1. Chapter 5 8051 Addressing Modes

2. 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。

3. Sections 5.1 Immediate and register addressing modes
5.2 Accessing memory using various address modes
5.3 Bit addresses for I/O and RAM
5.4 Extra-128-byte on-chip RAM in 8052
5.3 節 bit address的部份是原本 chapter 8 的部份.

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
We use MOV as an example. One can use any instruction as long as that instruction supports the addressing mode.
先不管這些名詞的意義, 先猜猜看有哪些方法可以存取記憶體 ROM 和 RAM.
我們只是用MOV來當範例, 只要 instruction 本身有支援的 addressing mode, programmer 都可以拿來用.
Ex: INC 包括 2,3,4 三種addressing mode: “INC R0”, “INC 20H”. 見 Table 10.

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
5. indexed － for on-chip ROM access
MOVC

8. Immediate Addressing Mode
The source operand is a constant.
The immediate value 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 614).
When the instruction is assembled, the operand comes immediately after the opcode.

9. Example of Immediate Mode（1/2）
MOV A,#25H ;A=25H
C 3E MOV R4,#62 ;R4=62=3EH
Instruction Opcodes in Table 11
Hex code Mnemonic Operands Byte
MOV A, #data
7C MOV R4, #data
Instruction Opcodes in Table 10
Mnemonic Oscillator Period
MOV A, #data
MOV Rn, #data
Rn - Register R7-R0 of the currently selected Register Bank. in ‘Table 10 page 418

10. Example of Immediate Mode（2/2）
MOV DPTR,#4521H
Instruction Opcodes in Tables 10, 11
Hex code Mnemonic Operands Byte Oscillator Period
MOV DPTR, #data
DPTR =DPH+DPL：
MOV DPL,#21H
MOV DPH,#45H
DPTR has high byte DPH (address 83) and low byte DPL (address 82)
However, 8051 CPU does not provide a special machine instructions for MOV DPH, #data or MOV DPL,#data,
Although 8051 Assembler provide the assembly instructions for DPH and DPL
要注意存入DTPR的值要小於65536.

11. EQU (1/2) The EQU directive is used in the immediate addressing mode.
ORG 0H
COUNT EQU 30
C 1E MOV R4,#COUNT
MOV DPTR,#MYDATA
ORG 200H
D MYDATA:DB "America"
END
Assembler will transfer the COUNT=30=1EH to the immediate value.
MYDATA is the address of “America”, point to “A”
Assembler will transfer the MYDATA to the address 200H.
ORG 0H
COUNT EQU 30
C 1E MOV R4,#COUNT ;R4=30
MOV DPTR,#MYDATA ;DPTR=200H
ORG 200H
D MYDATA DB "America"
END

12. EQU (2/2)
In Pass 1, assembler reads these directives and creates a table:
COUNT 1EH
MYDATA 200H
In Pass 2, assembler transfers COUNT and MYDATA into numbers and creates machine codes:
MOV R4,#COUNT → 7C 1E
MOV DPTR,#MYDATA →

13. Addressing Mode 2 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
5. indexed － for on-chip ROM access
MOVC

14. Register Addressing Mode
Register addressing mode involves the use of registers to hold the data.
Register means Rn, A & CY.
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！
Register 指的是 Rn, A, DPTR & CY 等常用的 register.
因為這些 register 的存取特別常用, 所以會有特別的 instructions 及相對應之 opcode. Opcode 會特別短, 因為不需要將 register 轉成相對應之 RAM address.

15. Example of Register Mode（1/2）
E MOV A,R0
FA MOV R2,A
D ADD A,R5
Instruction Opcodes in Table 11
Hex code Mnemonic Operands Byte
E MOV A,R
FA MOV R2,A
2D ADD A,R

16. Example of Register Mode（2/2）
E MOV A,R0
FA MOV R2,A
D ADD A,R5
Instruction Opcodes in Table 10
Mnemonic Oscillator Period
MOV A, Rn
MOV Rn, A
ADD A, Rn

17. Section 5.2 Accessing Memory Using Various Address Modes

18. Addressing Mode 3 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
5. indexed － for on-chip ROM access
MOVC

19. Direct Addressing Mode (1/2)
There are 128 bytes of RAM in the 8051.
The RAM has been assigned address FH.
00-1FH：the register banks and stack
20-2FH：bit-addressable space to save single-bit data
30-7FH：scratch pad RAM
There is no name for some RAM locations -- so we need to use direct address mode to access them.
If an number begins without a pound sign, ‘＃’, then Assembler think it as the RAM address.
It is most often used to access RAM location 30-7FH.
This is due to the fact that register bank locations are accessed by the register names of R0-R7 -- register mode.
There is no such name for other RAM locations. -- so use direct address mode.
“Direct” 直接的 means “data address”, 直接給予其記憶體位址.

20. Direct Addressing Mode (2/2)
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.
We can use the direct addressing mode to access
128-byte on chip RAM
Special Function Registers (they are RAM addresses too.)

21. Example of Direct Mode（1/2）
A MOV R0,40H
F MOV 56H,A
MOV DPTR,#4521H
MOV DPH,#45H
5 000A MOV DPL,#21H
Instruction Opcodes Table 11
Hex code Mnemonic Operands Bytes
A MOV R0, data addr
F MOV data addr., A
MOV data addr., #data
DPTR has high byte DPH (address 83) and low byte DPL (address 82)

22. Example of Direct Mode（2/2）
A MOV R0,40H
F MOV 56H,A
MOV DPTR,#4521H
MOV DPH,#45H
5 000A MOV DPL,#21H
Instruction Opcodes Table 10
Mnemonic Oscillator Period
MOV Rn, direct
MOV direct, A
MOV direct, #data

23. Register Bank（1/2）
If we use register bank 0, then the following instructions 2＆3 do the same works：
C 64 MOV R4,#100
E5 04 MOV A,4 ;direct mode
EC MOV A,R4 ;register mode
Initially, the 8051 uses the register bank 0.
R4 has RAM address 04H.

24. Register Bank（2/2）
If we use register bank 1, then the following instructions 3＆4 do the different works：
D2 D3 SETB RS0 ;RS0=1
C 64 MOV R4,#100
E5 04 MOV A,4 ;A=0
EC MOV A,R4 ;A=100=64H
RS1=PSW.4=0 ＆ RS0=PSW.3=1  register bank 0
The bit address of RS0 is D3.
Initially, the content of RAM is 00H.
R4 has RAM address 0CH.
RAM 0CH has the value 100.

25. SFR（Special Function Register）
There are many special functions registers in the 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 80H to FFH is used by the SFR.
The addresses 00-0FH are addresses of RAM memory inside the So SFR needs to use the addresses 80H to FFH.
The addresses 80H to FFH is not a really RAM.
The unused locations 80H to FFH are reserved and must not be used by the 8051 programmer.
We can access SFR by direct addressing mode.

26. 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
DPTR
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

27. 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)

28. ACC and Its Address ACC has SFR address 0E0H.
E0 55 MOV 0E0H,#55H
MOV A,#55H
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.
E0 55 MOV 0E0H,#55H : 2 bytes 2 MCs
MOV A,#55H : 2 bytes 1 MC
For address 0E0H, if the instruction is for a bit “SETB”, then it refers to A.0
if the instruction is for a byte “MOV”, then it refers to A.
SFR addr. 0E7 0E E5 0E E3 0E2 0E1 0E0
ACC
A.7
A.6
A.5
A.4
A.3
A.2
A.1
A.0

29. 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 = 90H; P2 address = A0H
MOV A,#55H ;A=55H
MOV 90H,A ;P1=55H
MOV 0A0H,A ;P2=55H

30. 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 or for popping from the stack.
That is “PUSH A” is invalid. ”PUSH 3” is pushing the content of RAM address 3H.

31. Example 5-2
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:
; Register bank 0
PUSH ;push R5 onto stack C0 05
PUSH ;push R6 onto stack C0 06
PUSH 0E0H ;push register A onto stack
POP 0F0H ;pop top of stack into register B
POP ;pop top of stack into R2
POP ;pop top of stack into R3
C PUSH R5
C PUSH R6
C0 E0 PUSH A
C PUSH 05
C PUSH 06
C0 E0 PUSH 0E0H

32. Addressing Mode 4 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
5. indexed － for on-chip ROM access
MOVC

33. 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 RAM.
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.
So it is called “indirect”.
If you forget sign, 8051 Assembler will use the content of Ri.
使用 indirect addressing mode 的時機, (1) 通常是因為要處理相連的一串 memory, (2) 改 pointer R0/R1 會比改整個程式中的 address 有彈性 (3) 8052 的 upper 128-byte RAM 只能用 indirect addressing mode 處理.

34. Example of Register Indirect Mode（1/2）
MOV 20H,#100
MOV R0,#20H
E MOV
RAM
1. put 64H to addr. 20H
2. let R0 be the data address
1E F :
R H
3. copy the content in addr. R0=20H to A
A H

35. Example of Register Indirect Mode（2/2）
F0 80 MOV B,#080H
MOV R1,#31H
A7 F0
RAM
3. copy B to the RAM location with addr. R1=31H
2F :
2. let R1 be the data address
R H
1. let B=80H
B H

36. Example 5-3 (1/3)
Write a program to copy the value 55H into RAM memory locations 40H to 44H 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
RAM
copy A to the RAM location of addr. 43H
A H

37. 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
;copy A to RAM location where R0
; points to
INC R ;increment pointer. Now R0=41H
INC R ;R0=42H
INC R ;R0=43H
INC R0
RAM
R H
A H

38. 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 R ;increment R0 pointer
DJNZ R2,AGAIN ;loop until counter = 0
在最後一回合, INC R0 得到 R0=45, DJNZ R2,AGAIN 中R2 由 1減為 0 就離開 loop, 故A沒有寫入45H.

39. 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. (Flexible!)
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.
Use looping is more flexible too.
R0&R1 是唯一可用於 indirect addressing mode的register, 所以若要存取 ROM, 則沒有辦法用R0或R1來表示, 也就是說 indirect addressing mode 無法使用.
If we need to access externally connected RAM or on-chip ROM, we need a 16-bit pointer. In such cases, the DPTR register is used. (必須用 indexed addressing mode, 不可以用 indirect addressing mode)

40. 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 R ;increment R1 pointer
DJNZ R7,AGAIN ;loop until counter = 0

41. 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 ;get a byte from source
MOV @R1,A ;copy it to destination
INC R ;increment source pointer
INC R ;increment destination
; pointer
DJNZ R3,BACK ;keep doing it 10 times

42. Addressing Mode 5 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
5. indexed － for on-chip ROM access
MOVC

43. Indexed Addressing Mode
Indexed addressing mode is used to access 8051 on-chip ROM.
It 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).
Ex: MOVC

44. MOVC MOVC: to access internal ROM
Transfer data between internal ROM and A.
The “C” means code
16-bit memory address is held by DPTR or PC
MOV DPTR,# ;ROM address
MOVC
;access the ROM data with address A+DPTR
MOVC
;access the ROM data with address A+PC
MOVC 的用法包括 MOVC 及 MOVC
除了可以用 DPTR 存 ROM address, another register used in indexed addressing mode is the PC. See Appendix A.
PC的值不用事先給定.

45. Example of MOVC Register Indexed addressing Mode：
MOV DPTR,#MYDATA
E CLR A
MOVC
F MOV R0,A
FE HERE: SJMP HERE
MYDATA: DB "USA"
DPTR=#MYDATA=0008H
A+DPTR=0008H
ROM E4
A 41
A H

46. MOVX MOVX: to access external RAM/ROM connected to the 8051
Transfer data between external memory and A.
See Chapter 14 in page 431.
The “X” means external (memory space must be implemented externally).
Note: RAM can be read and write.
16-bit external memory address is held by DPTR
MOVX or MOVX @DPTR,A
8-bit external memory address is held by R0/R1
MOVX or MOVX @Ri,A
只能用 register A.
External memory 可以是 (1) 64k data ROM, (2) 64k data ROM 加上 64k program ROM, (3) 64k data RAM, (4) 64k data/program ROM, (5) 64k data ROM 加上 64k program ROM 加上 64k program RAM (6) 256k Data RAM
Program ROM 中可以有 code 及 data, 但 data ROM 無法讓 program 共享.
要能存取 external memory, 除了用 MOVX, 硬體上需要 control pins 來控制讀取那一個 memory, 寫入或讀出.

47. 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 E4 :
DPTR H H
Clear A= A+DPTR= 0200H
U S A
R H
A H
R H
R H

48. Example 5-6 (2/2) ORG 0000H ;burn into ROM from 0
MOV DPTR,#200H ;DPTR=0200H
CLR A ;clear A(A=0)
MOVC ;get the char space
MOV R0,A ;save it in R0
INC DPTR ;DPTR=201
MOVC ;get the next char
MOV R1,A ;save it in R1
INC DPTR ;DPTR=202
MOV R2,A ;save it in R2
HERE:SJMP HERE ;stay here
ORG 200H
MYDATA: DB “USA”
END ;end of program

49. 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,#40H
MOV R2,#7
BACK: CLR A
MOVC
MOV @R0,A
INC DPTR
INC R0
DJNZ R2,BACK
HERE: SJMP HERE
ORG 250H
MYDATA: DB “AMERICA”
END
ROM
RAM : D D
AME R I C A
A 41
R0 40
DPTR

50. 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
JZ HERE ;if A=0
MOV @R0,A ;leave the block
INC DPTR
INC R0
SJMP BACK
HERE: SJMP HERE
ORG 250H
MYDATA: DB “AMERICA”,0 ;notice null char
;for end of string
END

51. Example 5-8
Write a program to get the x value from P1 and send x2 to P2,
continuously.
Solution:
ORG 0
MOV DPTR,#XSQR_TABLE
MOV A,#0FFH
MOV P1,A ;P1 as INPUT PORT
BACK: MOV A,P ;GET X
MOVC ;Count the addr.
MOV P2,A ;Issue it to P2
SJMP BACK
ORG 300H
XSQR_TABLE:
DB 0,1,4,9,16,25,36,49,64,81
END
Please trace the program by set A=0, then A=1, then A=2,...
BACK: 應放在 MOV A,#0FFH的位置才對.

52. Example 5-9 Answer the following questions for Example 5-8.
(a) Indicate the content of ROM locations H.
(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=10H
305 = (19) 5×5=25=19H
306 = (24) 6×6=36=24H
307 = (31) 308 = (40) 309 = (51)
(b) ROM Addr.=306H; the value is 24H=36
(c) P2 = B=51H=81 in decimal.

53. Example 5-10
Write a program to toggle P1 at total of 200 times. Use RAM location 32H to hold your counter value instead of registers R0-R7.
Solution:
MOV P1,#55H
MOV 32H,#200
LOP1: CPL P1
ACALL DELAY
DJNZ 32H,LOP1
Usually, we use R0-R7 of bank 0, 8-1FH for stack.
所以要儲存資料可能要用到scratch pad. 這個範例就是以 direct addressing mode 存取 32H 的範例.
可以不要教.

54. Section 5.3 Bit Addresses for I/O and RAM

55. Bit-addressable RAM
The bit-addressable RAM locations are (byte addresses) 20H to 2FH.
Only 16 bytes of RAM are bit-addressable.
16 * 8 bits = 128 bits (in decimal) = 80H bits (in hex)
They are addressed as 00 to 7FH
The internal RAM locations 20 to 2FH are both byte-addressable and bit-addressable.
We can use only the single-bit instructions in Table 5-2 to access these bits.
Only direct addressing mode is used here.
Note that the bit addresses 80H to F7H belong to SFR.

56. Figure 8-3. 128 Bytes of Internal RAM
Byte address 7F
General purpose RAM 30
Bit-addressable locations 2F 7F 7E 7D 7C 7B 7A 79 78 77 76 75 74 73 72 71 70 6F 6E 6D 6C 6B 6A 69 68 67 66 65 64 63 62 61 60 5F 5E 5D 5C 5B 5A 59 58 57 56 55 54 53 52 51 50 4F 4E 4D 4C 4B 4A 49 48 47 46 45 44 43 42 41 40 3F 3E 3D 3C 3B 3A 39 38 37 36 35 34 33 32 31 30 2F 2E 2D 2C 2B 2A 29 28 27 26 25 24 23 22 21 20 1F 1E 1D 1C 1B 1A 19 18 17 16 15 14 13 12 11 10 0F 0E 0D 0C 0B 0A 09 08 07 06 05 04 03 02 01 00 2E 2D 2C 2B 2A 29 28 27 26 25 24 23 22 21 20 1F
Bank 3 18 17
Bank 2 10 0F
Bank 1 08 07
Default register bank for R0 - R7 00

57. Table 5-2: Single-Bit Instructions
Function
SETB
bit
Set the bit (bit = 1)
CLR
Clear the bit (bit = 0)
CPL
Complement the bit (bit = NOT bit) JB
bit,target
Jump to target if bit = 1 (jump if bit)
JNB
Jump to target if bit = 0 (jump if no bit)
JBC
Jump to target if bit = 1, clear bit (jump if bit, then clear) 

58. Example 5-11
Find out to which byte each of the following bits belongs.
Give the address of the RAM byte in hex.
(a)SETB 42H;set bit 42H to (d)SETB 28H;set bit 28H to 1
(b)CLR 67H;clear bit 67H (e)CLR 12 ;clear bit 12 (decimal)
(c)CLR 0FH;clear bit 0FH (f)SETB 05
Solution:
(a) RAM bit address of 42H belongs to D2 of RAM location 28H
(b) RAM bit address of 67H belongs to D7 of RAM location 2CH
(c) RAM bit address of 0FH belongs to D7 of RAM location 21H
(d) RAM bit address of 28H belongs to D0 of RAM location 25H
(e) RAM bit address of belongs to D4 of RAM location 21H
(f) RAM bit address of belongs to D5 of RAM location 20H

59. Bit-addressable of SFR
Every SFR register is assigned a bye address and ports P0-P3 are parts of the SFR.
8051 I/O ports are all bit-addressable.
Only registers B, A, PSW, IP, IE, SCON, TCON are bit-addressable.

60. Figure 5-2. SFR RAM Address (Byte and Bit) (1/2)
Special Function Registers
Byte address
Bit address
F7 F6 F5 F4 F3 F2 F1 F0
E7 E6 E5 E4 E3 E2 E1 E0
D7 D6 D5 D4 D3 D2 D1 D0
A7 A6 A5 A4 A3 A2 A1 A0
AF AC AB AA A9 A8
B7 B6 B5 B4 B3 B2 B1 B0
BC BB BA B9 B8 FF F0 E0 D0 B8 B0 A8 A0 B
ACC
PSW IP P3 IE P2

61. Figure 5-2. SFR RAM Address (Byte and Bit) (2/2)
Special Function Registers
9F 9E 9D 9C 9B 9A
8F 8E 8D 8C 8B 8A
not bit addressable 99 98 90 8D 8C 8B 8A 89
SBUF
SCON
TH1
TL0 P1
TMOD
DPH 88 87 83 82 81 80
TH0
TL1
TCON
PCON
DPL SP P0 

62. I/O Bit Addresses
When “SETB P1.0” is assembled, it becomes “SETB 90H” since P1.0 has RAM address of 90H.
SETB P1.0
SETB 90H
The machine codes for SETB P1.0 and SETB 90H are both D2 90.

63. Table 5-3: Bit Addresses for All Ports
P0 Addr.
P1 Addr.
P2 Addr.
P3 Addr.
Port’s Bit P P
P2.0 A0
P3.0 B0 D0 P P
P2.1 A1
P3.1 B1 D1 P P
P2.2 A2
P3.2 B2 D2 P P
P2.3 A3
P3.3 B3 D3 P P
P2.4 A4
P3.4 B4 D4 P P
P2.5 A5
P3.5 B5 D5 P P
P2.6 A6
P3.6 B6 D6 P P
P2.7 A7
P3.7 B7 D7

64. Example 5-12
For each of the following instructions, state to which port the bit belongs. Use Figure 5-3.
(a) SETB 86H (b) CLR 87H (c) SETB 92H
(d) SETB 0A7H
Solution:
(a) SETB 86H is for SETB P0.6 (D2 86)
(b) CLR 87H is for CLR P0.7 (C2 87)
(c) SETB 92H is for SETB P1.2 (D2 92)
(d) SETB 0A7H is for SETB P2.7 (D2 A7)

65. Figure 5-3. Bits of the PSW Register
PSW (Program Status Word) register
Bit address D7 D D5 D D D2 D D0
Byte address CY AC --
RS1
RS0 OV P D0
RS1
RS0
Register Bank
Address
00H – 07H 1
08H – 0FH 2
10H – 17H 3
18H – 1FH

66. Example 5-13 Write a program to save the accumulator in R7 of bank 2.
Solution:
CLR PSW.3 ;C2 D3
SETB PSW.4 ;D2 D3
MOV R7,A ;FF

67. Example 5-14
While there are instructions such as JNC and JC to check the carry flag bit (CY), there are no such instructions for the overflow flag bit (OV).
How would you write code to check OV?
Solution:
The OV flag is PSW.2 of the PSW register.
We can use the following instruction to check the OV flag.
JB PSW.2,TARGET ;jump if OV=1
(machine code is 20 D2 rel-addr)
有bit-addressable 這樣的 access 方式, 就可以

68. Example 5-15
Write a program to see if the RAM location 37H contains an even number.
If so, send it to P2.
If not, make it even and then send it to P2.
Solution:
MOV A,37H ;RAM location 37H→A
JNB ACC.0,YES ;If A is even, jump to YES
INC A ;it is odd, make it even
YES:MOV P2,A ;send A to P2

69. Example 5-16
Assume that bit P2.3 is an input and represents the condition of a
door. If it goes high, it means that the door is open.
Monitor the bit continuously.
Whenever it goes high, send a low-to-high pulse to port P1.5 to turn on a buzzer.
Solution:
HERE:JNB P2.3,HERE ;keep monitoring for high
CLR P ;clear bit (P1.5=0)
ACALL DELAY
SETB P1.5
SJMP HERE
low-to-high pulse

70. Example 5-17
The states of bits P1.2 and P1.3 of I/O port P1 must be saved before
they are changed. Write a program to save the status of P1.2 in bit
location 06 and the status of P1.3 in bit location 07.
Solution:
CLR ;clear bit address 06
CLR ;clear bit address 07
JNB P1.2,OVER ;If P1.2=0,jump
SETB ;
OVER:JNB P1.3,NEXT ;If P1.3=0,jump
SETB ;
NEXT:...

71. Example 5-18
Write a program to save the status of bits P1.7 on RAM address
bit 05.
Solution:
CY is used as a bit buffer.
MOV C,P1.7 ;save status of P1.2 on CY
MOV 05,C ;save in RAM bit location 05
The machine code is
A2 97 for MOV C,P1.7
92 05 for MOV 05,C

72. Example 5-19
Write a program to get the status of bits P1.7 and send it to pin P2.0.
Solution:
CY is used as a bit buffer.
HERE: MOV C,P1.7 ;get bit and send to CY
MOV P2.0,C ;send bit to port P2.0
SJMP HERE

73. Using BIT and EQU Directive
Assign bit-addressable I/O and RAM locations
name BIT bit-address
bit-address  name
OVEN_HOT BIT P2.3
HERE:JNB OVEN_HOT,HERE
Examples 5-20 to 5-23
We can also use the EQU directive to assign addresses.
OVEN_HOT EQU P2.3
Examples 5-24 to 5-25

74. Example 5-20
Assume that bit P2.3 is an input and represents the condition of an
oven. If it goes high, it means that the oven is hot.
Monitor the bit continuously.
Whenever it goes high, send a low-to-high pulse to port P1.5 to turn on a buzzer.
Solution:
OVEN_HOT BIT P2.3
BUZZER BIT P1.5
HERE:JNB OVEN_HOT,HERE
CLR BUZZER
ACALL DELAY
CPL BUZZER
SJMP HERE
Similar to Ex5-16, except the use of BIT directive allows us to assign the OVEN_HOT and BUZZER bit to any port.

75. Example 5-21
An LED is connected to pin P1.7. Write a program to toggle the LED forever.
Solution:
LED BIT P1.7 ;using BIT directive
HERE: CPL LED ;toggle LED
LCALL DELAY ;delay
SJMP HERE ;repeat forever

76. Example 5-22
A switch is connected to pin P1.7 and an LED to pin P2.0. Write a program to get the status of SW and send it to the LED.
Solution:
SW BIT P1.7
LED BIT P2.0
HERE: SETB P ;make P1.0 an input
MOV C,SW ;save P1.0 to carry
MOV LED,C ;send to LED
SJMP HERE ;keep repeating
和 Example 4-7 很像
“MOV P2.7, P1.0” is wrong since such an instruction does not exists. 所以必須使用CY當作仲介. However, “MOV P2, P1” is a valid instruction.

77. Example 5-23 (1/2)
Assume that RAM bit location 12H holds the status of whether there has been a phone call or not.
If it is high, it means there has been a new call since it was checked the last time. Write a program to display “New Message” on an LCD if bit RAM 12H is high.
If it is low, the LCD should say “No New Message”.
Solution:
Use CY to hold the status.
Use JNC to check CY flag.
We use “LCALL DISPLAY” to display message (see Chap. 12)

78. Example 5-23 (2/2) PHONBIT BIT 12H
MOV C,PHONBIT ;copy bit location 12H to
JNC NO ;check to see if is high
MOV DPTR,#400H ;address of YES_MG
LCALL DISPLAY ;display
SJMP EXIT ;get out
NO: MOV DPTR,#420H
LCALL DISPLAY
EXIT:
;data to be displayed on LCD
ORG 400H
YES_MG: DB “New Message”
ORG 420H
NO_MG: DB “No New Message”
Display YES_MG
Test
Jump if CY＝0
Not Jump if CY ≠0
EXIT
MOV C, 12H
Display NO_MG
SJMP EXIT

79. Example 5-24 (1/2)
A switch (SW) is connected to pin P1.7. Write a program to check the status of SW and perform the following:
(a) If SW=0, send “No” to P2.
(a) If SW=1, send “YES” to P2.
Solution:
This program is to show how to use EQU by port names.
SW EQU P ;bit address
MYDATA EQU P ;byte address
和 Example 4-6 很像

80. Example 5-24 (2/2) SW EQU P1.7 ;bit address
MYDATA EQU P ;byte address
HERE: SETB SW ;make P1.7 an input
MOV C,SW ;save P1.7 to carry
JC OVER
MOV MYDATA,#’N’;SW=0
MOV MYDATA,#’O’
SJMP HERE
OVER: MOV MYDATA,#’Y’;SW=1
MOV MYDATA,#’E’
MOV MYDATA,#’S’
和 Example 4-6 很像

81. Example 5-25 (1/2)
A switch (SW) is connected to pin P1.7. Write a program to check the status of SW and perform the following:
(a) If SW=0, send “0” to P2.
(a) If SW=1, send “1” to P2.
Solution:
This program is to show how to use EQU by bit/byte addresses.
SW EQU 97H ;bit address
MYDATA EQU 0A0H ;byte address
和 Example 4-6 很像

82. Example 5-25 (2/2) SW EQU 97H ;P1.7 bit address
MYDATA EQU 0A0H ;P2 byte address
HERE: SETB SW ;make P1.7 an input
MOV C,SW ;save P1.7 to carry
JC OVER
MOV MYDATA,#’0’;SW=0, P2=30H
SJMP HERE
OVER: MOV MYDATA,#’1’;SW=1, P2=31H
和 Example 4-6 很像

83. Section 5.4 Extra 128-byte On-chip RAM in 8052

84. The Upper 128 Bytes of RAM in 8052
8052 has 256-bytes RAM.
Lower 128 bytes: addresses 00-7FH
Upper 128 bytes: addresses 80-FFH
However, 80-FFH has been used by SFR.
To send 55H to P1: MOV 90H,#55H
SFRs are accessed by direct addressing model
Where is the location of 90H?
P1? or the RAM location 90H at 8052?

85. Figure 5-4. 8052 On-chip RAM Address SpaceFF
Indirect Access of upper 128-byte RAM
MOV R0,#90H
Upper
128-byte RAM 80 7F
scratch pad 30
執行下面的指令:
MOV 90H,#0FH 組譯後為 F ; 會把 0FH 放入 P1, P1=0FH.
MOV R0,#90H 組譯後為 ; R0=90H
MOV E6 ; 會到 upper 128-byte RAM 中讀出位址為 90H 的值, 存入 A 中. 所以 A=00.
所以使用 register indirect addressing mode 時, 大於等於 80H 的位址都是被當作 upper 128 byte RAM. 2F
Direct Access for SFR
MOV 90H,#55H
MOV P1,#55H
bit-addressable 20 1F
Bank 3 18 17
Bank 2 10 FF 0F
Special Function Registers
Bank 1 P1
90H 08 07
Bank 0 80 00

86. How to Distinguish Them?
We use different addressing modes to access them:
1. To access the SFRs, we use direct addressing mode.
Ex: MOV 90H,#55H
2. To access the upper 128 bytes, we use the indirect addressing mode, which uses R0 and R1 as pointers.
Ex: MOV R0,#90H
R0 and R1 have address values of 80H or higher.
對於 8052, 若用 direct addressing mode, 一定是指 SFR, 如此才能與 8051 相容.
所以要存取 upper 128-byte RAM, 只能用 indirect addressing mode. 或者說 indirect addressing mode 不可以用於 SFR.

87. Example 5-26
Write a program for 8052 to put 55H into the upper RAM locations of 90-99H.
Solution:
MOV A,#55H
MOV R0,#90H
MOV R2,#10
BACK: MOV @R0,A ;indirect addressing mode
INC R ;increment R0 pointer
DJNZ R2,BACK ;loop until counter = 0
SJMP $
END

88. Example 5-27 (1/2) Assume that the on-chip ROM has a message.
Write a program to copy it from code space into the upper memory space starting at address 80H.
Also, as you place a byte in upper RAM, give a copy to P0.
Solution:
This program is similar to Example 5-3.
We use indirect addressing mode to access both the lower and upper 128-byte RAM.
We also use the indexed addressing mode.
和 Example 5-3 很像
對於 8051/8052, indirect addressing mode 都能用來存取 lower 128-byte RAM 及 upper 128-byte RAM,
看課本, 程式執行前 RAM 的值如 Figure 5-5; 程式執行後 RAM 的值如 Figure 5-6.

89. Example 5-27 (2/2) Solution : ORG 0000 MOV DPTR,#MYDATA
MOV R1,#80H ;upper 128-byte RAM
B1: CLR A
MOVC ;read from code
JZ EXIT
MOV @R1,A ;copy to upper RAM
MOV P0,A ;give a copy to P0
INC DPTR
INC R1
SJMP B1
EXIT: SJMP $
ORG 300H
MYDATA: DB “The promise of World Peace”,0
END
類似 Example 5-7.

90. You are able to (1/2)
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

91. You are able to (2/2)
Code 8051 instructions to manipulate a look-up table
Access RAM, I/O, and ports using bit addresses
Discuss how to access the extra 128 bytes of RAM space in the 8052