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
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stringlengths 21
1.43k
⌀ |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
857
|
Test
|
Computer Organization
|
Input/Output System
|
Multiple-choice
|
Knowledge
|
English
|
In the basic structure of the I/O interface, which part is connected to the system bus, and the data transfer mode can only be parallel transmission?()
|
Data Port
|
Control port
|
Internal Interface
|
External Interface
|
C
| null |
858
|
Test
|
Computer Organization
|
Input/Output System
|
Multiple-choice
|
Knowledge
|
English
|
Which addressing mode in the I/O interface does not require the CPU to set special input/output instructions to access the port?()
|
Unified Addressing Method
|
Independent Addressing Mode
|
Unified Read/Write Method
|
Memory Mapping Method
|
A
| null |
859
|
Test
|
Computer Organization
|
Input/Output System
|
Multiple-choice
|
Knowledge
|
English
|
Which of the functions of the I/O interface is used to temporarily store data between peripherals and the CPU?()
|
Perform data buffering
|
Implement communication and control between the host and peripherals.
|
Transmit control commands and status information.
|
Perform address decoding and device selection.
|
A
| null |
860
|
Test
|
Computer Organization
|
Input/Output System
|
Multiple-choice
|
Knowledge
|
English
|
Among the following statements, the correct one is ( )
|
In a computer, the internal code of a Chinese character occupies 4B in the main memory.
|
The output font code of 16x16 dot matrix occupies 32B in the buffer memory.
|
The output font code of 16x16 dot matrix occupies 16B in the buffer storage area.
|
None of the above statements are correct.
|
B
| null |
861
|
Test
|
Computer Organization
|
Input/Output System
|
Multiple-choice
|
Knowledge
|
English
|
Under unified addressing, the distinction between storage units and I/O devices is made by ( ).
|
Different address codes
|
Different address lines
|
Different control lines
|
Different data lines
|
A
| null |
862
|
Test
|
Computer Organization
|
Input/Output System
|
Multiple-choice
|
Knowledge
|
English
|
Among the following functions, the one that belongs to the I/O interface is ( ).
Ⅰ.Conversion of data format
Ⅱ.Error and status detection during I/O process
Ⅲ.Control and timing of I/O operations
Ⅳ.Communication with hosts and peripherals
|
Ⅰ, Ⅳ
|
Ⅰ, Ⅲ, Ⅳ
|
Ⅰ, Ⅱ, Ⅳ
|
Ⅰ, Ⅱ, Ⅲ, Ⅳ
|
D
| null |
863
|
Test
|
Computer Organization
|
Input/Output System
|
Multiple-choice
|
Knowledge
|
English
|
Among the following statements, the correct one is ( )
|
Only I/O instructions can access I/O devices.
|
Under unified addressing, I/O devices cannot be accessed directly.
|
Instructions for accessing memory must not access I/O devices.
|
Only in computers with specialized IO instructions can I/O devices be addressed independently.
|
D
| null |
864
|
Test
|
Computer Organization
|
Input/Output System
|
Multiple-choice
|
Knowledge
|
English
|
The disk drive adopts the ( ) method to write data to the disk platter tracks.
|
parallel
|
Serial
|
Parallel-Serial
|
Serial-Parallel
|
B
| null |
865
|
Test
|
Computer Organization
|
Input/Output System
|
Multiple-choice
|
Knowledge
|
English
|
Programmers making system calls to access devices use the ().
|
Logical address
|
Physical address
|
Primary device address
|
From the device address
|
A
| null |
866
|
Test
|
Computer Organization
|
Input/Output System
|
Multiple-choice
|
Knowledge
|
English
|
In a computer system equipped with channels, the interrupt caused when a user program requires input/output is ( ).
|
Visit Control Interrupt
|
I/O interrupt
|
Procedural Interruption
|
External Interrupt
|
A
| null |
867
|
Test
|
Computer Organization
|
Input/Output System
|
Multiple-choice
|
Knowledge
|
English
|
The following does not belong to program control instructions ()
|
Unconditional transfer refers to today
|
Conditional branch instruction
|
Interrupt Implicit Instruction
|
Loop instruction
|
C
| null |
868
|
Test
|
Computer Organization
|
Input/Output System
|
Multiple-choice
|
Knowledge
|
English
|
In the information transmission between the host and peripherals, ( ) is not a program-controlled method.
|
Direct Numerical Control (DNC)
|
Program Interruption
|
Direct Memory Access (DMA)
|
Channel Control
|
C
| null |
869
|
Test
|
Computer Organization
|
Input/Output System
|
Multiple-choice
|
Knowledge
|
English
|
When an interrupt occurs, the protection and updating of the program counter content are accomplished by ( ).
|
Hardware Automation
|
Push instruction and transfer instruction.
|
Memory Access Instruction
|
Interrupt Service Routine
|
A
| null |
870
|
Test
|
Computer Organization
|
Input/Output System
|
Multiple-choice
|
Knowledge
|
English
|
In the process of transferring data in DMA mode, since the content of the ( ) is not destroyed, the CPU can work normally (except for memory access).
|
Program Counter
|
Program Counter and Registers
|
Instruction Register
|
Stack register
|
B
| null |
871
|
Test
|
Computer Organization
|
Input/Output System
|
Multiple-choice
|
Reasoning
|
English
|
Among the following statements about I/O devices, the correct one is ( ). I. Keyboard, mouse, monitor, and printer are human-computer interaction devices II. In microcomputers, VGA represents a video transmission standard III. Printers can be classified from the typing principle perspective into dot matrix printers and character printers IV. The mouse is suitable for implementing input operations through the interrupt method.
|
II, III, IV
|
I, II, IV
|
I, II, III
|
I, II, III, IV
|
B
|
Only I, II, and IV are consistent with the actual situation.
|
872
|
Test
|
Computer Organization
|
Input/Output System
|
Multiple-choice
|
Reasoning
|
English
|
The resolution of the CRT is 1024x1024 pixels, and the number of colors per pixel is 256, then the word length per cell of the refresh memory is ( ), and the total capacity is ( ).
|
8B, 256MB
|
8bit, 1MB
|
8bit, 256KB
|
8B, 32MB
|
B
|
The refreshable memory has a word length of 8 bits per cell, with a total capacity of 1024x1024x8 bits = 1MB.
|
873
|
Test
|
Computer Organization
|
Input/Output System
|
Multiple-choice
|
Reasoning
|
English
|
The logic for interrupt priority determination is similar to the bus arbitration method, and the error statement among the following is ( ).
|
In the bus arbitration method, the independent request mode has the fastest response time, which comes at the cost of increasing the number of control lines.
|
In the bus arbitration method, the counter timing inquiry method uses a Bus Request (BR) line and a device address line. If the count starts from 0 each time, devices with smaller numbers have higher priority.
|
Bus arbitration generally refers to the method by which I/O devices contend for priority access to the bus, while interrupt priority refers to the method by which I/O devices contend for priority access to the CPU.
|
Interrupt priority logic can be implemented either through hardware or software.
|
B
|
The independent request method has the fastest response time in bus arbitration, while the counter timing inquiry method is used to determine the priority of devices in bus arbitration. Bus arbitration generally involves I/O devices contending for the bus, whereas interrupt priority resolution involves I/O devices contending for CPU priority. Interrupt priority logic can be implemented through hardware or software.
|
874
|
Test
|
Computer Organization
|
Input/Output System
|
Multiple-choice
|
Reasoning
|
English
|
The following statement is incorrect ( ).
|
Interrupt service routines are generally modules of the operating system.
|
The interrupt vector method can improve the identification speed of interrupt sources.
|
The interrupt vector address is the entry address of the interrupt service routine.
|
The phenomenon of handling interrupts in an overlapping manner is called interrupt nesting.
|
C
|
The interrupt vector address is not the entry address of the interrupt service routine, but the address of the entry address of the interrupt service routine.
|
875
|
Test
|
Computer Organization
|
Input/Output System
|
Multiple-choice
|
Reasoning
|
English
|
The incorrect statement about program interrupt mode and DMA mode is ( )I. The priority of DMA is higher than that of program interrupts II. Program interrupt mode requires context saving, while DMA mode does not require context saving III. The interrupt request in program interrupt mode is to report the end of data transfer to the CPU, whereas the interrupt request in DMA mode is solely for data transfer.
|
Only II
|
II, III
|
Only III
|
I, III
|
C
|
DMA (Direct Memory Access) mode does not require CPU pre-transfer operations, only borrowing a little CPU time at the beginning and end, without occupying any other CPU resources. Interrupt mode involves program switching, where each operation requires saving and restoring the context, so DMA has a higher priority than interrupt requests, thereby speeding up processing efficiency. Therefore, statement I is correct. From the analysis, it is known that program interrupts require interrupting the current program, hence the need to save the context so that after the interrupt is completed, it can return to the original point to continue the unfinished work. DMA mode does not need to interrupt the current program and does not require saving the context, so statement I is correct. Statement II is just the opposite.
|
876
|
Test
|
Computer Organization
|
Input/Output System
|
Multiple-choice
|
Reasoning
|
English
|
Among the following statements, the correct one is ( ).
|
The process of program interruption is accomplished jointly by hardware and the interrupt service routine (ISR).
|
During the execution of each instruction, the bus cycle must be checked once for any interrupt requests.
|
Check for DMA requests, typically arranged at the end of an instruction execution process.
|
The last instruction of the interrupt service routine is an unconditional branch instruction.
|
A
|
The process of program interruption is accomplished jointly by hardware and the interrupt service routine. During the execution of each instruction, interrupt requests are not checked during every bus cycle. DMA requests are typically not checked at the end of the instruction execution process. The last instruction of the interrupt service routine is usually an interrupt return instruction, rather than an unconditional branch instruction.
|
877
|
Test
|
Computer Organization
|
Input/Output System
|
Multiple-choice
|
Reasoning
|
English
|
Interrupt responses should be prioritized from high to low using ( ).
|
Access Control → Procedural → Machine Failure
|
Access Control → Procedural → Reboot
|
External → Access Control → Procedural
|
Procedural → I/O → Supervisor Call
|
B
|
The priority order of interrupt responses from high to low is supervisor call → program → restart.
|
878
|
Test
|
Computer Organization
|
Input/Output System
|
Multiple-choice
|
Reasoning
|
English
|
In the following situations, an interrupt request may not occur is ( ).
|
DMA operation completed
|
An instruction has been executed.
|
The machine has malfunctioned.
|
Execute the "soft interrupt" instruction.
|
B
|
The completion of an instruction will not trigger an interrupt request.
|
879
|
Test
|
Computer Organization
|
Input/Output System
|
Multiple-choice
|
Reasoning
|
English
|
The computer has 4 levels of interrupts, with priorities from high to low being 1-2-3-4. If the priority order is changed, with the interrupt mask for level 1 being 1101, level 2 being 0100, level 3 being 1111, and level 4 being 0101, then the modified priority order from high to low is ( ).
|
1-2-3-4
|
3-1-4-2
|
1-3-4-2
|
2-1-3-4
|
B
|
The modified priority order is 3-1-4-2, which means the priorities from highest to lowest are 3, 1, 4, 2, respectively.
|
880
|
Test
|
Computer Organization
|
Input/Output System
|
Multiple-choice
|
Reasoning
|
English
|
In the interrupt response cycle, the primary task accomplished by the CPU is ().
|
Enable interrupts, maintain breakpoints, send interrupt response signals and generate vector addresses.
|
Disable interrupts, save breakpoints, issue interrupt response signals, and generate vector addresses.
|
Upon interrupt, execute the interrupt service routine (ISR).
|
Enable interrupts, execute the interrupt service routine (ISR).
|
B
|
During the interrupt response cycle, the CPU primarily accomplishes the tasks of disabling interrupts, saving the context, sending an interrupt acknowledgment signal, and generating a vector address.
|
881
|
Test
|
Computer Organization
|
Input/Output System
|
Multiple-choice
|
Reasoning
|
English
|
The first step that the CPU completes when responding to an interrupt is ().
|
Enable Interrupts
|
Preserve breakpoints
|
Interrupt Handling
|
Enter the interrupt service routine.
|
C
|
The first step that the CPU completes when responding to an interrupt is to disable interrupts.
|
882
|
Test
|
Computer Organization
|
Input/Output System
|
Multiple-choice
|
Reasoning
|
English
|
Setting the interrupt mask flag can change ().
|
Priority of Interrupt Requests from Multiple Interrupt Sources
|
The priority order in which the CPU responds to multiple interrupt requests.
|
The order in which multiple interrupt service routines start execution
|
The order in which multiple interrupt service routines are executed
|
D
|
Setting the interrupt mask flag can change the order in which multiple interrupt service routines are executed.
|
883
|
Test
|
Computer Organization
|
Input/Output System
|
Multiple-choice
|
Reasoning
|
English
|
Among the following statements, ( ) is correct.
|
Both the program interrupt method and the DMA (Direct Memory Access) method require an interrupt request to implement data transfer.
|
DMA requests, non-maskable interrupts, and maskable interrupts can all be responded to only after the current instruction has finished.
|
Both the interrupt-driven method and DMA (Direct Memory Access) involve interrupt requests, but they serve different purposes.
|
DMA must wait until the current cycle ends before it can perform cycle stealing.
|
C
|
Both the interrupt-driven method and DMA (Direct Memory Access) involve interrupt requests, but their purposes differ.
|
884
|
Test
|
Computer Organization
|
Input/Output System
|
Multiple-choice
|
Reasoning
|
English
|
The following description about performing IO in DMA mode is correct ( ).
|
A complete DMA process is partially controlled by the DMA controller and partially by the CPU.
|
A complete DMA process, fully controlled by the CPU.
|
A complete DMA process is entirely controlled by the DMA controller, with no CPU intervention in any control.
|
A complete DMA process is fully controlled by the CPU using cycle stealing.
|
A
|
The DMA process is primarily controlled by the DMA controller, but requires CPU involvement, especially at the start and end of data transfer.
|
885
|
Test
|
Computer Organization
|
Input/Output System
|
Multiple-choice
|
Reasoning
|
English
|
Regarding external interrupts (excluding faults) and DMA, the correct statement among the following is ( ).
|
When a DMA request and an interrupt request occur simultaneously, the DMA request is responded to.
|
DMA requests, non-maskable interrupts, and maskable interrupts can only be responded to after the current instruction has finished.
|
Non-maskable interrupt requests have the highest priority, while maskable interrupt requests have the lowest priority.
|
If interrupts are not enabled, all interrupt requests cannot be responded to.
|
A
|
DMA requests have a higher priority than interrupt requests, so when DMA and interrupt requests occur simultaneously, the CPU will respond to the DMA request.
|
886
|
Test
|
Computer Organization
|
Input/Output System
|
Multiple-choice
|
Reasoning
|
English
|
The following statement about DMA mode is incorrect ( ).
|
In DMA mode, the DMA controller requests bus access from the CPU.
|
DMA mode can be used for data input from keyboards and mice.
|
During the data transfer phase, there is no need for CPU intervention; it is entirely controlled by the DMA controller.
|
DMA mode requires the use of interrupt handling.
|
B
|
DMA mode is generally not used for data input from keyboards and mice, as these devices require immediate CPU response and cannot wait for the DMA controller to control data transfer.
|
1,100
|
Test
|
Computer Network
|
Overview and Architecture
|
Multiple-choice
|
Knowledge
|
English
|
How many bytes do the source and destination addresses each occupy in a MAC frame?
|
4B, 4B
|
6B, 6B
|
8B, 8B
|
10B, 10B
|
B
| null |
1,101
|
Test
|
Computer Network
|
Overview and Architecture
|
Multiple-choice
|
Knowledge
|
English
|
What does the total length of an IP datagram refer to?
|
Length of the data section
|
Length of the header
|
Length of the header plus data section
|
Length of the header and tail
|
C
| null |
1,102
|
Test
|
Computer Network
|
Overview and Architecture
|
Multiple-choice
|
Knowledge
|
English
|
What is the reason for IP datagram fragmentation?
|
Header length too long
|
Data Link Layer MTU Limitations
|
The data segment is too small.
|
Destination Host Restriction
|
B
| null |
1,103
|
Test
|
Computer Network
|
Overview and Architecture
|
Multiple-choice
|
Knowledge
|
English
|
Given that the data segment is 3800B and the maximum fragment size is 1420B, what is the fragment offset of the second fragment?
|
175
|
200
|
225
|
250
|
A
| null |
1,104
|
Test
|
Computer Network
|
Overview and Architecture
|
Multiple-choice
|
Knowledge
|
English
|
What are bytes 5-12 of a TCP message segment?
|
Source Address and Destination Address
|
Serial number and confirmation number
|
Window Size and Checksum
|
Source port and destination port
|
B
| null |
1,105
|
Test
|
Computer Network
|
Overview and Architecture
|
Multiple-choice
|
Knowledge
|
English
|
Which of the following descriptions is correct?
|
Gateways operate at the network layer.
|
Switches operate at the application layer.
|
Routers operate at the data link layer.
|
Hubs operate at the transport layer.
|
A
| null |
1,106
|
Test
|
Computer Network
|
Overview and Architecture
|
Multiple-choice
|
Knowledge
|
English
|
What are the dynamic routing protocols in the routing table?
|
TCP, UDP
|
RIP, OSPF
|
ARP, ICMP
|
HTTP, FTP
|
B
| null |
1,107
|
Test
|
Computer Network
|
Overview and Architecture
|
Multiple-choice
|
Knowledge
|
English
|
What is the layer number of the transport layer in end-to-end communication?
|
The first layer
|
Second Layer
|
The third layer
|
Layer 4
|
D
| null |
1,108
|
Test
|
Computer Network
|
Overview and Architecture
|
Multiple-choice
|
Knowledge
|
English
|
What are the service characteristics of an IP datagram?
|
Connected, reliable delivery
|
Connectionless, best-effort delivery
|
Reliable but with high latency
|
Unreliable but with lower latency
|
B
| null |
1,109
|
Test
|
Computer Network
|
Overview and Architecture
|
Multiple-choice
|
Knowledge
|
English
|
If the data part is 3800B and the maximum fragment size is 1420B, what is the total length in bytes of the third fragment?
|
1020B
|
1120B
|
1220B
|
1320B
|
A
| null |
1,110
|
Test
|
Computer Network
|
Overview and Architecture
|
Multiple-choice
|
Knowledge
|
English
|
How many bytes do the sequence number and acknowledgment number each occupy in a TCP segment?
|
2B, 6B
|
4B, 4B
|
6B, 2B
|
8B, 8B
|
B
| null |
1,111
|
Test
|
Computer Network
|
Overview and Architecture
|
Multiple-choice
|
Knowledge
|
English
|
What happens when a switch forwards data and the destination MAC address is not in the forwarding table? ()
|
Discard data
|
Broadcast new destination address
|
Forward to all interfaces
|
Send an ARP request
|
C
| null |
1,112
|
Test
|
Computer Network
|
Overview and Architecture
|
Multiple-choice
|
Knowledge
|
English
|
What is the function of the ARP table? ()
|
IP-Port Mapping
|
IP-MAC mapping
|
MAC-Port Mapping
|
MAC-IP mapping
|
B
| null |
1,113
|
Test
|
Computer Network
|
Overview and Architecture
|
Multiple-choice
|
Knowledge
|
English
|
What are the protocols of the transport layer? ()
|
ARP, ICMP, IP, OSPF
|
PPP, HDLC, CSMA
|
TCP, UDP
|
DHCP, FTP, HTTP
|
C
| null |
1,114
|
Test
|
Computer Network
|
Overview and Architecture
|
Multiple-choice
|
Knowledge
|
English
|
What is the first layer of end-to-end communication? ()
|
Data Link Layer
|
Network Layer
|
Transport Layer
|
Application Layer
|
C
| null |
1,115
|
Test
|
Computer Network
|
Overview and Architecture
|
Multiple-choice
|
Knowledge
|
English
|
What is a broadcast domain? ()
|
The area between the sender and the receiver
|
Number of stations that can receive the signal
|
Broadcast range
|
Scope of the Transport Layer
|
C
| null |
1,116
|
Test
|
Computer Network
|
Overview and Architecture
|
Multiple-choice
|
Knowledge
|
English
|
What is the range of the total length of a MAC frame in bytes? ()
|
64B - 1518B
|
20B - 100B
|
46B - 1500B
|
18B - 500B
|
A
| null |
1,117
|
Test
|
Computer Network
|
Overview and Architecture
|
Multiple-choice
|
Knowledge
|
English
|
In IP datagram fragmentation, if MF is 0, what does it indicate? ()
|
Not the last fragment
|
The last fragment
|
Allow Fragmentation
|
Prohibit Fragmentation
|
B
| null |
1,118
|
Test
|
Computer Network
|
Overview and Architecture
|
Multiple-choice
|
Knowledge
|
English
|
How many bytes does the window size field of a TCP segment occupy? ()
|
2B
|
4B
|
6B
|
8B
|
A
| null |
1,119
|
Test
|
Computer Network
|
Overview and Architecture
|
Multiple-choice
|
Knowledge
|
English
|
What is the primary operational layer of a router?
|
Data Link Layer
|
Network Layer
|
Transport Layer
|
Application Layer
|
B
| null |
1,120
|
Test
|
Computer Network
|
Overview and Architecture
|
Multiple-choice
|
Knowledge
|
English
|
Computer networks can be understood as ( ).
|
Software module for executing computer data processing
|
A collective interconnected by autonomous computers
|
A tightly-coupled system implemented by multiple processors through shared memory.
|
Distributed system for collaboratively accomplishing a task
|
B
| null |
1,121
|
Test
|
Computer Network
|
Overview and Architecture
|
Multiple-choice
|
Knowledge
|
English
|
The most basic function of computer networks is ( ).
|
Data Communication
|
Resource Sharing
|
Distributed Processing
|
Information Synthesis Processing
|
A
| null |
1,122
|
Test
|
Computer Network
|
Overview and Architecture
|
Multiple-choice
|
Knowledge
|
English
|
The basic components of a computer network system are ().
|
Local Area Network (LAN) and Wide Area Network (WAN)
|
Local computer network and communication network
|
Communication Subnet and Resource Subnet
|
Servers and Workstations
|
C
| null |
1,123
|
Test
|
Computer Network
|
Overview and Architecture
|
Multiple-choice
|
Knowledge
|
English
|
In computer networks, what can be absent is ( ).
|
Client machine
|
server
|
Operating System
|
Database Management System (DBMS)
|
D
| null |
1,124
|
Test
|
Computer Network
|
Overview and Architecture
|
Multiple-choice
|
Knowledge
|
English
|
The primary resources of computer networks mainly refer to ( ).
|
Servers, routers, communication lines, and user computers
|
Computer operating systems, databases, and application software
|
Computer hardware, software, and data
|
Web server, database server, and file server
|
C
| null |
1,125
|
Test
|
Computer Network
|
Overview and Architecture
|
Multiple-choice
|
Knowledge
|
English
|
The differences between Local Area Network (LAN) and Wide Area Network (WAN) lie not only in the different scopes they cover but also primarily in their ()
|
The medium used is different.
|
The protocol used is different.
|
The supported communication capacity varies.
|
The services provided differ.
|
B
| null |
1,126
|
Test
|
Computer Network
|
Overview and Architecture
|
Multiple-choice
|
Knowledge
|
English
|
A large number of computers are now connected to wide area networks through local area networks such as Ethernet, and the interconnection between local area networks and wide area networks is achieved through ().
|
Router
|
Resource Subnet
|
Bridge Device
|
Repeater
|
A
| null |
1,127
|
Test
|
Computer Network
|
Overview and Architecture
|
Multiple-choice
|
Knowledge
|
English
|
The following () are disadvantages of packet-switched networks.
|
Low channel utilization
|
Additional information incurs high overhead.
|
Propagation delay is large.
|
Terminals of different specifications have difficulty communicating with each other.
|
B
| null |
1,128
|
Test
|
Computer Network
|
Overview and Architecture
|
Multiple-choice
|
Knowledge
|
English
|
( ) is not the objective of layering network models.
|
Provide standard language
|
Definition of the method for function execution
|
Define the standard interface
|
Increase the independence between functions
|
B
| null |
1,129
|
Test
|
Computer Network
|
Overview and Architecture
|
Multiple-choice
|
Knowledge
|
English
|
The advantages of dividing user data into blocks for transmission do not include ( ).
|
Reduce latency time
|
Improve error control efficiency
|
Enable multiple applications to use a shared communication medium more fairly.
|
The proportion of effective data in the Protocol Data Unit (PDU) is larger.
|
D
| null |
1,130
|
Test
|
Computer Network
|
Overview and Architecture
|
Multiple-choice
|
Knowledge
|
English
|
A protocol refers to the rules or agreements for communication between ().
|
The upper and lower layers of the same node
|
Different nodes for peer entities
|
Adjacent entities
|
Different nodes
|
D
| null |
1,131
|
Test
|
Computer Network
|
Overview and Architecture
|
Multiple-choice
|
Knowledge
|
English
|
Regarding computer networks and their structural models, the following statements are incorrect ( )
|
The world's first computer network was ARPAnet.
|
The Internet originated from ARPAnet.
|
The International Organization for Standardization (ISO) has developed the OSI/RM reference model, which is the standard implemented in practice.
|
The TCP/IP reference model is divided into 4 layers.
|
C
| null |
1,132
|
Test
|
Computer Network
|
Overview and Architecture
|
Multiple-choice
|
Knowledge
|
English
|
( ) are the three main concepts of the OSI reference model in computer networks.
|
Services, Interfaces, Protocols
|
Structure, model, exchange
|
Subnet, hierarchy, port
|
Wide Area Network (WAN), Metropolitan Area Network (MAN), Local Area Network (LAN)
|
A
| null |
1,133
|
Test
|
Computer Network
|
Overview and Architecture
|
Multiple-choice
|
Knowledge
|
English
|
When data is transmitted from end system A to end system B, the entity that does not participate in data encapsulation is ( ).
|
Physical Layer
|
Data Link Layer
|
Network Layer
|
Presentation Layer
|
A
| null |
1,134
|
Test
|
Computer Network
|
Overview and Architecture
|
Multiple-choice
|
Reasoning
|
English
|
Among the following statements about broadcast networks, the incorrect one is ( )
|
Shared Broadcast Channel
|
There is no route selection problem.
|
You can do without the network layer.
|
No need for service access point
|
D
|
Broadcast networks share a broadcast channel (such as a bus), which is commonly a communication method for local area networks (LANs operating at the data link layer). Therefore, the network layer is not required, and consequently, there is no issue of routing selection. However, the data link layer must access the services of the physical layer through Service Access Points (SAPs).
|
1,135
|
Test
|
Computer Network
|
Overview and Architecture
|
Multiple-choice
|
Reasoning
|
English
|
Which of the following best describes the function of the Data Link Layer of the OSI reference model? ( )
|
Provide an interface for users and the network.
|
Signal processing through medium transmission
|
Control messages are routed through the network.
|
Ensure the correct sequence and integrity of data.
|
D
|
The functions of the data link layer include establishing, dismantling, separating, framing, and synchronizing link connections, as well as error detection, etc. A represents the functions of the application layer, B represents the functions of the physical layer, C represents the functions of the network layer, and D is indeed the functions of the data link layer.
|
1,136
|
Test
|
Computer Network
|
Overview and Architecture
|
Multiple-choice
|
Reasoning
|
English
|
Among the following statements, the one that correctly describes the data encapsulation process in the OSI reference model is ().
|
The data link layer only adds the source physical address and destination physical address to the packet.
|
The network layer encapsulates the data generated by higher-level protocols into packets and adds third-layer addressing and control information.
|
The transport layer encapsulates the data stream into frames and adds reliability and flow control information.
|
The presentation layer segments the data generated by higher-level protocols into data segments and adds the corresponding source and destination port information.
|
B
|
The Data Link Layer, in addition to adding source and destination physical addresses to packets, also adds control information. The PDU of the Transport Layer is not called a frame; the Presentation Layer is not responsible for dividing the data generated by higher-level protocols into segments, and it is the Transport Layer that is responsible for adding the corresponding source and destination port information. Option B correctly describes the data encapsulation process in the OSI reference model; after passing through the Network Layer, only the third layer PCI is added.
|
1,137
|
Test
|
Computer Network
|
Physical Layer
|
Multiple-choice
|
Knowledge
|
English
|
What are the advantages of using one-hot encoding? ()
|
Time base, high 1 low 0
|
Time base, high 0 low 1
|
High 1, low 0, without considering time.
|
High 0 Low 1, without considering time
|
A
| null |
1,138
|
Test
|
Computer Network
|
Physical Layer
|
Multiple-choice
|
Knowledge
|
English
|
What are the characteristics of Manchester encoding? ()
|
The first half is high and the second half is low, with a value of 1.
|
The first half is low, and the second half is high, with a value of 1.
|
High 1 Low 0 Alternating
|
High 0 Low 1 Alternating
|
A
| null |
1,139
|
Test
|
Computer Network
|
Physical Layer
|
Multiple-choice
|
Knowledge
|
English
|
What is the definition of baud rate? ()
|
How many bits are transmitted per second?
|
How many bauds are transmitted per second?
|
How many bytes are transmitted per second
|
How many frames are transmitted per second
|
A
| null |
1,140
|
Test
|
Computer Network
|
Physical Layer
|
Multiple-choice
|
Knowledge
|
English
|
What are the characteristics of circuit switching? ()
|
Dedicated physical communication channel
|
No need to establish a dedicated line.
|
The transmission unit is packet.
|
The transmission unit is a frame.
|
A
| null |
1,141
|
Test
|
Computer Network
|
Physical Layer
|
Multiple-choice
|
Knowledge
|
English
|
What is the difference between packet switching and message switching? ()
|
Size of the transmission unit
|
Is there a need to establish a dedicated line?
|
Transmission rate
|
Is there noise?
|
A
| null |
1,142
|
Test
|
Computer Network
|
Physical Layer
|
Multiple-choice
|
Knowledge
|
English
|
In computer networks,what does the voltage range refer to in electrical characteristics? ()
|
Transmission rate
|
Maximum current value
|
Minimum current value
|
The range of voltage variation
|
D
| null |
1,143
|
Test
|
Computer Network
|
Physical Layer
|
Multiple-choice
|
Knowledge
|
English
|
What is the main function of a repeater? ()
|
Digital Regeneration
|
Data Storage
|
Signal Amplification
|
Data Filtering
|
C
| null |
1,144
|
Test
|
Computer Network
|
Physical Layer
|
Multiple-choice
|
Knowledge
|
English
|
What is the relationship between baud rate and bit rate? ()
|
Bit rate = baud rate X number of binary bits corresponding to a single modulation state
|
Irrelevant
|
The baud rate is twice the bit rate.
|
The baud rate is half of the bit rate.
|
A
| null |
1,145
|
Test
|
Computer Network
|
Physical Layer
|
Multiple-choice
|
Knowledge
|
English
|
What is the unit of the signal-to-noise ratio in Shannon's theorem? ()
|
dB
|
Hz
|
b/s
|
B/s
|
A
| null |
1,146
|
Test
|
Computer Network
|
Physical Layer
|
Multiple-choice
|
Knowledge
|
English
|
What does store-and-forward refer to in message switching? ()
|
First store the entire message, then send it.
|
Store-and-Forward
|
First send the header, then send the data.
|
Do not store, directly send.
|
A
| null |
1,147
|
Test
|
Computer Network
|
Physical Layer
|
Multiple-choice
|
Knowledge
|
English
|
What does process characteristic focus on? ()
|
Voltage range
|
Timing and Work Procedures
|
Physical connection characteristics
|
Transmission rate
|
B
| null |
1,148
|
Test
|
Computer Network
|
Physical Layer
|
Multiple-choice
|
Knowledge
|
English
|
Repeaters must ensure that both segments of the network use the same what? ()
|
protocol
|
Physical Layer
|
Data Link Layer
|
Transmission rate
|
B
| null |
1,149
|
Test
|
Computer Network
|
Physical Layer
|
Multiple-choice
|
Knowledge
|
English
|
What does bit rate refer to? ()
|
Baud rate per second
|
Number of transmitted frames
|
Bytes transmitted per second
|
Number of transmitted bits
|
D
| null |
1,150
|
Test
|
Computer Network
|
Physical Layer
|
Multiple-choice
|
Knowledge
|
English
|
Encoding without synchronization information is ( ).
|
Non-return-to-zero encoding
|
Manchester encoding
|
Differential Manchester Encoding
|
Non-return-to-zero coding, Differential Manchester encoding
|
A
| null |
1,151
|
Test
|
Computer Network
|
Physical Layer
|
Multiple-choice
|
Knowledge
|
English
|
In the network, to transmit voice and computer-generated numbers, text, graphics, and images simultaneously, the voice signal must be digitized first. The following technology can digitize voice signals is ( ).
|
Manchester encoding
|
QAM
|
Differential Manchester Encoding
|
Pulse Code Modulation (PCM)
|
D
| null |
1,152
|
Test
|
Computer Network
|
Physical Layer
|
Multiple-choice
|
Knowledge
|
English
|
The measured data baud rate of an Ethernet is 40MBaud, so its data rate is ( ).
|
10Mb/s
|
20Mb/s
|
40Mb/s
|
80Mb/s
|
B
| null |
1,153
|
Test
|
Computer Network
|
Physical Layer
|
Multiple-choice
|
Knowledge
|
English
|
Among the following data exchange methods, the one where data experiences long and variable network transmission delays, and cannot be used for voice data transmission is ( ).
|
Circuit Switching
|
Message Switching
|
Datagram Switching
|
Virtual Circuit Switching
|
B
| null |
1,154
|
Test
|
Computer Network
|
Physical Layer
|
Multiple-choice
|
Knowledge
|
English
|
The main benefit of using shielding technology in cables is ( ).
|
Reduce signal attenuation
|
Reduce electromagnetic interference radiation
|
Reduce physical damage
|
Reduce the impedance of the cable
|
B
| null |
1,155
|
Test
|
Computer Network
|
Physical Layer
|
Multiple-choice
|
Knowledge
|
English
|
The principle of multimode fiber optic transmission of light signals is ( ).
|
Refractive properties of light
|
Emission characteristics of light
|
Total internal reflection of light
|
Diffraction characteristics of light
|
C
| null |
1,156
|
Test
|
Computer Network
|
Physical Layer
|
Multiple-choice
|
Knowledge
|
English
|
The following statement about single-mode fiber is correct ( )
|
The thicker the optical fiber, the higher the data transmission rate.
|
If the diameter of the optical fiber is reduced to the size of just one wavelength of light, then the light will propagate in a straight line.
|
The light source is a light-emitting diode (LED) or laser.
|
Fiber optics are hollow.
|
B
| null |
1,157
|
Test
|
Computer Network
|
Physical Layer
|
Multiple-choice
|
Knowledge
|
English
|
The following statement about satellite communication is incorrect ( )
|
Satellite communication features long distances and wide coverage areas.
|
Satellite communication facilitates the implementation of broadcast communication and multiple-address communication.
|
The advantage of satellite communication is that it is not affected by weather conditions and has a very low bit error rate.
|
High communication costs and significant latency are the shortcomings of satellite communication.
|
C
| null |
1,158
|
Test
|
Computer Network
|
Physical Layer
|
Multiple-choice
|
Knowledge
|
English
|
The network specifies at the physical layer that a signal level of +10V to +15V represents binary 0, and -10V to -15V represents binary 1, with a cable length limited to within 15m, which reflects the ( ) of the physical layer interface.
|
Mechanical properties
|
Functional Characteristics
|
Electrical Characteristics
|
Procedure Characteristics
|
C
| null |
1,159
|
Test
|
Computer Network
|
Physical Layer
|
Multiple-choice
|
Knowledge
|
English
|
The function of a repeater is ().
|
Amplify signal
|
Forwarding Frame
|
Storage Frame
|
Addressing
|
A
| null |
1,160
|
Test
|
Computer Network
|
Physical Layer
|
Multiple-choice
|
Knowledge
|
English
|
When two network segments are interconnected at the physical layer, it is required that ().
|
The data transfer rate and the data link layer protocols can vary.
|
The data transfer rate and the data link layer protocol must be the same.
|
The data transfer rate must be consistent, but the data link layer protocols can differ.
|
The data transfer rate can vary, but the data link layer protocol must be the same.
|
C
| null |
1,161
|
Test
|
Computer Network
|
Physical Layer
|
Multiple-choice
|
Knowledge
|
English
|
Ethernet follows the IEEE 802.3 standard. When networking with thick coaxial cable, the length of each segment must not exceed 500m. Segments exceeding 500m must be divided, and the segments are connected using ( ).
|
Network Adapter
|
Repeater
|
modem
|
Gateway
|
B
| null |
1,162
|
Test
|
Computer Network
|
Physical Layer
|
Multiple-choice
|
Knowledge
|
English
|
A collection of workstations connected by a hub ( )
|
Belong to the same collision domain and also belong to the same broadcast domain.
|
Belong to different collision domains, but the same broadcast domain.
|
Belong to different collision domains and also belong to different broadcast domains.
|
Belong to the same collision domain, but not to the same broadcast domain.
|
A
| null |
1,163
|
Test
|
Computer Network
|
Physical Layer
|
Multiple-choice
|
Knowledge
|
English
|
If 5 computers are connected to a 10Mb/s hub, the average bandwidth allocated to each computer is ( ).
|
2Mb/s
|
5Mb/s
|
10Mb/s
|
50Mb/s
|
A
| null |
1,164
|
Test
|
Computer Network
|
Physical Layer
|
Multiple-choice
|
Reasoning
|
English
|
The following statement about Manchester encoding is correct ( )
|
The starting edge of each signal serves as a clock signal, which is beneficial for synchronization.
|
Incorporate both the clock and data values into the signal.
|
This encoding mechanism for analog signals is particularly suitable for transmitting sound.
|
A transition in the middle of each bit indicates that the signal value is 0.
|
B
|
Manchester encoding divides each bit into two equal intervals. The first interval is a high voltage level and the second interval is a low voltage level to represent bit 1, and it is the opposite for bit 0. The opposite convention can also be adopted, hence option D is incorrect. The transition in the middle of the bit serves as the clock signal, and the voltage level of each bit serves as the data signal, therefore option B is correct. Manchester encoding includes both the clock and data within the data stream, transmitting the clock synchronization signal along with the code information to the other party, thus option A is incorrect. Sound is analog data, while Manchester encoding is best suited for transmitting binary digital signals, therefore option C is incorrect.
|
1,165
|
Test
|
Computer Network
|
Physical Layer
|
Multiple-choice
|
Reasoning
|
English
|
The baud rate of a certain channel is 1000 Baud, and if its data transmission rate reaches 4kb/s, then the number of effective discrete values taken by a single signal element is ( ).
|
4
|
16
|
2
|
8
|
B
|
Bit rate = Baud rate x log₂(number of effective discrete values), the number of effective discrete values per symbol = Bit rate / Baud rate = 4000 / 1000 = 4.
|
1,166
|
Test
|
Computer Network
|
Physical Layer
|
Multiple-choice
|
Reasoning
|
English
|
It is known that the signal transmission rate of a certain channel is 64kb/s, and a carrier signal symbol has 4 effective discrete values, then the baud rate of this channel is ( ).
|
64kBaud
|
128kBaud
|
16kBaud
|
32kBaud
|
D
|
If a symbol can take 2^n different discrete values, it contains n bits of information. In this question, a symbol contains 2 bits of information. Since the baud rate numerically equals the bit rate divided by the number of bits per symbol, the baud rate is (64 / 2) k = 32kBaud.
|
1,167
|
Test
|
Computer Network
|
Physical Layer
|
Multiple-choice
|
Reasoning
|
English
|
There is a noiseless 8KHz channel with each signal containing 8 levels, sampled at 24k times per second, then the maximum transmission rate that can be obtained is ( ).
|
24kb/s
|
32kb/s
|
48kb/s
|
72kb/s
|
C
|
A noiseless signal should satisfy the Nyquist theorem, which states that the maximum data transmission rate = 2W log₂V bits/second. Substituting the data from the question, we get an answer of 48k bits/second, or 48kb/s. Note that the given sampling rate of 24k samples per second in the question is meaningless, as it exceeds the upper limit of the baud rate, 2W = 16kBaud, thus answer D is incorrect.
|
1,168
|
Test
|
Computer Network
|
Physical Layer
|
Multiple-choice
|
Reasoning
|
English
|
Binary signals transmitted over a 4kHz channel with a signal-to-noise ratio of 127:1 can achieve a maximum data transfer rate of ( ).
|
28000b/s
|
8000b/s
|
4000b/s
|
infinity
|
B
|
According to Shannon's theorem, the maximum data rate = W log₂(1 + S/N) = 4000 × log₂(1 + 127) = 28000 b/s. It is easy to mistakenly choose A for this question. However, note the restriction of "binary signals" in the question. According to Nyquist's theorem, the maximum data transfer rate = 2W log₂V = 2 × 4000 × log₂2 = 8000 b/s. Among the two upper limits, we take the smaller one, therefore the correct choice is B.
|
1,169
|
Test
|
Computer Network
|
Physical Layer
|
Multiple-choice
|
Reasoning
|
English
|
Using a QAM modulation method with 8 different phases, each with two amplitude levels, the achievable data transmission rate at a signal transmission rate of 1200 Baud is ( ).
|
2400b/s
|
3600b/s
|
9600b/s
|
4800b/s
|
D
|
Each signal has 8 × 2 = 16 variations, and each symbol carries log₂¹⁶ bits of information, thus the information transmission rate is 1200 × 4 = 4800 b/s.
|
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