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Principle of computer composition - interview questions for postgraduate entrance examination (review outline, key points and reference)
2022-07-02 02:52:00 【Jatine】
List of articles
- The first 1 Chapter : Introduction to Computer Systems
- 1、 What are the two parts of the computer system ? What does the performance of a computer system depend on ?
- 2、 Computer system 5 From bottom to top, the layer hierarchy is composed of five layers ? Which are physical machines , Which are virtual machines ?
- 3、 In computer system architecture , What is translation ? What is explanation ?
- 4、 What is computer architecture ? What is computer composition ? Take multiplication instruction as an example to illustrate the difference between the two .
- 5、 The main features of von Neumann machine ?
- 6、 Draw the composition block diagram of modern computer .
- 7、 What is a storage unit 、 Storage word 、 Storage word length 、 memory bank ?
- 8、 In main memory , What is? MAR, What is? MDR, What determines the maximum capacity of memory ?
- 9、 What is machine word length , What is a long stored word ?
- 10、 hypothesis MAR The number of bits in the register is 16 position ,MDR The number of bits in the register is 16 position , What is the maximum capacity of memory ?
- The third chapter The system bus
- 1、 Why use a bus ?
- 2、 What are the two basic characteristics of the bus ?
- 3、 The system bus is different according to the transmission information , What are the categories ? Is one-way , Or two-way ?
- 4、 What is bus width 、 Bus bandwidth 、 Bus multiplexing 、 Number of signal lines ?
- 5、 Suppose the working frequency of the bus is 33MHz, The bus width is 32 position , Then what is its maximum transmission rate ?
- 6、 Briefly explain the concept and shortcomings of single bus structure ?( Why should modern computers adopt multi bus structure ?)
- 7、 What are the three ways of centralized bus optimization control , The priority of which method cannot be changed ?
- 8、 What is a bus cycle , What are the stages ?
- 9、 What is bus communication control , What kinds of bus communication control ?
- 10、 What is synchronous communication ? Its advantages and disadvantages ?
- 11、 What is asynchronous communication ? What are the types of asynchronous communication ?
- 12、 What is baud rate ? What is bit rate ?( You need to master how to calculate baud rate 、 Bit rate )
- 13、 Asynchronous communication , What are the parameters that need to be set in general ?
- 14、 Briefly introduce the basic principle of semi synchronous communication .
- 15、 Briefly introduce the basic principle of separated communication .
- 16、 Can parity check correct errors ? Hamming code can correct code ?
- Chapter four Memory
- 1、 Memory is accessed by , Can be divided into four categories ? Which belong to random access memory , Which belong to serial access memory ?
- 2、 Which three indicators are used to measure memory ? register 、 cache 、 In main memory , Which is the fastest ? Which is the cheapest ?
- 3、 What are the two common storage system hierarchies ? How transparent ? What problems are they used to solve ?
- 4、 There are two ways to store words in a storage unit , Large end mode and small end mode . What does each mean ?x86 Which storage method is used ?
- 5、 Three main technical indicators of main memory
- 6、 What is access time ? What is an access cycle ? Which is the big one ?
- 7、 What is memory bandwidth ?( Learn how to calculate memory bandwidth )
- 8、 What are the two ways of semiconductor memory chip decoding driver , Please give a brief description of .
- 9、 What are the two main types of ram ? Which needs refreshing ? Please start from the speed 、 Capacity 、 Price and other aspects are briefly compared .
- 10、 What kinds of read-only memory ?
- 11、 The capacity of a single memory chip is limited , It is difficult to meet the actual needs , Therefore, several memory chips must be connected together to form a memory with sufficient capacity .
- 12、 Carefully grasp the expansion of memory , Including address space allocation 、 Connection of address line 、 The connection of the data line 、 Generation and connection of chip selection signal, etc ;
- 13、 Suppose the binary code to be detected is n position , In order to make it have 1 Bit error correction capability , Need to add K Bit detection bit , form n+k Bit code . ask , How many detection bits should be added ?
- 14、 For Hamming code , You should master the coding method of Hamming code ( According to the principle of spouse or matchmaker ), And give Hamming code , Get the original information to be transmitted ( Including the error correction process ).
- 15、 Three ways to improve the speed of memory access .
- 16、 Briefly describe the working principle of single multi word storage system , And its advantages .
- 17、 What are the two addressing methods for multi-body parallel systems ? Please briefly explain the addressing method and its advantages .
- 18、 In four bit low bit interleaved addressing , Assume that the access cycle is T, The bus transmission cycle is τ, In order to realize pipeline storage , What conditions should be met ? If you read four words consecutively , How much time does it take ?
- 20、 You need to master the high-order cross addressing of multi-body parallel memory ( Sequential storage ) And low bit cross addressing ( Cross storage ) Under the circumstances , How memory bandwidth is calculated .
- 22、 What is the principle of program locality .
- 23、Cache shooting 、 Calculation of average access time and access efficiency .
- 24、Cache There are two ways to write ?
- 25、 Map the main memory address to Cache Address is called address mapping , common Cache What are the mapping methods ?
- 26、 Advantages and disadvantages of direct mapping ?
- 27、 Advantages and disadvantages of fully associative mapping ?
- 28、 You need to master various mapping methods , Write out the main memory address format 、cache Address format , And the main memory address to cache Address translation .
- 29、Cache What are the commonly used replacement algorithms ? Which has the highest hit rate ?
- 30、 What does the three address structure of the disk include ?
- The fifth chapter I / O system
- 1、I/O The development of the system can be roughly divided into 4 Stages ?
- 2、I/O There are two ways to address devices ? What are the advantages and disadvantages of each ?
- 3、I/O What are the contact methods between the equipment and the host ?
- 4、I/O What are the four types of buses ?
- 5、I/O The equipment usually uses D trigger ( Complete trigger ) and B trigger ( Work trigger ) To identify the state of the device .
- 6、 The basic working principle of program query .
- 7、 What is interruption ?
- 8、 What are the four parts of the basic flow of the interrupt service program ?
- 9、 What are single interrupt and multiple interrupt ?
- 10、CPU Time to respond to interruption ?
- 11、 What is? DMA?
- 12、 stay DMA In the way , because DMA Interface and CPU Shared main memory , There may be a conflict between the two competing for main memory , To resolve conflicts ,DMA When exchanging data with main memory , Which three working methods are usually used ?
- 13、DMA What are the three parts of the work process ?
- Chapter six The operation of a computer
- 1、 Master the original code calculation method of signed numbers , And get the truth value through the original code ;
- 2、 Master the method of complement calculation , And find the original code through complement , Then the method of finding the truth .
- 3、 In original code 0 Yes 2 There are two ways to express ( Positive zero and negative zero ), In complement 0 There is only one way to express ( Positive zero and negative zero are expressed in the same way )
- 4、 Suppose the number of signed digits is 8( Include sign bits ), The range of true values that complement can represent ?
- 5、 Master the method of inverse code and code shift .
- 6、 What is fixed-point representation ? What is floating point representation ?
- 7、 Representation of floating point numbers in machines , What are the components of ?
- 8、 Master the representation range of normalized floating-point numbers ( Maximum positive number 、 Minimum positive number 、 Maximum negative number 、 Minimum negative number ) The calculation method of .
- 9、IEEE754 What are the components of floating-point numbers specified in the standard ?
- 10、IEEE754 Floating point numbers specified in the standard , What form is the order code and mantissa expressed ?
- 11、float How many seats ?double How many seats ?
- 12、 Arithmetically shift positive numbers , When the positive number adopts the source code 、 Complement code 、 Inverse code time , When moving left or right , What code is added in low or high order ?
- 13、 Arithmetically shift negative numbers , When the negative number adopts the source code 、 Complement code 、 Inverse code time , When moving left or right , What code is added in low or high order ?
- 14、 What is logical shift ?
- 15、 When adding and subtracting , Under what circumstances may overflow occur ? How to simply judge the overflow ?
- 16、 Can fixed-point multiplication be realized by addition and shift ?
- 17、 What are the basic steps of floating-point addition and subtraction ?
- 18、 How to judge whether the floating-point operation result overflows ?
- Chapter vii. Command system
- 1、 What are machine instructions ? What is an instruction system ?
- 2、 What are the two main parts of an instruction ? Please briefly explain the functions of each part .
- 3、 There are two types of opcode length: fixed length and variable length , What are the advantages of each ?
- 4、 The address in the address code in the instruction can be the address of which device ?
- 5、 The number of addresses in the instruction can be several ?
- 6、 Suppose there are four addresses in the instruction 、 Three addresses 、 Two addresses and one address , Each needs to visit and deposit several times ?
- 7、 When the register is used instead of the address code field in the instruction word , What are the advantages ?
- 8、 When data is stored in memory , Why align by boundary ?
- 9、 What are the two types of addressing methods ?
- 10、 What is a formal address ? What is a valid address ?
- 11、 Understand the concept of various addressing methods and the way to form effective addresses according to formal addresses .
- 12、 What is? RISC? What is? CISC?
《 The principle of computer organization 》 Jiang Benshan edition
The first 1 Chapter : Introduction to Computer Systems
1、 What are the two parts of the computer system ? What does the performance of a computer system depend on ?
The computer system is made up of “ Hardware ” and “ Software ” form . Measuring the performance of a computer is based on a number of technical indicators , It includes various performance indicators of hardware , It also includes various functions of the software .
1) The computer system consists of hardware and software .
2) The performance of a computer system is determined by both hardware and software .
2、 Computer system 5 From bottom to top, the layer hierarchy is composed of five layers ? Which are physical machines , Which are virtual machines ?
1) Microprogramming machines 、 Traditional machines 、 Operating system machine 、 Assembly language machine 、 High level language machine
2) Microprogrammed machines and traditional machines are physical machines , Others are virtual machines .
3、 In computer system architecture , What is translation ? What is explanation ?
1) translate : Translate all programs written in one language into another language , And then execute ;
2) explain : Translate a statement of a program written in one language into one or more statements in another language , And then execute , After executing this language , Explain the next .
4、 What is computer architecture ? What is computer composition ? Take multiplication instruction as an example to illustrate the difference between the two .
1) Computer architecture refers to the properties of computers that can be seen by programmers . Such as instruction set 、 Data type, etc ;
2) Computer composition refers to how to realize the attributes embodied in the computer architecture ;
3) Take the multiplication instruction as an example , Does the computer have multiplication instructions , It belongs to the problem of architecture . The multiplication instruction adopts a special multiplier , Still use adder and shifter , It belongs to the problem of computer composition .
5、 The main features of von Neumann machine ?
1) A computer consists of an arithmetic unit 、 Memory 、 controller 、 Input device and output device are composed of five parts ;
2) Instructions and data are stored in memory , And can be accessed by address ;
3) Instructions and data are represented in binary ;
4) Instructions consist of opcodes and address codes , The opcode indicates the nature of the operation , The address code indicates the location of the operand in memory ;
5) Instructions are stored in order in memory , It is usually taken out and executed in automatic order ;
6) The machine is centered on the arithmetic unit ,I/O The data exchange between the device and the memory should also be through the arithmetic unit .( therefore , Later, there was a computer structure centered on memory )
6、 Draw the composition block diagram of modern computer .
P10, chart 1.9
7、 What is a storage unit 、 Storage word 、 Storage word length 、 memory bank ?
Storage unit : A storage unit that stores a storage word and has a specific storage address ;
Storage word : All binary data stored in a storage unit , Access binary data obtained from a storage unit according to a certain address .
Storage word length : The number of bits of binary data in a stored word , That is, the number of bits of binary data obtained by accessing a storage unit according to a certain address ;
memory bank : A storage device consisting of multiple storage units .
8、 In main memory , What is? MAR, What is? MDR, What determines the maximum capacity of memory ?
1)MAR: Store address register , Save the address of the storage unit to be accessed . Reflect the number of storage units .
2)MDR: Storage data register , Cache read / Data written to the storage unit . Reflect the storage word length .
3) The maximum capacity of the memory consists of MAR The sum of the bits of the register MDR The number of bits of the register determines .
9、 What is machine word length , What is a long stored word ?
Machine word length :CPU The number of bits of binary data that can be processed at one time .
Storage word length : The number of bits of binary data obtained by accessing a storage unit according to a certain address .
10、 hypothesis MAR The number of bits in the register is 16 position ,MDR The number of bits in the register is 16 position , What is the maximum capacity of memory ?
1)MAR The number of bits in the register is 16 position , The number of addresses that can be represented is 2 Of 16 Power , by 64K;
2)MDR The number of bits in the register is 16 position , The storage word length is 16 position , That is to say 2 Bytes ;
3) The maximum capacity of the memory is 64K * 2B = 128K Byte
The third chapter The system bus
1、 Why use a bus ?
In the von Neumann structure , There are separate wires between each component , Not only are there many lines , And it leads to expansion I/O The equipment is not easy . That is, expand a I/O equipment , Many wires need to be connected .
therefore , Bus connection mode is introduced , Connect multiple devices on the same set of buses , Constitute a common transmission channel between devices .
2、 What are the two basic characteristics of the bus ?
1) share : Multiple components are connected to the same set of buses , All components exchange data through the bus .
2) time-sharing : At the same time , Only information sent by one component can be transmitted on the bus ;
3、 The system bus is different according to the transmission information , What are the categories ? Is one-way , Or two-way ?
1) Divided into data bus 、 Address bus and control bus .
2) data bus : Data information is transmitted between various functional components , Two way transmission ;
3) Address bus : Used to indicate on the data bus , The address of the main memory unit where the source data or destination data is located . A one-way : from CPU issue
4) Control bus : Used to send various control signals . For a single wire in the control bus , Is one-way , That is, it can only be sent from one part to another . In a group of control buses , There are inputs and outputs , therefore , The control bus can also be regarded as bidirectional .
4、 What is bus width 、 Bus bandwidth 、 Bus multiplexing 、 Number of signal lines ?
1) Bus width : Number of data buses , It's usually 8 Multiple . It is an important index to measure the performance of computer system ;
2) Bus bandwidth : That is, the bus data transmission rate , The maximum number of bytes that can be transferred per second on the bus .
3) Bus multiplexing : Time sharing transmission of two signals on a signal line . For example, time-sharing multiplexing of data bus and address bus ;
4) Number of signal lines : Address bus 、 The sum of the number of lines of data bus and control bus .
5、 Suppose the working frequency of the bus is 33MHz, The bus width is 32 position , Then what is its maximum transmission rate ?
33 * (32/8) = 132 MB/s
6、 Briefly explain the concept and shortcomings of single bus structure ?( Why should modern computers adopt multi bus structure ?)
In a single bus architecture , All the parts (CPU、 Main memory 、I/O equipment ) Are connected to a set of buses .
But all information is transmitted through this group of buses , At the same time, only one component can send information to the bus , As a result, the bus becomes the bottleneck of the system .
therefore , Developed a multi bus structure , The basic idea is to connect devices with similar speed to the same group of buses , Buses are connected through bus controllers .
for example CPU and Cache Between 、I/O Equipment, etc .
7、 What are the three ways of centralized bus optimization control , The priority of which method cannot be changed ?
1) Chain query 、 Counter timing query 、 And independent requests .
2) The priority of chain query cannot be changed , The priority nearest to the controller is the highest .
8、 What is a bus cycle , What are the stages ?
1) Bus cycle : Two components on the bus complete a complete and reliable data transmission time ;
2) There are four stages :
Application allocation stage : Apply for bus
Addressing phase : Send address and relevant orders
Data transmission stage : Data exchange
end : Remove the signal from the bus , Give up the bus
9、 What is bus communication control , What kinds of bus communication control ?
1) Bus communication control : Solve how the communication parties know the start and end of transmission , And how to coordinate ;
2) Synchronous communication 、 asynchronous communication 、 Semi synchronous communication 、 Separate communication
10、 What is synchronous communication ? Its advantages and disadvantages ?
1) Synchronous communication : Each component on the bus is controlled by a unified clock signal ; In the bus cycle , There are clear regulations on how each component operates in each clock cycle .
2) advantage : Fast , The cooperation between each module is simple
3) shortcoming : Design a common clock with the slowest component on the bus , Affect bus efficiency .
11、 What is asynchronous communication ? What are the types of asynchronous communication ?
1) asynchronous communication : There is no unified clock standard for all components on the bus , Adopt reply communication ;( After the main module sends a request , Wait until the response signal is fed back from the module before starting communication )
2) No interlock 、 Semi interlocking 、 Full interlock .( Need to understand the meaning of various ways )
12、 What is baud rate ? What is bit rate ?( You need to master how to calculate baud rate 、 Bit rate )
Baud rate : The number of bits of binary data transmitted in unit time , Company bps
Bit rate : The number of valid binary digits transmitted per unit time .
13、 Asynchronous communication , What are the parameters that need to be set in general ?
Baud rate 、 Stop bit (1/2/1.5)、 Check bit ( Odd check 、 Even check 、 No verification )
14、 Briefly introduce the basic principle of semi synchronous communication .
Semi synchronous communication combines synchronous communication and asynchronous communication .
Synchronous communication : Adopt a unified clock , It stipulates what to do in a certain clock cycle ;
asynchronous communication : If the slave module is not ready , Add one more “ Waiting response ” The signal .
15、 Briefly introduce the basic principle of separated communication .
After the main module sends the address and command , Abandon the bus , During data preparation from the module , So that the bus can be used by other devices . Improve bus utilization .
however , This kind of control is complicated .
16、 Can parity check correct errors ? Hamming code can correct code ?
1) Parity can only detect errors , Can't correct .
2) Hamming code can correct errors .
Chapter four Memory
1、 Memory is accessed by , Can be divided into four categories ? Which belong to random access memory , Which belong to serial access memory ?
1) It can be divided into random access memory 、 Sequential memory and direct memory ;
2) Ram , The access time is independent of the physical address ;
3) Sequential memory ( Typically, such as tape ) And direct memory ( Typical examples are disks ) It belongs to serial memory , That is, the access time is related to the physical address .
2、 Which three indicators are used to measure memory ? register 、 cache 、 In main memory , Which is the fastest ? Which is the cheapest ?
1) Speed 、 Capacity 、 Bit price .
2) Register speed is the fastest , Main memory is the cheapest .
3、 What are the two common storage system hierarchies ? How transparent ? What problems are they used to solve ?
1) cache - Main memory level : To relieve CPU The problem of mismatch with the main memory speed , Done by hardware , Completely transparent to all programmers .
2) Main memory - Secondary storage level : Used to solve the problem of insufficient main memory capacity , It is completed by the operating system and hardware , Transparent to application designers , Opaque to system programmers .
( Now the general memory can be accessed by word , It can also be accessed by bytes , therefore , When addressing memory , Each byte has an independent address .)
4、 There are two ways to store words in a storage unit , Large end mode and small end mode . What does each mean ?x86 Which storage method is used ?
1) Big end way : The low order of the word is stored in the high address of the memory , The high bit of the word is stored in the low address of the memory ;
2) The small end way : The low order of the word is stored in the low address of the memory , The high bit of the word is stored in the high address of the memory .
3)x86CPU The small end mode is adopted .
5、 Three main technical indicators of main memory
storage capacity 、 Access speed and storage bandwidth
6、 What is access time ? What is an access cycle ? Which is the big one ?
1) Access time : Start the memory once to complete this operation ( Read or write ) Time required ;
2) Access cycle : The minimum interval required to start the memory twice in succession ;
3) Access cycle includes access time ;
7、 What is memory bandwidth ?( Learn how to calculate memory bandwidth )
The amount of information accessed by the memory per unit time ;
8、 What are the two ways of semiconductor memory chip decoding driver , Please give a brief description of .
1) Line selection : All address chips are decoded by a decoder , Choose a storage unit , Suitable for chips with small storage capacity ;
2) To be legal again : Divide the addresses into two groups , Each group is decoded by a decoder , Select rows or columns , That's ok 、 The intersection of columns is the storage bit to be accessed .
9、 What are the two main types of ram ? Which needs refreshing ? Please start from the speed 、 Capacity 、 Price and other aspects are briefly compared .
1) static state RAM: The principle of latch is adopted to realize ;
2) dynamic RAM: The principle of capacitance is adopted to realize , Need to refresh .
3) Compared to dynamic RAM, static state RAM Fast 、 Small capacity 、 high price , Generally used for caching , And dynamic RAM Generally used for memory .
10、 What kinds of read-only memory ?
1) Mask ROM(MROM): The content cannot be changed after leaving the factory .
2)PROM: Programmable read only memory , It can be programmed once ;
3)EPROM: Erasable read-only ROM, Irradiate with ultraviolet light ;
4)EEPROM: Electrically erasable read-only ROM.
6)FLash Memory: use EEPROM Nonvolatile memory .
11、 The capacity of a single memory chip is limited , It is difficult to meet the actual needs , Therefore, several memory chips must be connected together to form a memory with sufficient capacity .
Memory expansion usually includes bit expansion and word expansion , What is word expansion , What is bit extension ? Please give a brief example
1) Bit expansion : Increase the word length of memory , For example, two 1K * 4 Bit memory chip structure 1 individual 1K*8 Bit memory ;
2) Word extension : Increase the number of words in the memory , For example, two 1K * 8 Bit memory chip structure 1 individual 2K * 8 Bit memory ;
Usually, word expansion and bit expansion are mixed .
12、 Carefully grasp the expansion of memory , Including address space allocation 、 Connection of address line 、 The connection of the data line 、 Generation and connection of chip selection signal, etc ;
see P94 page , example 4.1
13、 Suppose the binary code to be detected is n position , In order to make it have 1 Bit error correction capability , Need to add K Bit detection bit , form n+k Bit code . ask , How many detection bits should be added ?
The number of detection bits that should be added :2 Of k The power is greater than or equal to n+k+1.
Because to make it have 1 Bit detection capability , You have to use k Bits to illustrate n+k Which one is wrong ,k The number of potential expressions is 2 Of k Power , and n+k Who is it
Something went wrong or it was all right , share n+k+1 Species status , therefore ,k The value of needs to meet :2 Of k The power is greater than or equal to n+k+1
14、 For Hamming code , You should master the coding method of Hamming code ( According to the principle of spouse or matchmaker ), And give Hamming code , Get the original information to be transmitted ( Including the error correction process ).
15、 Three ways to improve the speed of memory access .
1) Adopt high-speed components ;
2) Adopt storage hierarchy :cache- Main memory structure ;
3) Adjust the main memory structure : Including single multi word , There are two ways of multi-body parallel .
16、 Briefly describe the working principle of single multi word storage system , And its advantages .
1) A single multi word storage system accesses and takes out multiple at a time CPU word , That is, the stored word is CPU The word n times ( Suppose a memory access takes out n individual cpu word ).
2) Advantage is : Significantly improve memory bandwidth .
17、 What are the two addressing methods for multi-body parallel systems ? Please briefly explain the addressing method and its advantages .
1) High bit cross addressing : The address mode of the storage bank is sequential storage , That is, when a bank is full , Then save to the next ; The high order of the address of the storage unit is the number of the storage bank .
High bit cross addressing does not improve the speed of a single memory access , But it can make multiple applications access memory in parallel , Improve the concurrency of the system .
2) Low bit cross addressing : The addressing method of the memory bank is cross storage . That is, the program is continuously stored in adjacent storage . The low order of the address of the storage unit is the number of the storage bank .
Low bit cross addressing can significantly improve the speed of a single memory access .
18、 In four bit low bit interleaved addressing , Assume that the access cycle is T, The bus transmission cycle is τ, In order to realize pipeline storage , What conditions should be met ? If you read four words consecutively , How much time does it take ?
1)T= 4τ
2) Read four words in succession , The time required is T + (4-1)τ
Be careful : Suppose it is not low bit cross addressing , But high-order cross addressing , The time required to read four words continuously is still 4T.
20、 You need to master the high-order cross addressing of multi-body parallel memory ( Sequential storage ) And low bit cross addressing ( Cross storage ) Under the circumstances , How memory bandwidth is calculated .
21、 stay CPU And memory cache Why .
1) avoid cpu Empty wait I/O Visiting and depositing ;
2) relieve CPU The problem of mismatch with the main memory speed .
22、 What is the principle of program locality .
CPU Access instructions or data from the master , In a certain amount of time , Only access to the local address area of main memory .
23、Cache shooting 、 Calculation of average access time and access efficiency .
24、Cache There are two ways to write ?
1) Write direct : The write operation writes Cache Write to main memory again ;
2) Write back : Write data only Cache Instead of writing to main memory , When Cache The data in is written into main memory after being replaced .
25、 Map the main memory address to Cache Address is called address mapping , common Cache What are the mapping methods ?
Direct mapping 、 All associative mapping 、 Group associative mapping .
26、 Advantages and disadvantages of direct mapping ?
advantage : Address conversion is fast . shortcoming :cache Utilization rate is not high , Block collision rate is high ;
27、 Advantages and disadvantages of fully associative mapping ?
advantage :cache High utilization , Low block collision rate . shortcoming : Address translation is complicated , Need more hardware .
28、 You need to master various mapping methods , Write out the main memory address format 、cache Address format , And the main memory address to cache Address translation .
29、Cache What are the commonly used replacement algorithms ? Which has the highest hit rate ?
1) fifo 、 Recently, the least used algorithm and random replacement algorithm ;
2) The highest hit rate is the least recently used algorithm ;
30、 What does the three address structure of the disk include ?
cylinder 、 Head number and sector number
The fifth chapter I / O system
1、I/O The development of the system can be roughly divided into 4 Stages ?
1) In the early ( Decentralized connection 、 Serial operation 、 Program query )
2) Interface module and DMA Stage ( Bus connection 、 Parallel work 、 Interruptions and DMA)
3) Channel stage ( A channel is a processor with special functions )
4)I/O Processor stage
I/O In fact, the development of the system will gradually CPU From heavy I/O The process of being liberated from work ;
2、I/O There are two ways to address devices ? What are the advantages and disadvantages of each ?
1) Unified addressing method : And memory are uniformly addressed ,I/O Address as part of memory address ; No special I/O Instructions , But it takes up memory space .
2) Independent addressing mode : Address separately from the storage address , Special I/O Instructions .
3、I/O What are the contact methods between the equipment and the host ?
I/O When exchanging information between devices and hosts, you must understand each other's status . according to I/O Different working speeds of equipment , Can be divided into 3 class :
1) Respond immediately : Regardless of its state ( Think it's always ready ), Suitable for slow-moving equipment .
2) Answer signal : Interact by answering signals ;
3) Synchronization time scale : Adopt unified clock signal .
4、I/O What are the four types of buses ?
cable 、 Equipment selection line 、 State line 、 Command line
5、I/O The equipment usually uses D trigger ( Complete trigger ) and B trigger ( Work trigger ) To identify the state of the device .
D=0,B=0: Pause state ;
D=0,B=1: Ready to go
D=1,B=0: Ready state
6、 The basic working principle of program query .
cpu Keep checking I/O Equipment status , Lead to CPU and I/O The device works serially .
7、 What is interruption ?
The computer is executing a program , When there is an abnormal emptying or special request , The computer stops the running of the current program , Turn to handle these abnormal emptying or special requests , After processing , Then return to the gap of the current procedure , Continue with the original procedure , That is, interrupt .
8、 What are the four parts of the basic flow of the interrupt service program ?
1) Protection site
2) Interruption of service
3) Restore the scene
4) Interrupt return
9、 What are single interrupt and multiple interrupt ?
1) Single interrupt : It is not allowed to interrupt the current interrupt service program ;
2) Multiple interrupts : Allow higher level interrupt sources to interrupt the current interrupt service program , Also known as interrupt nesting ;
10、CPU Time to respond to interruption ?
After the current instruction is executed ,cpu Send an interrupt query signal , in other words , The interrupt response must be carried out after the execution of each instruction , It is impossible to respond to interrupts during instruction execution .
11、 What is? DMA?
DMA: Direct memory access . In main memory and I/O Establish independent bus connection between devices .
12、 stay DMA In the way , because DMA Interface and CPU Shared main memory , There may be a conflict between the two competing for main memory , To resolve conflicts ,DMA When exchanging data with main memory , Which three working methods are usually used ?
1) stop it CPU Access main memory :DMA Memory access priority is high ;
2) Periodic misappropriation ( steal ):DMA Misappropriate storage or steal bus access rights for one or more main memory access cycles ;
3)DMA and CPU Alternate access : take CPU The work cycle is divided into two parts , Part of it is for DMA Visiting and depositing , Part of it is for CPU Visiting and depositing .
13、DMA What are the three parts of the work process ?
1) Preprocessing
2) The data transfer
2) post-processing
Chapter six The operation of a computer
1、 Master the original code calculation method of signed numbers , And get the truth value through the original code ;
2、 Master the method of complement calculation , And find the original code through complement , Then the method of finding the truth .
1) Find the complement through the original code : The sign bits remain the same , You take the opposite , Last place plus 1;
2) Find the original code through complement : The sign bits remain the same , You take the opposite , Last place plus 1;
3、 In original code 0 Yes 2 There are two ways to express ( Positive zero and negative zero ), In complement 0 There is only one way to express ( Positive zero and negative zero are expressed in the same way )
4、 Suppose the number of signed digits is 8( Include sign bits ), The range of true values that complement can represent ?
The range of true values that complement can represent is -128~+127( See complement definition )
5、 Master the method of inverse code and code shift .
6、 What is fixed-point representation ? What is floating point representation ?
1) Fixed point representation : The number whose decimal point is fixed at a certain position is the fixed number ;
2) Floating point means : The number whose decimal point position can float .
7、 Representation of floating point numbers in machines , What are the components of ?
From the mantissa 、 Numeral character 、 Order code 、 The level symbol consists of four parts .
8、 Master the representation range of normalized floating-point numbers ( Maximum positive number 、 Minimum positive number 、 Maximum negative number 、 Minimum negative number ) The calculation method of .
9、IEEE754 What are the components of floating-point numbers specified in the standard ?
By number sign 、 Order code ( Rank containing sign ) And mantissa .
10、IEEE754 Floating point numbers specified in the standard , What form is the order code and mantissa expressed ?
Step codes are represented by shift codes , Its offset is 2^(n-1), The mantissa is expressed in the original code .
11、float How many seats ?double How many seats ?
float Is a short real number , Occupy 32 position , Among them, the order code 8 position , mantissa 23 position .
double Is a long real number , Occupy 64 position , Among them, the order code accounts for 11 position , Mantissa 52 position .
12、 Arithmetically shift positive numbers , When the positive number adopts the source code 、 Complement code 、 Inverse code time , When moving left or right , What code is added in low or high order ?
For positive numbers , Its source 、 Complement code 、 The inverse code is equal to the truth , Shift left , Low complement 0, When you move right , Top up 0.
13、 Arithmetically shift negative numbers , When the negative number adopts the source code 、 Complement code 、 Inverse code time , When moving left or right , What code is added in low or high order ?
For source code , When moving left or right , Both low and high positions are supplemented 0;
For complements : Shift left , Low complement 0, When moving to the right, the high position is added 1
For inverse code : When moving left or right , Both low and high positions are supplemented 1;
14、 What is logical shift ?
Logical shift is the shift of unsigned numbers , Because there is no sign bit in unsigned numbers , Shift left , High displacement loss , Make up zero for low position . When you move right , Low displacement , Zero up for high position .
15、 When adding and subtracting , Under what circumstances may overflow occur ? How to simply judge the overflow ?
1) Positive numbers plus positive numbers , Positive minus negative , Negative numbers plus negative numbers , Negative minus positive , Overflow may occur .
2) If the symbols of two numbers participating in the operation are the same ( Addition converted to complement ), The result is different from the source operand symbol , Overflow .
3) If the complement is 1 Bit sign bit , If the carry of the most significant bit is different from that of the sign bit , Overflow occurs .
16、 Can fixed-point multiplication be realized by addition and shift ?
Sure .
17、 What are the basic steps of floating-point addition and subtraction ?
1) Antithetic order : Align decimal points ;
2) Sum the mantissa : Sum the two mantissa after the order according to the fixed-point addition and subtraction operation rules ;
3) normalized : Mantissa normalization ;
4) Round off : Mantissa right gauge , Missing value bits ;
5) Overflow judgment : Judge whether the result overflows .
18、 How to judge whether the floating-point operation result overflows ?
Whether the order code is beyond its representation range .( Use 2 Symbol bit judgment overflow )
Chapter vii. Command system
1、 What are machine instructions ? What is an instruction system ?
1) Machine instructions : Every machine language statement ;
2) Command system : A collection of all machine instructions .
2、 What are the two main parts of an instruction ? Please briefly explain the functions of each part .
1) opcode : Indicates the operation to be completed by the instruction ;
2) Address code : Indicates the data or data source to be operated by the instruction ;
3、 There are two types of opcode length: fixed length and variable length , What are the advantages of each ?
1) Fixed length : Convenient for hardware design , Instruction decoding time is short ;
2) Variable length : Compressed the average length of opcodes ;
4、 The address in the address code in the instruction can be the address of which device ?
It can be the main memory address 、 Register address or I/O Address of the device ;
5、 The number of addresses in the instruction can be several ?
Four addresses 、 Three addresses 、 2. Address 、 One address and zero address .
6、 Suppose there are four addresses in the instruction 、 Three addresses 、 Two addresses and one address , Each needs to visit and deposit several times ?
1) Four addresses : Visiting and depositing 4 Time ;
2) Three addresses : Visiting and depositing 4 Time ;
3) Two addresses : Visiting and depositing 3 Time ;
4) An address : Visiting and depositing 2 Time ;
7、 When the register is used instead of the address code field in the instruction word , What are the advantages ?
1) Expand the addressing range of the instruction word ;
2) Shorten instruction word length ;
3) Reduce the number of deposit visits
8、 When data is stored in memory , Why align by boundary ?
Reduce the number of deposit visits .
9、 What are the two types of addressing methods ?
1) Command addressing : The instruction address of the next instruction to be executed ;
2) Data addressing : Determine the operand address of this instruction .
10、 What is a formal address ? What is a valid address ?
1) Formal address : The address code field of the instruction usually does not represent the real address of the operand , Become a formal address , Write it down as A;
2) Valid address : The real address of the operand , Write it down as EA, It is determined by addressing characteristics and formal address ;
11、 Understand the concept of various addressing methods and the way to form effective addresses according to formal addresses .
Address immediately 、 Direct addressing 、 Implicit addressing 、 Indirect addressing 、 Register addressing 、 Register indirection 、 Base addressing ( Implicit or explicit )、 Addressing 、 Relative addressing 、 Stack addressing
12、 What is? RISC? What is? CISC?
RISC: Reduced instruction set ;
CISC: Complex instruction set ;
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