chapter 5 8051 addressing modes -...

1

Chapter 5

8051 Addressing Modes

2

Sections

5.1 Immediate and register addressing modes

5.2 Accessing memory using various address modes

3

Objective

• 程式中的資料可能是放在 Register 中，或在RAM 中某一位址上，或在 ROM 一塊特殊區域放置資料，或者是指令中直接給予定值。

• 設計 8051 IC 的人們，提供這些存取資料的方式。這些方式便叫作 Addressing Mode。

– 中文稱為“定址模式”：決定參數位址的模式

– 也許不同家的 Assembler 會有不同的指令寫法，但基本上 addressing mode 都是一樣的。

4

Section 5.1

Immediate and Register Addressing

Modes

5

What is Addressing Mode

• The CPU can access data in various ways.

• The data could be in a register, or in memory

（RAM or ROM）, or be provided as an

immediate value.

• These various ways of accessing data are called

addressing mode.

6

Addressing Mode in the 8051

• Five addressing mode in the 8051：

1. immediate

2. register

3. direct

4. register indirect

5. indexed

7

Addressing Mode 1

1. immediate － the operand is a constant

MOV A,#01FH

2. register － the operand is in a register

MOV A,R0

3. direct － access the data in the RAM with address

MOV A,01FH

4. register indirect － the register holds the RAM

address of the data

MOV A,@R0

5. indexed － for on-chip ROM access

MOVC A,@A+DPTR

8

Immediate Addressing Mode

• The source operand is a constant.

• When the instruction is assembled, the operand

comes immediately after the opcode.

• The immediate vale can be loaded into any of the

registers.

– The immediate data must be preceded by the pound

sign, ‘＃’.

– The immediate value is bounded by the size of register.

– Please use the simulation tools to find the the machine

code and the content of registers after execution.

– See Tables 10 ＆11 (page 418).

9

Example of Immediate Mode（（（（1/2））））

• Immediate Mode：1 0000 74 25 MOV A,#25H ;A=25H

2 0002 7C 3E MOV R4,#62 ;A=62=3EH

3 0004 90 45 21 MOV DPTR,#4521H

• Instruction Opcodes in Table 11

Hex code Mnemonic Operands Byte

74 MOV A, #data 2

7C MOV R4, #data 2

90 MOV DPTR, #data 3

10

Example of Immediate Mode（（（（2/2））））

• Immediate Mode：1 0000 74 25 MOV A,#25H ;A=25H

2 0002 7C 3E MOV R4,#62 ;A=62=3EH

3 0004 90 45 21 MOV DPTR,#4521H


Mnemonic Oscillator Period

MOV A, #data 12

MOV Rn, #data 12

MOV DPTR, #data 24

11

EQU

• The EQU directive is used in the immediate

addressing mode.

1 0000 ORG 0H

2 0000 COUNT EQU 30

3 0000 7C 1E MOV R4,#COUNT

4 0002 90 02 00 MOV DPTR,#MYDATA

5 0200 ORG 200H

6 0200 41 6D 65 72 69 MYDATA DB "America"

7 0207 END

12

Addressing Mode 2


MOV A,#01FH


MOV A,R0


MOV A,01FH


address of the data

MOV A,@R0


MOVC A,@A+DPTR

13

Register Addressing Mode

• Register addressing mode involves the use of

registers to hold the data.

– The source and destination registers must match in size.

– The movement of data between Rn registers is not

allowed. “MOV R4,R7” is illegal.

• You can find that the opcode in register addressing

mode is short！

14

Example of Register Mode（（（（1/2））））

• Register Mode：1 0000 E8 MOV A,R0

2 0001 FA MOV R2,A

3 0002 2D ADD A,R5


Hex code Mnemonic Operands Byte

E8 MOV A,R0 1

FA MOV R2,A 1

2D ADD A,R5 1

15

Example of Register Mode（（（（2/2））））

• Register Mode：1 0000 E8 MOV A,R0

2 0001 FA MOV R2,A

3 0002 2D ADD A,R5



MOV A, Rn 12

MOV Rn, A 12

ADD A, Rn 12

16

Section 5.2

Accessing Memory Using Various

Address Modes

17

Addressing Mode 3


MOV A,#01FH


MOV A,R0


MOV A,01FH


address of the data

MOV A,@R0


MOVC A,@A+DPTR

18

Direct Addressing Mode

• There are 128 bytes of RAM in the 8051.

• The RAM has been assigned address 00 - 7FH.

– 00-1FH：the register banks and stack

– 20-2FH：bit-addressable space to save single-bit data

– 30-7FH：scratch pad RAM

• In direct addressing mode, the data is in a RAM

memory location whose address is known, and this

address is given as a part of the instruction.

– If an number begins without a pound sign, ‘＃’, then

Assembler think it as the RAM address.

19

Example of Direct Mode（（（（1/2））））

• Direct Mode：1 0000 A8 40 MOV R0,40H

2 0002 F5 56 MOV 56H,A

3 0004 90 45 21 MOV DPTR,#4521

4 0007 75 83 45 MOV DPH,#45H

5 000A 75 82 21 MOV DPL,#21H

• Instruction Opcodes Table 11

Hex code Mnemonic Operands Bytes

A8 MOV R0, data addr. 2

F5 MOV data addr., A 2

75 MOV data addr., #data 3

20

Example of Direct Mode（（（（2/2））））

• Direct Mode：1 0000 A8 40 MOV R0,40H

2 0002 F5 56 MOV 56H,A

3 0004 90 45 21 MOV DPTR,#4521

4 0007 75 83 45 MOV DPH,#45H

5 000A 75 82 21 MOV DPL,#21H

• Instruction Opcodes Table 10


MOV Rn, direct 24

MOV direct, A 12

MOV direct, #data 24

21

Register Bank（（（（1/2））））

• If we use register bank 0, then the following

instructions 2＆3 do the same works：

1 0000 7C 64 MOV R4,#100

2 0002 E5 04 MOV A,4 ;direct mode

3 0004 EC MOV A,R4 ;register mode

– Initially, the 8051 uses the register bank 0.

– R4 has RAM address 04H.

22

Register Bank（（（（2/2））））

• If we use register bank 1, then the following

instructions 3＆4 do the different works：1 0000 D2 D3 SETB RS0 ;RS0=1

2 0002 7C 64 MOV R4,#100

3 0004 E5 04 MOV A,4 ;A=0

4 0006 EC MOV A,R4 ;A=100=64H

– RS1=PSW.4=0 ＆ RS0=PSW.3=1 ⇒ register bank 0

– Initially, the content of RAM is 00H.

– R4 has RAM address 0CH.

– RAM 0CH has the value 100.

23

SFR（（（（Special Function Register））））

• There are many special functions registers in the

8051. We call them SFR.

– Example：A, B, PSW, and DPTR

• The 8051 Assembler provides that the SFR can be

accessed by their name or by their addresses.

• See Table 5-1 for SFR addresses

• The SFR have addresses between 80H and FFH.

• Not all the address space of 80 to FF is used by the

SFR.

24

Table 5-1: Special Function Register (SFR) Addresses（1/2）

Symbol Name Address

ACC* Accumulator 0E0H

B* B register 0F0H

PSW* Program status word 0D0H

SP Stack pointer 81H

DOTR Data pointer 2 bytes

DPL Low byte 82H

DPH High byte 83H

P0* Port 0 80H

P1* Port 1 90H

P2* Port 2 0A0H

P3* Port 3 0B0H

IP* Interrupt priority control 0B8H

IE* Interrupt enable control 0A8H

TMOD Timer/counter mode control 89H

25

Table 5-1: Special Function Register (SFR) Addresses （2/2）

Symbol Name Address

TCON* Timer/counter control 88H

T2CON* Timer/counter 2 control 0C8H

T2MOD Timer/counter mode control 0C9H

TH0 Timer/counter 0 high byte 8CH

TL0 Timer/counter 0 low byte 8AH

TH1 Timer/counter 1 high byte 8DH

TL1 Timer/counter 1 low byte 8BH

TH2 Timer/counter 2 high byte 0CDH

TL2 Timer/counter 2 low byte 0CCH

RCAP2H T/C 2 capture register high byte 0CBH

RCAP2L T/C 2 capture register low byte 0CAH

SCON* Serial control 98H

SBUF Serial data buffer 99H

PCON Power control 87H

*bit addressable (discussed further in Chapter 8)

26

ACC and Its Address

• ACC has SFR address 0E0H.

1 0000 75 E0 55 MOV 0E0H,#55H

2 0003 74 55 MOV A,#55H

3 0005 D2 E1 SETB A.1

– Compare their code size and execution time.

• “ACC＊”, ＊ means this register is bit

addressable. You can access each bit of ACC

independently.

A.6 A.5 A.4 A.3 A.2 A.1A.7 A.0ACC

SFC addr. 0E7 0E6 0E5 0E4 0E3 0E2 0E1 0E0

27

Example 5-1

Write code to send 55H to ports P1 and P2, using

(a) their names

(b) their addresses.

Solution:

(a) MOV A,#55H ;A=55H

MOV P1,A ;P1=55H

MOV P2,A ;P2=55H

(b) From Table 5-1, P1 address = 80H; P2 address = A0H

MOV A,#55H ;A=55H

MOV 80H,A ;P1=55H

MOV 0A0H,A ;P2=55H

28

Stack

• Another major use of direct addressing mode is

the stack.

• In the 8051 family, only direct addressing mode is

allowed for pushing onto the stack.

29

Example 5-2（（（（1/2））））

Show the code to push R5, R6, and A onto the stack and then pop

them back them into R2, R3, and B.

We want：B = A, R2 = R6, and R3 = R5.

Solution:

PUSH 05 ;push R5 onto stack

PUSH 06 ;push R6 onto stack

PUSH 0E0H ;push register A onto stack

POP 0F0H ;pop top of stack into register B

POP 02 ;pop top of stack into R2

POP 03 ;pop top of stack into R3

30

Example 5-2（（（（1/2））））

• Different assembler provide different instruction

for the stack.

• In our simulation tools, they are the same：1 0000 C0 05 PUSH R5

2 0002 C0 06 PUSH R6

3 0004 C0 E0 PUSH A

1 0000 C0 05 PUSH 05

2 0002 C0 06 PUSH 06

3 0004 C0 E0 PUSH 0E0H

31

Addressing Mode 4


MOV A,#01FH


MOV A,R0


MOV A,01FH


address of the data

MOV A,@R0


MOVC A,@A+DPTR

32

Register Indirect Addressing Mode

• In the register indirect addressing mode, a register

is used as a pointer to the data.

– That is, this register holds the RAM address of the data.

• Only registers R0 and R1 can be used to hold the

address of an operand located in RA.

– Usually, R0 and R1 are denoted by Ri.

• When R0 and R1 hold the addresses of RAM

locations, they must be preceded by the “@” sign.

33

Example of Register Indirect Mode（（（（1/2））））

• Register Indirect Mode：1 0000 75 20 64 MOV 20H,#100

2 0003 78 20 MOV R0,#20H

3 0005 E6 MOV A,@R0

R0 20H

A 64H

RAM

1E 00

1F 00

20 64

21 00

22 00

23 :

1. put 64H to

addr. 20H 2. let R0 be the

data address

3. copy the content in

addr. R0=20H to A

34

Example of Register Indirect Mode（（（（2/2））））

• Register Indirect Mode：1 0000 75 F0 80 MOV B,#080H

2 0003 79 31 MOV R1,#31H

3 0005 A7 F0 MOV @R1,B

R1 31H

B 80H

RAM

2F 00

30 00

31 80

32 00

33 00

34 :1. let B=80H

2. let R0 be the

data address

3. copy B to the RAM

location with addr.

R1=31H

35

Example 5-3 (1/3)

Write a program to copy the value 55H into RAM memory

locations 40H to 45H using

(a) direct addressing mode,

(b) register indirect addressing mode without a loop,

(c) with a loop.

Solution of (a) :

MOV A,#55H

MOV 40H,A

MOV 41H,A

MOV 42H,A

MOV 43H,A

MOV 44H,A

A 55H

RAM

40 55

41 55

42 55

43 55

44 00

45 50

copy A to the RAM

location of addr. 43H

36

Example 5-3 (2/3)

Solution of (b) register indirect addressing mode without a loop

MOV A,#55H ;load A with value 55H

MOV R0,#40H ;load the pointer. R0=40H

MOV @R0,A ;copy A to RAM location where R0

; points to

INC R0 ;increment pointer. Now R0=41H

MOV @R0,A

INC R0 ;R0=42H

MOV @R0,A

INC R0 ;R0=43H

MOV @R0,A

INC R0

MOV @R0,A

R0 43H

A 55H

RAM

40 55

41 55

42 55

43 00

44 00

45 00

37

Example 5-3 (3/3)

Solution of (c) with a loop:

MOV A,#55H ;A=55H

MOV R0,#40H ;load pointer. R0=40H,

MOV R2,#05H ;load counter, R2=5

AGAIN: MOV @R0,A ;copy 55 to RAM location

; R0 points to

INC R0 ;increment R0 pointer

DJNZ R2,AGAIN ;loop until counter = 0

38

Advantage of Register Indirect Addressing

Mode

• One of the advantages of register indirect

addressing mode is that it makes accessing data

dynamic rather than static.

• Solution (c) in Example 5-3 is the most efficient

and is possible only because of register indirect

addressing mode.

– Looping is not possible in direct addressing mode.

– See Examples 5-4, 5-5, too.

• Their use is limited to accessing any information

in the internal RAM.

39

Example 5-4

Write a program to clear 16 RAM locations starting at RAM address

60H.

Solution:

CLR A ;A=0

MOV R1,#60H ;load pointer. R1=60H

MOV R7,#16 ;load counter, R7=10H

AGAIN: MOV @R1,A ;clear RAM location R1

; points to

INC R1 ;increment R1 pointer

DJNZ R7,AGAIN ;loop until counter = 0

40

Example 5-5

Write a program to copy a block of 10 bytes of data from RAM locations starting at 35H to RAM locations starting at 60H.

Solution:

MOV R0,#35H ;source pointer

MOV R1,#60H ;destination pointer

MOV R3,#10 ;counter

BACK: MOV A,@R0 ;get a byte from source

MOV @R1,A ;copy it to destination

INC R0 ;increment source pointer

INC R1 ;increment destination

; pointer

DJNZ R3,BACK ;keep doing it 10 times

41

Addressing Mode 5


MOV A,#01FH


MOV A,R0


MOV A,01FH


address of the data

MOV A,@R0


MOVC A,@A+DPTR

42

Indexed Addressing Mode

• Indexed addressing mode is widely used in

accessing data elements of look-up table entries

located in the program ROM space of the 8051.

– A look-up table is a ROM block where the data is given

previously (then you can access it frequently).

– The instruction used for this purpose is MOVC.

• DPTR can be used to access memory externally

connected to the 8051. See Chapter 14.

• Another register used in indexed addressing mode

is the PC. See Appendix A.

43

MOVC

• Copy the source operand to the destination

operand.

MOVC A, @A+DPTR

– The “C” means code (program code in on-chip ROM).

– A+DPTR is the address of the data element stored in

on-chop ROM.

– Put the ROM value to A.

44

Example of MOVC

• Register Indexed addressing Mode：1 0000 90 00 06 MOV DPTR,#MYDATA

2 0003 E4 CLR A

3 0004 93 MOVC A,@A+DPTR

4 0005 F8 MOV R0,A

5 0006 80 FE HERE: SJMP HERE

5 0008 55 53 41 MYDATA: DB "USA“

– DPTR=#MYDATA=0008H

– A+DPTR=0008H

ROM

00 90

01 00

02 06

03 E4

:

08 55

09 53

0A 41

A 55H

45

Example 5-6 (1/2)

In this program, assume that the word “USA” is burned into ROM

locations starting at 200H, and that the program is burned into

ROM locations starting at 0. Analyze how the program works and

state where “USA” is stored after this program is run.Solution:

ROM

0000 90

0001 02

0002 00

0003 E4

:

0200 55

0201 53

0202 41

A 55H

A+DPTR= 0200H

DPTR 02H 00H

R0 55H U

S

AR1 00H

R2 00H

46

Example 5-6 (2/2)

ORG 0000H ;burn into ROM from 0MOV DPTR,#200H ;DPTR=200HCLR A ;clear A(A=0)MOVC A,@A+DPTR ;get the char spaceMOV R0,A ;save it in R0INC DPTR ;DPTR=201 CLR A ;clear A(A=0)MOVC A,@A+DPTR ;get the next charMOV R1,A ;save it in R1INC DPTR ;DPTR=202 CLR A ;clear A(A=0)MOVC A,@A+DPTR ;get the next charMOV R2,A ;save it in R2

HERE:SJMP HERE ;stay hereORG 200H

MYDATA: DB “USA”END ;end of program

47

Example 5-7 (1/2)

Assuming that ROM space starting at 250H contains “America”, write

a program to transfer the bytes into RAM locations starting at 40H.

Solution of (a) This method uses a counter：：：：ORG 0000

MOV DPTR,#MYDATA ;Initialization MOV R0,#40HMOV R2,#7

BACK: CLR A MOVC A,@A+DPTR

MOV @R0,A

INC DPTR

INC R0

DJNZ R2,BACKHERE: SJMP HERE

ORG 250H

MYDATA: DB “AMERICA”

END

ROM

0000 90

0001 02

0002 50

0003 78

:

0250 41

0251 4D

0252 45

0253 52

A

M

E

R

I

C

A

N

RAM

40 41

41 4D

42 45

43 52

44 49

45 43

46 41

47 4E

A 41

R0 40

DPTR 02 50

48

Example 5-7 (2/2)

Solution of (b) This method uses null char for end of string：：：：ORG 0000

MOV DPTR,#MYDATA

MOV R0,#40H ;No “MOV R2,#7”

BACK: CLR A

MOVC A,@A+DPTR

JZ HERE ;if A=0

MOV @R0,A ;leave the block

INC DPTR

INC R0

SJMP BACKHERE: SJMP HERE

ORG 250H

MYDATA: DB “AMERICA”,0 ;notice null char

;for end of string

END

49

Example 5-8

Write a program to get the x value from P1 and send x2 to P2,

continuously.

Solution:

ORG 0MOV DPTR,#XSQR_TABLE MOV A,#0FFH MOV P1,A ;P1 as INPUT PORT

BACK: MOV A,P1 ;GET XMOVC A,@A+DPTR ;Count the addr.MOV P2,A ;Issue it to P2SJMP BACK ORG 300H

XSQR_TABLE:DB 0,1,4,9,16,25,36,49,64,81END

50

Example 5-9

Answer the following questions for Example 5-8.

(a) Indicate the content of ROM locations 300-309H.

(b) At what ROM location is the square of 6, and what value should

be there?

(c) Assume that P1 has a value of 9: what value is at P2 (in binary)?

Solution:

(a) All values are in hex.300 = (00) 301 = (01) 302 = (04) 303 = (09)

304 = (10) 4×4=16=10H305 = (19) 5×5=25=19H306 = (24) 6×6=36=24H307 = (31) 308 = (40) 309 = (51)

(b) ROM Addr.=306H; the value 24H=36

(c) P2 = 01010001B=51H=81 in decimal.

51

You are able to

• List the 5 addressing modes of the 8051

microcontroller

• Contrast and compare the addressing modes

• Code 8051 Assembly language instructions using

each addressing mode

• List the SFR（special function registers）address

• Discuss how to access the SFR

• Manipulate the stack using direct addressing mode

• Code 8051 instructions to manipulate a look-up

table

52

Homework

• Chapter 5 Problems：2,3,8,11,12,13

• Note:

– Please write and compile the program of Problems

8,11,12,13.

chapter 5 8051 addressing modes -...

Documents