Unit practice experiment 8 - using cmstudio to design microprogram instructions based on basic model machine (1)

One 、 The experiment purpose

1. • Master the use of simulation software ⽤
2. • Master the design of machine instructions ⽅ Law

Two 、 Experimental equipment

1. Hardware ： nothing
2. Software ：CMSstudio

3、 ... and 、 Experimental content

1. • For the following instructions
   – INL DR – OUTH SR
   – STA [ADDR], SR
   – LD DR, [ADDR]
   – JMP ADDR
2. • Write the machine instruction format
3. • Draw the microinstruction flow chart
4. • send ⽤ The above instruction system is programmed , Realization
   – INPUT read ⼊ The data is saved to ADDR,
   – take ADDR The content is OUTPUT display ⽰.
   – cycle
   – Be careful ：INPUT read ⼊ To memory and memory output to OUTPUT What you have done ⽤ The register requirements of are different ⼀ individual

Four 、 Experimental principle

1. • The program and data are in the same ⼀ address space • AR To refer to as an address pointer • increase PC send AR Micromanipulation
2. Programs and data are stored separately , Program segment and data segment , And stack segments • Fingering ：PC • Get the opcode and operand • Data access ：AR
3. • Stack pointer ：SP • The stack data – Stored in data segments • Stack data access ： – SP send AR – adopt AR To achieve
4. • INL DR • OUTL SR • STA [ADDR], SR • LD DR, [ADDR] • JMP ADDR

5、 ... and 、 The experimental steps

For the following instructions
– INL DR
– OUTL SR
– JMP ADDR
– ADD DR, SR
1. • Write the machine instruction format
2. • Draw the microinstruction flow chart （ Including current address and next address ）
3. • send ⽤ The above instruction system is programmed , Realization
– from INPUT read ⼊ Data to DR
– DR=SR+DR
– DR The data is in OUTPUT display ⽰
– cycle

6、 ... and 、 Commissioning process 、 Results and Analysis

1. Reference materials , Combine with your own reality , Record and analyze the commissioning process .
   practice
   • For the following instructions
   – INL DR
   – OUTL SR
   – JMP ADDR
   – ADD DR, SR

Write the machine code format

The setting rules are as follows

Explain why it is designed in this way ：

1. INL RD
   It's only about RD, therefore Rs The essence is to design casually
2. JMP ADDR,ADDR The title is designed as 16 position , So the correspondence is ARRRL and ADDRH Two layers of
3. IR7-IR4 Is used to jump the address

Finally, when filling in the program area and the instruction system, the above IR7-IR0 The part of should be transformed into 16 Base display

such as 0010 0000 Turn it into 20

Draw the microinstruction flow chart （ Including current address and next address ）

The experimental instruction is based on p85, We can get a similar design

【 The above figure is an example given in the experiment instruction , The following figure shows the implementation of this topic 】

It should be noted that , here 002 Some of the boxes are drawn incorrectly , Should be RAM->BUS,BUS->IR Realize jump （ That is to say IR = RAM[AR]
here RAM Represents the data area ,AR It refers to the data area “ The pointer ”）

3. send ⽤ The above instruction system is programmed , Realization

– from INPUT read ⼊ Data to DR
– DR=SR+DR
– DR The data is in OUTPUT display ⽰
– cycle

; Source program 
#LOAD "work8.IS"         ; Pre call instruction system / Microprograms 
data    segment         ; Load all the programs into the data storage 
        assume ds:data

start: 	INL R0
		OUTL R0
		ADD R0,R1
		JMP start

; Instruction system part 
; Mnemonic symbol   Operands                Instruction code   length 
;-----------------------------------------------------
INL     R0                   20     1                 ;input Enter a number into the general-purpose register R0
OUTL    R0                   40     1                 ; General purpose register R0 The value of is output to output Low byte of 
JMP     *                    60     3                 ; take ADDR The address of is transmitted to PC
ADD     R0,R1                81     1                 ; General purpose register R0 and R1 The result of adding the internal data is saved to the general register R0

Explain why these are 20 1,40 1,60 3,81 1

1. According to my previous design ,2 Represents the converted address （0001） The address is specified as 0010（IR7-IR4 part ） Register due to Appoint RD Use R0,IR1 and IR0 Namely 00, And the rest of the IR3 and IR2 It's an influence , I'm going to set it to 00（ It's actually a RS set R0, But subsequent microinstructions don't use this RS）, So the design is 0010 0000, Turn into 16 Hexadecimal is 20, But actually , You set it to 24 28 2A It also meets the requirements Of
2. ditto , Here is RD In less than , So set it to 41 42 43 It's OK, too
3. ADDR Take up two places , So the length 2 Plus the original 1, Is the total 3,RSRD I can't use it, so here 60-6F Can be set
4. 81,8 Represents the address to be converted .RD yes R1 01,RS yes R0,00, So is 0001, That is to say 1

Microinstruction settings

Figure 2 Forwarding address 20 The situation of Need me （ Low address ） legend （rs

Pay attention to the actual operation , If you want to be right REG（ register ） To operate , Remember RS accidentally RD p.

Figure 3 Forwarding address 40 The situation of accidentally （ Low address rd） Transmission couple

Figure 4 Forwarding address 60 The situation of

Figure 5 Forwarding address 80 The situation of Add it up

Full body （ The rest of the variants understand themselves ）

 Command system ：
; Mnemonic symbol   Operands                Instruction code   length 
;-----------------------------------------------------
INL     R0                   20     1                 ;input Enter a number into the general-purpose register R0, use Rd, Use low ( accidentally ）
INL     R1                   21     1                 ; ditto   The register is R1
INL     R2                   22     1                 ; ditto   The register is R2
INL     R3                   23     1                 ; ditto   The register is R3

OUTL    R0                   44     1                 ; General purpose register R0 The value of is output to output Low byte of , Because in Rs, So use high order （ p. ）
OUTL    R1                   45     1                 ; ditto   The register is R1
OUTL    R2                   46     1                 ; ditto   The register is R2
OUTL    R3                   47     1                 ; ditto   The register is R3
JMP     *                    60     3                 ; take ADDR The address of is transmitted to PC
ADD     R0,R0                80     1                 ; General purpose register R0 and R0 The result of adding the internal data is saved to the general register R0
ADD     R0,R1                81     1                 ; General purpose register R0 and R1 The result of adding the internal data is saved to the general register R0
ADD     R0,R2                82     1                 ;R0+=R2
ADD     R0,R3                83     1                 ;R0+=R3
ADD     R1,R0                84     1                 ;R1+=R0
ADD     R1,R1                85     1                 ;R1+=R1
ADD     R1,R2                86     1                 ;R1+=R2
ADD     R1,R3                87     1                 ;R1+=R3
ADD     R2,R0                88     1                 ;R2+=R0
ADD     R2,R1                89     1                 ;R2+=R1
ADD     R2,R2                8A     1                 ;R2+=R2
ADD     R2,R3                8B     1                 ;R2+=R3
ADD     R3,R0                8C     1                 ;R3+=R0
ADD     R3,R1                8D     1                 ;R3+=R1
ADD     R3,R2                8E     1                 ;R3+=R2
ADD     R3,R3                8F     1                 ;R3+=R3

 Source program 
; Basic model machine program 

#LOAD "work8.IS"         ; Pre call instruction system / Microprograms 
#SET RAM 0260h = 5500h  ; Data area 0260H Unit preset data 5500h, There is no use here 

data    segment         ; Load the program into the data store 
        assume ds:data

        org   0
        
start:  INL R0
        ADD R0,R1
        OUTL R0
        JMP start

data    ends
        end   start