ID
int64 1
1.96k
| Split
stringclasses 1
value | Domain
stringclasses 4
values | SubDomain
stringclasses 24
values | Format
stringclasses 1
value | Tag
stringclasses 2
values | Language
stringclasses 1
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
⌀ |
---|---|---|---|---|---|---|---|---|---|---|---|---|---|
1,920 |
Test
|
Operating System
|
Input/Output Management
|
Multiple-choice
|
Knowledge
|
English
|
To enable concurrent processes to perform input and output efficiently, it is best to adopt a () structure buffering technique.
|
Buffer Pool
|
Circular Buffer
|
Single Buffering
|
Double Buffering
|
A
| null |
1,921 |
Test
|
Operating System
|
Input/Output Management
|
Multiple-choice
|
Knowledge
|
English
|
If the time spent on I/O is much shorter than the CPU processing time, then the buffer ( ).
|
Most effective
|
Almost ineffective
|
Equilibrium
|
None of the above answers are correct.
|
B
| null |
1,922 |
Test
|
Operating System
|
Input/Output Management
|
Multiple-choice
|
Knowledge
|
English
|
The key issue to consider in buffer management is ().
|
Select the size of the buffer
|
Determine the number of buffers
|
Implement synchronization for process access to the buffer.
|
Limit the number of processes
|
C
| null |
1,923 |
Test
|
Operating System
|
Input/Output Management
|
Multiple-choice
|
Knowledge
|
English
|
The following () is not part of the device management data structures,
|
PCB
|
DCT
|
COCT
|
CHCT
|
A
| null |
1,924 |
Test
|
Operating System
|
Input/Output Management
|
Multiple-choice
|
Knowledge
|
English
|
The following () is not a method of device allocation.
|
Exclusive Allocation
|
Shared Allocation
|
Virtual Allocation
|
Partition Allocation
|
D
| null |
1,925 |
Test
|
Operating System
|
Input/Output Management
|
Multiple-choice
|
Knowledge
|
English
|
The device among the following that is a shared device is ().
|
printer
|
Tape drive
|
disk
|
Tape drives and disks
|
C
| null |
1,926 |
Test
|
Operating System
|
Input/Output Management
|
Multiple-choice
|
Knowledge
|
English
|
Virtual devices are implemented through the use of () technology.
|
channel
|
Buffering
|
SPOOLing
|
controller
|
C
| null |
1,927 |
Test
|
Operating System
|
Input/Output Management
|
Multiple-choice
|
Knowledge
|
English
|
In systems that utilize SPOOLing technology, the user's print results are first sent to ().
|
Disk Fixed Region
|
Memory Fixed Region
|
Terminal
|
printer
|
A
| null |
1,928 |
Test
|
Operating System
|
Input/Output Management
|
Multiple-choice
|
Knowledge
|
English
|
In a computer system that employs SPOOLing technology, peripheral computers need ().
|
A machine
|
multiple
|
At least one
|
0 units
|
D
| null |
1,929 |
Test
|
Operating System
|
Input/Output Management
|
Multiple-choice
|
Knowledge
|
English
|
The following statement about SPOOLing is incorrect ().
|
SPOOLing systems do not require exclusive access to devices.
|
The SPOOLing system has accelerated the speed of job execution.
|
The SPOOLing system turns exclusive devices into shared devices.
|
The SPOOLing system has improved the utilization of dedicated devices.
|
A
| null |
1,930 |
Test
|
Operating System
|
Input/Output Management
|
Multiple-choice
|
Knowledge
|
English
|
() is a technique used in operating systems to trade space for time.
|
SPOOLing technology
|
Virtual Memory Technology
|
Covering and Switching Techniques
|
Channel Technology
|
A
| null |
1,931 |
Test
|
Operating System
|
Input/Output Management
|
Multiple-choice
|
Knowledge
|
English
|
In systems that utilize SPOOLing technology, the user's print data is first sent to the ().
|
Disk Fixed Region
|
Memory Fixed Region
|
Terminal
|
printer
|
A
| null |
1,932 |
Test
|
Operating System
|
Input/Output Management
|
Multiple-choice
|
Knowledge
|
English
|
The disk is a sharable device, but at each moment, () job starts it.
|
The content can consist of an arbitrary number of
|
can define multiple
|
At least one
|
At most one
|
D
| null |
1,933 |
Test
|
Operating System
|
Input/Output Management
|
Multiple-choice
|
Knowledge
|
English
|
Media that can be accessed both randomly and sequentially include () Ⅰ. CD Ⅱ. Tape Ⅲ. USB flash drive Ⅳ. Disk.
|
II, III, IV
|
I, III, IV
|
III, IV
|
Only IV
|
B
| null |
1,934 |
Test
|
Operating System
|
Input/Output Management
|
Multiple-choice
|
Knowledge
|
English
|
The purpose of disk scheduling is to reduce the () time.
|
Seeking the Way
|
Latency
|
Transmission
|
Start
|
A
| null |
1,935 |
Test
|
Operating System
|
Input/Output Management
|
Multiple-choice
|
Knowledge
|
English
|
Files on the disk are read/written in units of ().
|
block
|
Record
|
cylindrical surface
|
track
|
A
| null |
1,936 |
Test
|
Operating System
|
Input/Output Management
|
Multiple-choice
|
Knowledge
|
English
|
Among the following times for reading data from a disk, the most significant impact is ().
|
Processing Time
|
Latency Time
|
Transmission Time
|
Seeking Time
|
D
| null |
1,937 |
Test
|
Operating System
|
Input/Output Management
|
Multiple-choice
|
Knowledge
|
English
|
The boot sector of the operating system on the hard drive is generated during ().
|
When partitioning a hard drive
|
When performing a low-level format on a hard disk
|
The hard drive comes with factory settings.
|
When performing high-level formatting on a hard disk
|
D
| null |
1,938 |
Test
|
Operating System
|
Input/Output Management
|
Multiple-choice
|
Knowledge
|
English
|
Assuming the disk rotates at a speed of 3000 revolutions per minute, and the disk surface is divided into 10 sectors, then the time to read one sector is ().
|
20ms
|
5ms
|
2ms
|
1ms
|
C
| null |
1,939 |
Test
|
Operating System
|
Input/Output Management
|
Multiple-choice
|
Knowledge
|
English
|
Among the following statements about Solid State Drives (SSD), the incorrect one is ().
|
Flash-based storage technology
|
The random read/write performance is significantly higher than that of a disk.
|
Random writes are relatively slow.
|
Resistant to wear
|
D
| null |
1,940 |
Test
|
Operating System
|
Input/Output Management
|
Multiple-choice
|
Reasoning
|
English
|
The correct statement about the device attributes is ().
|
The fundamental characteristic of character devices is that they are addressable by byte, meaning they can specify the source address for input or the destination address for output.
|
Shared devices must be addressable and capable of random access.
|
Shared devices refer to equipment that allows multiple processes to access simultaneously at the same time.
|
Deadlocks may occur in processes when allocating both shared and exclusive devices.
|
B
|
Addressability is a fundamental characteristic of block devices, so option A is incorrect; shared devices refer to devices that allow multiple processes to access them simultaneously over a period of time, thus option C is incorrect. Allocating shared devices does not lead to process deadlock, so option D is incorrect.
|
1,941 |
Test
|
Operating System
|
Input/Output Management
|
Multiple-choice
|
Reasoning
|
English
|
Regarding the relationship between channels, device controllers, and devices, the correct statement is ().
|
Device controllers and channels can control devices independently.
|
For the same set of input/output commands, the device controller, channel, and device can work in parallel.
|
Channel controller controls device controller, device controller controls device operation.
|
None of the above answers are correct.
|
C
|
The control relationship among the three is hierarchical, with only option C being correct.
|
1,942 |
Test
|
Operating System
|
Input/Output Management
|
Multiple-choice
|
Reasoning
|
English
|
I/O interrupt is a means for the CPU to coordinate with the channel, so an interrupt is generated during ().
|
The CPU executes the "Start I/O" instruction but is rejected by the channel.
|
The channel has received the boot request from the CPU.
|
The channel has completed the execution of the channel program.
|
During the execution of the channel program
|
C
|
When the CPU initiates a channel, regardless of whether the initiation is successful, the channel must respond to the CPU. During the execution of the channel program (used for data transfer), the CPU and the channel operate in parallel. Once the channel completes the execution of the channel program, it sends an I/O interrupt to report to the CPU.
|
1,943 |
Test
|
Operating System
|
Input/Output Management
|
Multiple-choice
|
Reasoning
|
English
|
The work of translating system call parameters into device operation commands is done by ().
|
User-Level I/O
|
Device-independent operating system software
|
Interrupt Handling
|
Device driver
|
B
|
System calls are the general interface provided by the operating system to user programs, which do not change with specific devices. Device drivers, on the other hand, are responsible for executing I/O commands issued by the operating system, and they vary with different devices.
|
1,944 |
Test
|
Operating System
|
Input/Output Management
|
Multiple-choice
|
Reasoning
|
English
|
A typical text print page has 50 lines, with 80 characters per line. Assuming a standard printer can print 6 pages per minute, the time to write a character to the printer's output register is very short and can be considered negligible. If each character printed requires 50μs of interrupt handling time (including all services), when operating this printer using interrupt-driven I/O, the percentage of CPU overhead due to interrupts is ().
|
0.02
|
0.05
|
0.2
|
0.5
|
A
|
This printer prints 50x80x6=24000 characters per minute, which is 400 characters per second. Each character print interrupt requires 50μs of CPU time, so the system overhead for interrupts per second is 400x50μs=20ms. If interrupt-driven I/O is used, the remaining 980ms of CPU time can be used for other processing, with the interrupt overhead accounting for 2% of the CPU. Therefore, it is meaningful to operate this printer using interrupt-driven I/O.
|
1,945 |
Test
|
Operating System
|
Input/Output Management
|
Multiple-choice
|
Reasoning
|
English
|
The primary purpose of introducing a high-speed cache is ().
|
Increase the utilization of the CPU
|
Increase the utilization of I/O devices.
|
Improving the mismatch between CPU and IO device speeds
|
Save Memory
|
C
|
The execution speeds of the CPU and I/O devices are usually unequal, with the former being fast and the latter slow. High-speed buffering technology is used to mitigate the mismatch between the two.
|
1,946 |
Test
|
Operating System
|
Input/Output Management
|
Multiple-choice
|
Reasoning
|
English
|
Assuming the time taken to transfer a block of data from the disk to the buffer is 80μs, the time taken to transfer data from the buffer to the user area is 40μs, and the time taken by the CPU to process a block of data is 30μs. If multiple blocks of data need to be processed and single-buffer transfer of disk data is used, then the total time taken to process one block of data is ().
|
120μs
|
110μs
|
150μs
|
70μs
|
A
|
When using a single buffer for data transfer, the operations on the buffer by the device and the processor are serial. During the i-th disk read operation to transfer data to the buffer, the system can simultaneously read the (i-1)-th data from the user area for computation. These two operations can be parallel, and are serial with the data transfer from the buffer to the user area. Therefore, the total time taken by the system to process a block of data is max(80μs, 30μs) + 40μs = 120μs.
|
1,947 |
Test
|
Operating System
|
Input/Output Management
|
Multiple-choice
|
Reasoning
|
English
|
The operating system uses double buffering to transfer data from the disk. Assuming the time taken to transfer data from the disk to the buffer is T_1, and the time taken to transfer data from the buffer to the user space is T_2 (assuming T_2 is much less than T_1), and the time taken by the CPU to process the data is T_3, then the total time taken by the system to process the data is ().
|
T_1 + T_2 + T_3
|
max(T_2,T_3) + T_1
|
max(T_1, T_3) + T_2
|
max(T_1, T_3)
|
D
|
The total time taken to process the data assumes that buffer 1 has already transferred the data to the user area by default. Then, we discuss the scenarios: if T_3 > T_1, meaning the CPU processes the data block slower than the data transfer, it implies that the I/O device can input continuously. The disk transfers data to the buffer and then to the user area, which can be considered as parallel processing with the CPU data processing. The time spent depends on the maximum time consumed by the CPU, so the total time used by the system is T_3. If T_3 < T_1, meaning the CPU processes data faster than the data transfer, the CPU does not need to wait for the I/O device. The disk transfers data to the buffer, and the transfer of data from the buffer to the user area and the CPU data processing can be considered as parallel operations. Thus, the time spent depends on the time T_1 taken by the disk to transfer data to the buffer. Therefore, the correct choice is D.
|
1,948 |
Test
|
Operating System
|
Input/Output Management
|
Multiple-choice
|
Reasoning
|
English
|
Consider the following I/O operations on a single-user computer, which require the use of buffering techniques (). I. Using a mouse under a graphical user interface II. Tape drive in a multitasking operating system (assuming no device pre-allocation) III. Disk drive containing user files IV. Graphics card directly connected to the bus using memory-mapped I/O.
|
I, III
|
II, IV
|
II, III, IV
|
Select All
|
D
|
When the mouse moves, if a high-priority operation occurs, buffering technology must be used to record the mouse activity, Ⅰ is correct. Due to the different throughput between disk drives and the target or source I/O devices, buffering technology must be adopted, Ⅱ is correct. In order to transfer data from the user job space to the disk or from the disk to the user job space, buffering technology must be used, Ⅲ is correct. For the convenience of accessing multiple graphics and improving performance, buffering technology can be used, especially when displaying the current graphic while retrieving the next one, double buffering technology should be adopted, Ⅳ is correct. In summary, the correct answer to this question is D.
|
1,949 |
Test
|
Operating System
|
Input/Output Management
|
Multiple-choice
|
Reasoning
|
English
|
Using spooling technology, a portion of the disk is used as a common buffer to replace the printer. The user's operations on the printer are actually storage operations on the disk, and the part that replaces the printer is completed by ().
|
Exclusive Device
|
Shared device
|
Virtual Device
|
General physical equipment
|
C
|
Printers are dedicated devices, and by utilizing SPOOLing technology, they can be transformed into virtual devices that can be shared by multiple users.
|
1,950 |
Test
|
Operating System
|
Input/Output Management
|
Multiple-choice
|
Reasoning
|
English
|
The following statement about dedicated devices and shared devices is incorrect ().
|
Printers, scanners, and the like are considered exclusive devices.
|
Static allocation is often used for exclusive devices.
|
Shared devices refer to those that can be used by another job before the current one has vacated, but at any given moment, only one job can use the device.
|
Static allocation is often used for shared devices.
|
D
|
Exclusive devices use static allocation, while shared devices employ dynamic allocation.
|
1,951 |
Test
|
Operating System
|
Input/Output Management
|
Multiple-choice
|
Reasoning
|
English
|
Among the following statements about rotational latency, the incorrect one is ().
|
The magnitude of rotational latency is independent of the disk scheduling algorithm.
|
The magnitude of rotational latency depends on the disk's free space allocation program.
|
The magnitude of rotational latency is related to the physical structure of the file.
|
The processing time for sector data has a significant impact on rotational latency.
|
D
|
Disk scheduling algorithms are designed to reduce seek time. The processing time of sector data mainly affects the transfer time. Options B and C are both related to rotational latency. The physical structure of a file corresponds to the method of disk space allocation, including contiguous allocation, linked allocation, and indexed allocation. In disks with contiguous allocation, the physical addresses of files are consecutive; whereas in disks with linked allocation, the physical addresses of files are non-consecutive, thus both are related to rotational latency.
|
1,952 |
Test
|
Operating System
|
Input/Output Management
|
Multiple-choice
|
Reasoning
|
English
|
Among the following algorithms, the one used for disk scheduling is ().
|
Round Robin Scheduling Algorithm
|
LRU Algorithm
|
Shortest Seek Time First Algorithm
|
Priority-based Preemptive Algorithm
|
C
|
Option A is a process scheduling algorithm; Option B is a page replacement algorithm; Option D can be used for both process scheduling and job scheduling. Only Option C is a disk scheduling algorithm.
|
1,953 |
Test
|
Operating System
|
Input/Output Management
|
Multiple-choice
|
Reasoning
|
English
|
The following algorithm () may experience "starvation" in its execution.
|
Elevator Scheduling
|
Shortest Seek Time First
|
Cyclic Scanning Algorithm
|
First Come, First Served
|
B
|
In the Shortest Seek Time First algorithm, the continuous fulfillment of new disk access requests that are closer to the disk head may lead to the indefinite postponement of requests that are farther away, resulting in a "starvation" phenomenon.
|
1,954 |
Test
|
Operating System
|
Input/Output Management
|
Multiple-choice
|
Reasoning
|
English
|
The rotational speed of a disk is 7200 RPM, with 160 sectors per track and 512B per sector, then ideally, its data transfer rate is ().
|
7200x160KB/s
|
7200KB/s
|
9600KB/s
|
19200KB/s
|
C
|
The disk's rotational speed is 7200 RPM (revolutions per minute) = 120 RPM (revolutions per second), passing through 160 sectors per revolution, with each sector being 512B. Therefore, the data transfer rate = 120 x 160 x 512 / 1024 KB/s = 9600 KB/s.
|
1,955 |
Test
|
Operating System
|
Input/Output Management
|
Multiple-choice
|
Reasoning
|
English
|
Assuming the recording density of the tape is 400 characters/inch (1in=0.0254m), each logical record is 80 characters, the inter-block gap is 0.4 inches, and there are 3000 logical records to be stored, the length of tape required to store these records is (), and the tape utilization is ().
|
1500 inches, 33.3%
|
1500 inches, 43.5%
|
1800 inches, 33.3%
|
1800 inches, 43.5%
|
C
|
The length of magnetic tape occupied by one logical record is 80/400 = 0.2 inches, thus the length of tape required to store 3000 logical records is (0.2+0.4)x3000 = 1800 inches, with a utilization rate of 0.2/(0.2+0.4) = 33.3%.
|
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