Wang Shuang assembly language detailed learning notes 3: registers (memory access)

Courseware source

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Preface

In the 2 In the chapter , We mainly come from CPU From the perspective of how to execute instructions 8086CPU Logical structure of 、 The way to form a physical address 、 Related registers and some instructions .
In this chapter , Let's continue to learn a few registers from the perspective of accessing memory .

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3.1 Storage of words in memory

1.CPU in , use 16 Bit register to store a word .

2. high 8 Bits hold the high byte , low 8 Bits hold the lower bytes .

3. When stored in memory , Since the memory unit is a byte unit （ A unit holds a byte ）, A word is stored in two consecutive memory units , The low order byte of this word is stored in the low address unit , The high byte is stored in the high address unit .

For example 0 The address begins to store 20000（4E20H）

In the diagram above :
1. We use it 0、1 Two memory cells hold data 20000（4E20H）.

2.0、1 Two memory units are used to store a word , These two units can be regarded as a starting address 0 Word unit of （ A memory unit for one word , from 0、1 Two byte units make up ）.

3. For this word unit ,0 Unit No. is a low address unit ,1 Unit number is a high address unit , Then the font data 4E20H The lower bits of are stored in 0 Unit 2 , The high bits are stored in 1 Unit 2 .

4. In the same way 2、3 The number unit is regarded as a word unit , It starts at 2. Store data in this word unit 18（0012H）, It's in 2 Unit number contains the lower byte 12H, stay 3 Unit number contains the high byte 00H.

3.1.1 Word unit

Word unit , That is to store a font data （16 position ） The memory unit of , It consists of two memory units with consecutive addresses . The high byte of font data in a high address memory unit , The low byte of font data in a low address memory unit .

In the future study , We will start at N The word unit of is abbreviated as N Address word unit . For example, a word unit consists of 2、3 Two memory units , Then the starting address of this word unit is 2, We can say this is 2 Address word unit .

3.1.2 Problem analysis

3.1.3 Summary

Conclusion ：
Any unit of memory with two consecutive addresses ,N Unit number one and N+1 Unit No , Think of them as two memory units , It can also be regarded as an address N The high byte unit and the low byte unit in the word unit of .

3.2 DS and [address]

1.
2.

mov al,[0] explain You can also use mov The instruction sends the contents of a memory unit into a register . From which memory unit to which register ？ Must be specified in the instruction . Registers are indicated by register people , The address of the memory unit is used to indicate . obviously , here mov The format of the instruction should be ：mov Register name , Memory unit address .

“[…]” Represents a memory unit ,“[…]” Medium 0 Represents the offset address of the memory unit . We know , Only offset address can't locate a memory unit , So what is the segment address of the memory unit ？ Instruction execution ,8086CPU Pick up automatically ds The data in is the segment address of the memory unit .

3.2.1 Problem analysis

3.3 The transmission of words

3.3.1 Problem analysis 1

After single step tracking , give the result as follows
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3.3.2 Problem analysis 2

3.4 mov、add、sub Instructions

1.mov、add、sub Instructions will affect the contents of registers

2. and cmp It's a comparison command ,cmp The function of is equivalent to the subtraction instruction , Just don't save the results .cmp After the command is executed , Will affect flag register . Other related instructions know the comparison result by identifying these affected flag register bits .

1.

Pictured above , use A Command at a preset address 073F：0100 It's about , In the form of a compilation mov ax,ds Write instructions , Reuse T Command execution , You can see the result of execution , Segment register ds The value in is sent to the register ax in . Through verification, we know ,“mov register , Segment register ” Is the correct instruction .

2.

The experimental results on your computer are shown in the figure below
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3.



10000H All the data in are 0,DS It should also be for 0,0 The data at the address is shown in the figure above .

The answer is that they cannot operate on segment registers .

3.5 Data segment

3.5.1 Problem analysis

3.1~3.5 Summary

1. When a word is stored in memory , To use two consecutive address memory units to store , The low order byte of a word is stored in a low address unit , The upper byte is stored in the higher address unit .
2. use mov Instruction to access memory cells , Can be in mov Only the address given in the instruction unit is offset , here , The segment address defaults to DS In the register .
3. [address] Represents an offset address of address The memory unit of .
4. When transferring font data between memory and register , High address units and high 8 Bit register 、 Low address units and low 8 Bit registers correspond to .
5. mov、add、sub Is an instruction with two operands .jmp Is an instruction that has an operation object .
6. You can speculate on your own , stay Debug New format of experimental instructions in .
   
   
   
   
   
   

Detection point 3.1

Be careful ,8086CPU All of the registers are 16 position , If you exceed it, you should round off the high position .

The answer is here

3.6 Stack

3.7 CPU Stack mechanism provided

Today's CPU All have stack design ,8086CPU No exception .8086CPU Provides instructions to access memory space by stack . It means , Based on 8086CPU When programming , You can use a piece of memory as a stack .

3.7.1 Two questions

It can be seen that ,8086CPU in , When entering the stack , The top of the stack grows from high address to low address .

Problem analysis 3.6

The answer is as follows ：

pop The execution of instructions

pop ax The process of implementation and push ax Just the opposite

Be careful

3.8 The problem of stack top out of bounds

The following two figures describe the execution push、pop After the instruction , When the top of the stack exceeds the stack space .

3.8.1 Conclusion

3.9 push、pop Instructions

problem 3.7

problem 3.8

problem 3.9

problem 3.10

Conclusion

Overview of stack

3.10 Stack segment

problem 3.11

problem 3.12

Summary of paragraph

Detection point 3.2

The answer is here

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