Lecture 5. EEE3410 Microcontroller Applications Department of Electrical Engineering Assembly Language Programming (1)

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1 Department of Electrical Engineering Lecture Assembly Language Programming (1) 1

2 In this Lecture 8051 programming model Assembly language syntax Operation codes and operands Machine instructions How 8051 interprets binary data Example of Assembly language program 8051 Instruction Set Instruction time calculation Data movement instructions 2

3 Introduction You are now learning to be a programmer (a person who writes programs) You should know the internal structure of the 8051 to do good programming Programming model how programmers see the 8051 In the model, programmers can access the registers and accumulators with programs 3

4 8051 Programming Model 7F 4

5 Assembly Language Syntax Syntax = format/rule [label:] mnemonic [operands][;comment] Items in square brackets are optional 5

6 Mnemonic/ Operation Code (Opcode( Opcode) ( 操作碼 ) Program ( 電腦程式 ) = a set of instructions ( 指令 ) All computers work according to the program All instructions contain a verb ( 動詞 ), which tells the computer what to do This verb is called mnemonic/operation code e.g. MOV R0, #12h MOV is the opcode 6

7 Operand ( 操作元 ) Apart from Opcode, an instruction also includes object) to act on. The object is called an operand. Operand is optional. Instructions can have one, two or no operands. e.g. MOV R0, #12h --- R0 and #12h are two operands INC R1 --- R1 is the only one operand NOP --- no operand follows 7

8 Mnemonic / Machine Instructions Primitive ( 基本的 ) operations of the 8051 e.g. ADD A, #34 Each microprocessor has its unique instruction set Machine instructions = opcode + operand(s) Unary operand: 1 operand, e.g. CLR A Binary operand: 2 operands, e.g. ADD A, #10 8

9 Binary nature of machine instruction Unlike human, computers do not know verbal instructions; they only know 0s and 1s Binary data: program should be in a stream of 0s and 1s It is also called Machine Language For convenient purpose, machine instructions are usually expressed in hexadecimal (i.e. base-16) format, called machine codes. e.g. Mnemonic : ADD A, #10h Equivalent Machine codes: 24h 10h (hexadecimal) 9

10 How 8051 Interprets Binary Data EEE3410 Microcontroller Applications Machine instructions can be 3 bytes (24 bits), 2 bytes (16 bits) or 1 byte (8 bits) long The 1st byte (8 bits) is the operation code (opcode) The remaining byte(s) is/are the supplement data for the operation code 1-byte instruction: Contain the opcode only. Actions do not need supplement data. e.g. Mnemonic NOP ADD A, R0 INC A Equivalent Machine codes 00h (hexadecimal) 28h 04h 10

11 How 8051 Interprets Binary Data 2-byte instruction: The 1st byte is the opcode. The 2nd byte may be either an immediate data (a number) or the low-order byte of an address e.g. Mnemonic ADD A, #30h ADD A, 30h Equivalent Machine codes 24h 30h (hexadecimal) 25h 30h 3-byte instruction: The 1st byte is the opcode. The 2nd and the 3rd byte are the high-order byte and the low-order byte of an 16-bit memory address e.g. Mnemonic Equivalent Machine codes LJMP 0130h 02h 01h 30h (hexadecimal) 11

12 Content of the List file of an assembly language program LOC OBJ LINE SOURCE COUNT EQU 25H ;COUNT = 25H ORG 0H ;start (origin) at location AD25 3 MOV R5, COUNT ;load 25H into R F34 4 MOV R7, #34H ;load 34H into R MOV A, #0 ;load 0 into A D 6 ADD A, R5 ;add contents of R5 to A, now A = A + R F 7 ADD A, R7 ;add contents of R7 to A, now A = A + R ADD A, #12H ;add to A value 12H, now A = A + 12H 000A 00 9 NOP ;no operation 000B 80FE 10 HERE: SJMP HERE ;stay in this loop ORG 20H DATA1: DB 39H ; D DATA2: DB "America ; END ;end of assembly source file SYMBOL TABLE LISTING N A M E T Y P E V A L U E ATTRIBUTES COUNT N NUMB 0025H A DATA1 C ADDR 0020H A DATA2 C ADDR 0021H A HERE C ADDR 000BH A REGISTER BANK(S) USED: 0 ASSEMBLY COMPLETE. 0 WARNING(S), 0 ERROR(S) Source program in Assembly Language (Mnemonics) Machine codes stored in memory 12

13 Pseudo-instructions/Directives Beside mnemonics, directives are used to define variables and memory locations where the machine codes are stored. These directives are interpreted by assembler during the conversion of the assembly language program into machine codes. - ORG (origin) indicates the beginning of the address of the instructions. The number that comes after ORG can be either hex or decimal. - END indicates to the assembler the end of the source assembly instructions. 13

14 Pseudo-instructions/Directives - EQU (equate) used to define a constant without occupying a memory location. It does not set aside storage for a data item but associates a constant value with a data label so that when the label appears in the program. Its constant value will be substituted for the label. - DB (define byte) DB (define byte) used to define 8-bit data and store them in assigned memory locations. Define data can be in decimal, binary, hex, or ASCII formats. 14

15 8051 Instruction Set There are roughly 241 instructions in 8051 instruction set. They can be grouped into 6 groups. - Data movement instructions - Arithmetic operation instructions - Logic and byte operation instructions - Bit operation instructions - Program branching instructions - Special instruction Refer to the 8051 instruction set summary for details 15

16 Instruction Time Calculation Depends on the clock frequency of oscillator 1 machine cycle = 12 oscillator cycle For a MHz oscillator, the time for 1 machine cycle is: 1 machine cycle =.(12) = µ s T 6 16

17 Example 5-15 For an 8051 system of MHz, find how long it takes to execute each of the following instructions. (a) MOV R3, #55 (b) DEC R3 (c) DJNZ R2, target (d) LJMP (e) SJMP (f) NOP (g) MUL AB 17

18 Solution to Example 5-15 Instruction (a) MOV R3, #55 (b) DEC R3 (c) DJNZ R2, target (d) LJMP (e) SJMP (f) NOP (g) MUL AB Machine cycles Time to execute 1 x 1.085µs = µs 1 x 1.085µs = µs 2 x 1.085µs = 2.17 µs 2 x 1.085µs = 2.17 µs 2 x 1.085µs = 2.17 µs 1 x 1.085µs = µs 4 x 1.085µs = 4.34 µs 18

19 Example 5-25 EEE3410 Microcontroller Applications Find the execution time for the following program segment, assuming a crystal frequency of MHz. ORG MOV MOV MUL PUSH MOV PUSH ADD MOV SWAP END 00h A, #100 B, #2 AB 01 A, B 02 A, R1 30h, A A Machine Cycle

20 Solution to Example 5-25 Total machine cycles required, [ ] = 16 machine cycles The execution time of the program segment is : Execution time = 16 x 1.085µs = 17.36µs # 20

21 Data Movement Instructions Moving data from a source to a destination MOV MOVX MOVC PUSH POP XCH XCHD 21

22 The MOV Instruction MOV destination, source Used for data movement inside the 8051 No change in original (source) data: a copy of it is made and the copy is then moved to the destination Source can be data/register/memory locations Destination can be register/memory locations Note that the destination cannot be an immediate data 22

23 Examples of MOV Instruction MOV A, 80h - Copy data from 80h (port 0) to register A EEE3410 Microcontroller Applications MOV 80h, A - Copy data from register A to RAM address 80h (port 0) MOV 3Ah, #3Ah - Copy immediate data 3Ah to RAM location 3Ah MOV R0, 12h - Copy data from RAM location 12h to R0 MOV 5Ch, A - Copy data from register A to RAM location 5Ch MOV 08Ah, 77h - Copy data from RAM address 77h to 08Ah (IE register) 23

24 The MOVX Instruction MOVX destination, source X means the data movement is external to the 8051 data movement is between the external RAM and the (internal) register A All MOVX instructions must involve register A All MOVX instructions must use indirect addressing mode Operation is similar to the MOV instruction 24

25 Examples of MOVX Instruction A - Copy data from A to the 16-bit address in DPTR A - Copy data from A to the 8-bit address in R0 MOVX - Copy data from the 8-bit address in R1 to A MOVX - Copy data from the 16-bit address in DPTR to A 25

26 The MOVC Instruction MOVC destination, source C means the data movement is from the source address in code ROM to register A Used with data transfer between internal/external ROM and register A, e.g. reading a table from the program memory 26

27 Examples of MOVC Instruction MOVC (Copy the code byte to A) - This code byte is found at the ROM address formed by adding A and the DPTR MOVC (Copy the code byte to A) - This code byte is found at the ROM address formed by adding A and the PC Note that the PC is incremented by 1 before added to A to form the final address of the code byte 27

28 Examples of MOVX and MOVC Instructions 8051 Read Write Read Write Read A register Data Data R0 or R1 DPTR External RAM Internal and External ROM DPTR + A PC + A MOVC A,@A + DPTR MOVC A,@A + PC 28

29 The Stack EEE3410 Microcontroller Applications Stack: an area of internal RAM for fast data storage and retrieval Stack operation follows the first-in-last-out (FILO), or equivalently, last-in-first-out (LIFO) logic (i.e. the stack grows up as data is stored) Stack pointer (SP): a register that stores the address of the stack s top item Adding data to the stack: SP increments (+1) and then data is stored on the stack Getting data from the stack: Data is read from the stack and then SP decrements ( 1) 29

30 The PUSH Instruction PUSH source Copy data from the source address to the stack SP is incremented by 1 before source data is copied to the stack Too many PUSH operations may overflow the stack, (i.e. stack runs out of memory) 30

31 The POP Instruction POP destination Copy data from the stack to the destination address SP is decremented by 1 ( 1) after data is copied from the stack 31

32 Other Remarks SP register is set to 07h when the 8051 is reset To prevent the stack from running out of memory, programmer needs to initialize the SP to a value above the highest address likely to be used by the program This value is usually above the register banks The first PUSH operation writes data to R0 in bank 1(08h) SP rolls over to 00h after it reaches FFh PUSH > 7Fh will result in error as the RAM ends at address 7Fh 32

33 Summary of PUSH and POP SP + 2 Push Y Y Pop Y SP SP + 1 Push X X Pop X SP - 1 SP SP - 2 Increment Before PUSHing Internal RAM Decrement After POPing 33

34 Examples MOV 81h, #30h ; Copy the immediate data 30h to SP MOV R0, #0ACh ; Copy the immediate data ACh to R0 ; (i.e. 00h) PUSH 00h ; SP=31h, address 31h contains the ; number ACh PUSH 00h ; SP=32h, address 32h contains the ; number ACh POP 01h ; SP=31h, address R1 (i.e. 01h) ; contains the number ACh POP 80h ; SP=30h, port 0 latch (i.e. 80h) ; contains the number ACh 34

35 Data Exchanging XCH and XCHD are data exchange instructions data movement is bi-directional (i.e. source destination) Data exchanging operations are internal to the 8051 All data exchanging operations must use register A 35

36 The XCH and XCHD Instructions XCH XCHD destination, source destination, source XCH: Data exchange between register A and the addressed byte XCHD: Data exchange between the lowernibble of A and the addressed byte (Upper-nibble of A remains unchanged) 36

37 Examples XCH A, R7 Exchange bytes between registers A and R7 XCH A, 0F0h Exchange bytes between registers A and B XCH Exchange bytes between register A and address in R1 XCHD Exchange lower-nibble in register A and the address in R1 37

38 Read reference The 8051 Microcontroller and Embedded Systems - Using Assembly and C, Mazidi Chapter 2 P.37 P.63 38

39 Department of Electrical Engineering END of Lecture Assembly Language Programming (1) 1

Assembly Language programming (1)

EEE3410 Microcontroller Applications LABORATORY Experiment 1 Assembly Language programming (1) Name Class Date Class No. Marks Familiarisation and use of 8051 Simulation software Objectives To learn how