (). | Reduce the amount of I/O when reading files | Reduce the amount of I/O when writing files. | Reduce the amount of I/O when searching for files | Reduce the amount of I/O data when copying files | C | null |
1,889 | Test | Operating System | File Management | Multiple-choice | Knowledge | English | Among the following statements, () falls within the category of a file's logical structure. | Continuous file | system file | Linked file | Streaming file | D | null |
1,890 | Test | Operating System | File Management | Multiple-choice | Knowledge | English | When using magnetic tape as a file storage medium, files can only be organized in (). | Sequential file | Linked file | Index file | Directory file | A | null |
1,891 | Test | Operating System | File Management | Multiple-choice | Knowledge | English | The following is an inappropriate method for direct access allocation of external storage: (). | Continuous Distribution | Link Allocation | Index Allocation | All of the above answers are suitable. | B | null |
1,892 | Test | Operating System | File Management | Multiple-choice | Knowledge | English | If the physical structure of a file in the file system adopts a contiguous structure, the information about the physical location of the file in the FCB should include () Ⅰ. Starting block address Ⅱ. File length Ⅲ. Index table address | Only Ⅰ | I, II | II, III | I, III | B | null |
1,893 | Test | Operating System | File Management | Multiple-choice | Knowledge | English | The file system creates an index table () for each file, which stores the disk locations of the file's data blocks. | Open the file table | Bitmap | Index Table | Free block linked list | C | null |
1,894 | Test | Operating System | File Management | Multiple-choice | Knowledge | English | The relative path name of a file starts from (), and is a string composed of all the subdirectory names along the entire path, tracing step by step through each level of subdirectories, finally leading to the specified file. | Current directory | root directory | Multi-level Directory | Secondary Directory | A | null |
1,895 | Test | Operating System | File Management | Multiple-choice | Knowledge | English | The purpose of the file system adopting a multi-level directory structure is (). | Reduce system overhead | Save storage space | Resolve naming conflicts | Shorten the transmission time | C | null |
1,896 | Test | Operating System | File Management | Multiple-choice | Knowledge | English | Among the following options, () is not an object type defined by the Linux implementation of the Virtual File System (VFS). | Superblock object | inode object | File object | Data object | D | null |
1,897 | Test | Operating System | File Management | Multiple-choice | Reasoning | English | In the file system, the following is not a method of file protection (). | Password | Access Control | User Permissions Table | After reading and writing, use the close command. | D | In file systems, passwords, access control, and user permission tables are common methods of file protection. |
1,898 | Test | Operating System | File Management | Multiple-choice | Reasoning | English | The following statement about the index table is () correct. | Each record in the index table can have multiple index entries. | When accessing an indexed file, it is necessary to first search the index table. | The index table contains data of the index files and their physical addresses. | One of the purposes of creating an index is to reduce storage space. | B | An index file consists of a logical file and an index table. When accessing an index file, the index table must be searched first. Each index entry contains only the length of the record and its starting position in the logical file. Since each record must have an index entry, this increases the storage cost. |
1,899 | Test | Operating System | File Management | Multiple-choice | Reasoning | English | In the physical structure of the following document, what is not conducive to the dynamic growth of file length is (). | Continuous Structure | Linkage structure | Index structure | Hash structure | A | The requirement for continuous storage space means that the size of the file must be known in advance, so that a sufficiently large storage area can be found in the storage space based on its size. If the file grows dynamically, the space it occupies will become increasingly large. Even if the final size of the file is... |
1,900 | Test | Operating System | File Management | Multiple-choice | Reasoning | English | Assuming a log file that uses linked allocation, with a fixed logical record length of 100B, and employs record clustering technique for storage on disk. The disk block size is 512B. If the directory entry for the file has already been loaded into memory, then after modifying the 22nd logical record, the disk is access... | 3 | 4 | 5 | 6 | D | The 22nd logical record is stored in the 5th physical block (22x100/512=4, remainder 152). Since the file uses a linked file physical structure, it is necessary to start reading from the first physical block pointed to by the directory entry, which requires initiating the disk 5 times. After modification, a write-back ... |
1,901 | Test | Operating System | File Management | Multiple-choice | Reasoning | English | The correct statement about index files is (). | In the index file, each entry in the index table contains the key of the corresponding record and the physical address where the record is stored. | When retrieving a sequential file, the first step is to read the first block number of the file from the FCB; whereas, when retrieving an indexed file, one should first read the starting address of the file index block from the FCB. | For a file with a three-level index, accessing a record typically requires three disk accesses. | When dealing with large files, whether performing sequential access or random access, using an indexed file method is usually the fastest. | B | The entries in the index table contain the key of the corresponding record and the logical address where the record is stored; a three-level index requires four disk accesses; index files are fast for random access, while sequential files are fast for sequential access. |
1,902 | Test | Operating System | File Management | Multiple-choice | Reasoning | English | In a file system, where the FCB occupies 64B and the block size is 1KB, a single-level directory is used. Assuming there are 3200 directory entries in the file directory, the average number of disk accesses required to locate a file is (). | 50 | 54 | 100 | 200 | C | The number of disk blocks occupied by 3200 directory entries is 3200x64B/1KB=200 blocks. Since the average number of disk accesses for a single-level directory is 1/2 the number of blocks (sequentially searching through all the directory entries in the directory table, with each entry being an FCB), the average number ... |
1,903 | Test | Operating System | File Management | Multiple-choice | Reasoning | English | The correct statement about directory indexing is (). | Due to the faster retrieval speed of hashing, modern operating systems have replaced the traditional sequential search method with it. | When using the sequential search method, the file's pathname should be used for the tree directory, and the search should start from the root directory and proceed level by level. | When using the sequential search method, the search should be stopped as soon as one component of the pathname is not found. | After completing the search using the sequential search method, the physical address of the file can be obtained. | C | To implement user access to files by name, the system first uses the filename provided by the user to form a search path and searches the directory. In sequential search, if a component of the pathname is not found, it indicates that a certain directory or file in the pathname does not exist, and there is no need to co... |
1,904 | Test | Operating System | File Management | Multiple-choice | Reasoning | English | Among the following statements, the incorrect one is (). Ⅰ. A duplicate file of the same file on different storage media within the same system should use the same physical structure Ⅱ. Access to a file is often restricted by both user access permissions and user priority Ⅲ. After adopting a tree directory structure fo... | II | I, III | I, III, IV | Select All | D | Files on tape are usually stored sequentially, while this method is not commonly used on hard disks, and random access is used in memory, Statement I is incorrect. Access control to files is often restricted by both user access rights and file attributes, Statement II is incorrect. In a tree directory structure, files ... |
1,905 | Test | Operating System | File Management | Multiple-choice | Reasoning | English | If a bitmap composed of 8 words (word length of 32 bits) is used to manage memory, assuming that a user returns a memory block with a block number of 100, its corresponding position in the bitmap is (). | Font size is 3, bit position is 5. | Font size is 4, bit position is 4. | Font size is 3, bit position is 4. | Font size is 4, bit position is 5. | B | First, determine the row number where block number 100 is located. Blocks 1~32 are in row number 1, 33~64 in row number 2, 65~96 in row number 3, and 97~128 in row number 4. Therefore, block number 100 is in row number 4. Next, determine the column of block number 100 in row number 4. The first column in row number 4 i... |
1,906 | Test | Operating System | Input/Output Management | Multiple-choice | Knowledge | English | The buffer pool in buffering technology is in (). | Main Memory | external storage | Read-Only Memory | Register | A | null |
1,907 | Test | Operating System | Input/Output Management | Multiple-choice | Knowledge | English | In the SPOOLing system, the user process is actually allocated to the (). | The peripherals requested by the user | External storage area, i.e., virtual device. | A portion of the device's storage area | Part of the equipment's space | B | null |
1,908 | Test | Operating System | Input/Output Management | Multiple-choice | Knowledge | English | Virtual devices refer to (). | Allow users to use more devices than the physical devices present in the system. | Allow users to utilize physical devices in a standardized manner. | Transform a physical device into multiple corresponding logical devices. | Allow user programs to use devices in the system without having to load entirely into main memory. | C | null |
1,909 | Test | Operating System | Input/Output Management | Multiple-choice | Knowledge | English | To facilitate the development of higher-level software, device controllers usually need to provide (). | Control registers, status registers, and control commands. | I/O Address Register, Operating Mode Status Register, and Control Commands | Interrupt Register, Control Register, and Control Command | Control registers, programming space, and control logic registers. | A | null |
1,910 | Test | Operating System | Input/Output Management | Multiple-choice | Knowledge | English | In device management, the role of the Device Mapping Table (DMT) is (). | Manage physical devices | Manage Logical Devices | Implement input/output | Establish the correspondence between logical devices and physical devices. | D | null |
1,911 | Test | Operating System | Input/Output Management | Multiple-choice | Knowledge | English | DMA mode establishes a direct data path between (). | I/O devices and main memory | Two I/O devices | I/O devices and CPU | CPU and main memory | A | null |
1,912 | Test | Operating System | Input/Output Management | Multiple-choice | Knowledge | English | Channels, also known as I/O processors, are used to implement information transmission between (). | Memory and Peripherals | CPU and peripherals | Memory and External Storage | CPU and external storage | A | null |
1,913 | Test | Operating System | Input/Output Management | Multiple-choice | Knowledge | English | In operating systems, () refers to a type of hardware mechanism. | Channel Technology | Buffer Pool | SPOOLing technology | Memory Overwriting Technique | A | null |
1,914 | Test | Operating System | Input/Output Management | Multiple-choice | Knowledge | English | In computer systems, what does not belong to the DMA controller is (). | Command/Status Register | Memory Address Register (MAR) | Data Register | Stack Pointer Register | D | null |
1,915 | Test | Operating System | Input/Output Management | Multiple-choice | Knowledge | English | () is used to connect a large number of low-speed or medium-speed I/O devices. | Data Selection Channel | Byte Multiplexing Channel | Data Multiplexing Channel | I/O processor | B | null |
1,916 | Test | Operating System | Input/Output Management | Multiple-choice | Knowledge | English | In the following question, () is not an issue that should be considered in device allocation. | Timeliness | Inherent attributes of the equipment | Device Independence | Safety | A | null |
1,917 | Test | Operating System | Input/Output Management | Multiple-choice | Knowledge | English | The incorrect statement regarding device management is (). | Channels are software that handle input/output. | The startup operations for all devices are uniformly managed by the system. | The I/O interrupt events from the channel are handled by the device manager. | The well-prepared channel program is stored in the main memory. | A | null |
1,918 | Test | Operating System | Input/Output Management | Multiple-choice | Knowledge | English | A computer system is configured with 2 plotters and 3 printers; to properly drive these devices, the system should provide () device drivers. | 5 | 3 | 2 | 1 | C | null |
1,919 | Test | Operating System | Input/Output Management | Multiple-choice | Knowledge | English | The independence of the equipment refers to (). | The device is independent of the computer system. | The system's management of the device is independent. | The device used by the user during programming is independent of the actual device in use. | Each device has a unique identification number. | C | null |
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