Principle of computer composition - interview questions for postgraduate entrance examination (review outline, key points and reference)

《 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 ;