ME4447/6405 Microprocessor Control of Manufacturing Systems and Introduction to Mechatronics Instructor: Professor Charles Ume LECTURE 7
Reading Assignments Reading assignments for this week and next week Read Chapters 5-8 in Basic Microprocessors and the 6800, by Ron Bishop. Chapter 5 Microcomputers-What Are They? Chapter 6 Programming Concepts Chapter 7 Addressing Modes Chapter 8 M6800 Software There will be questions and answers the rest of this week and next week based on your reading assignment.
Why use Assembly Language? HC12S CPU can only understand instructions written in binary called Machine Language. Writing programs in Machine Language is extremely difficult Mnemonics are simple codes, usually alphabetic, that is representative of instruction it represents (example: LDAA [LoaD Accumulator A]) A program written using Mnemonic Instructions is called Assembly Language program An Assembler can be used to translate Assembly Language program to Machine Language Program
Assembly Language Notations Address: Common term for memory location. Always written in hexadecimal. $4000 is the 4000 th 16 Memory Location (Note: $ signifies hexadecimal) Literal Value: A number used as data in program indicated by #. Can be represented in the following ways: #$FF = hexadecimal number FF #%1011 = binary number 1011 (Note: % signifies binary) #123 = decimal number 123 A Literal Value can be stored in an address Example: Literal Value #$FF is stored in address $4000 Example 2: Literal Value #$FE0A is stored in address $2000 (Note: #$FE is stored in address $2000 #$0A is stored in address $2001)
Assembly Language Directives Directives: Instructions from the programmer to Assembler NOT to microcontroller Example 1: ORG <address> Store translated machine language instructions in sequence starting at given address for any mnemonic instructions that follow Example 2: END Stop translating mnemonics instructions until another ORG is encountered (Note: More will be covered in later lectures)
Left margin of assembly program A Tab (8 white spaces) or Label (Note: A Label is another Assembly Directive and will be covered in later lectures) Assembly Language Format Assembly Directive Or Mnemonic Instruction (Note: Last three options are called Operands) Data that the Assembly Directive uses Or Blank if Mnemonic Instruction does not need Data Or Offset Address used to modify Program Counter by a Mnemonic Instruction Or Data that Mnemonic Instruction uses Or Address where the Data that Mnemonic Instruction will use is stored
In the previous slide, there were several options for the operand: Blank if Mnemonic Instruction does not need Data Offset Address used to modify Program Counter by a Mnemonic Instruction Data that Mnemonic instruction uses Address were Data that Mnemonic instruction uses is stored Which option a programmer uses is defined by the following addressing modes: Inherent Immediate Extended Addressing Modes Direct Indexed Relative Indexed Indirect (Note: All instructions are not capable of all addressing modes. Example: BLE [Branch if Less than or Equal to Zero] is only capable of Relative addressing mode)
Example: Programming Reference Guide Page 6
Blank if Mnemonic Instruction does not need Data If Mnemonic Instruction does not need data then it uses Inherent Addressing Mode Example: Write a program to clear accumulator A. Start programming at address $1000 Solution: ORG $1000 CLRA SWI END CLRA [ CLeaR accumulator A] is an instruction using Inherent Addressing (NOTE: SWI [SoftWare Interrupt] is a mnemonic instruction which tells the 9SC32 to store the content of cpu registers on the stack. Sets the I bit (the interrupt bit) on the CCR. Loads the program counter with the address stored in the SWI interrupt vector, and resume program execution at this location. If no address is stored in the SWI vector, the main program will stop execution at this point. Used The George in this W. course Woodruff to return School control of Mechanical to Mon12 Engineering Program)
Offset Address used to modify Program Counter by a Mnemonic Instruction If Mnemonic Instruction uses operand to modify Program counter then it uses Relative Addressing Mode Usually used in conjunction with Labels (Note: will be explained further in later lectures)
Data that Mnemonic instruction uses The Mnemonic instruction is using Immediate Addressing mode if the operand is Data used by the instruction Example: Write a program to load accumulator A with #$12. Start programming at address $1000 Solution: ORG $1000 LDAA #$12 SWI END LDAA is an instruction using Immediate Addressing mode in this example
Address were Data that Mnemonic instruction uses is stored The following addressing modes apply if the operand is an Address containing Data used by Mnemonic instruction : Direct Data is contained in Memory locations $00 to $FF Address is given as a single byte address between $00 to $FF Instructions using Direct addressing has fastest access to memory Example: LDAA $00 Loads accumulator A with Data value stored at memory location $00 Extended Data is contained in Memory locations $0100 to $FFFF Address is given as a two byte address between $0100 to $FFFF Example: LDAA $2000 Loads accumulator A with Data value stored at memory location $2000
Example Problem 1 Example : Write a program to add the numbers 10 10 and 11 10. Solution ORG $1000 LDAA #$0A *Puts number $0A in acc. A LDAB #$0B *Puts number $0B in acc. B ABA *Adds acc. B to acc. A STAA $00 *Stores results in address $00 SWI *Software interrupt END LDAB and LDAA use immediate addressing mode STAA uses direct addressing mode
Address were Data that Mnemonic instruction uses is stored (Continued) Indexed: Data is located within Memory locations $00 to $FFFF Address is given as a address offset between $00 to $FF plus content of the X or Y register Example: LDX #$2000 LDAA $03,X Loads accumulator A with Data value stored at memory location $2003 X + $03 = $2000 + $03 = $2003 (Note: LDX [ LoaD index register X])
Why is Indexed Addressing Mode needed? Example: Store Data Value #$20 into memory locations $2000 to $3000 Without Indexed Addressing Mode ORG $1000 LDAA #$20 STAA $2000 STAA $2001... STAA $3000 SWI END With Indexed Addressing Mode ORG $1000 LDAA #$20 LDX #$2000 LOOP STAA $00,X INX CPX #$3001 BNE LOOP SWI END Program on the Left is much longer than program on Right
ORG $1000 LDAA #$20 LDX #$2000 LOOP STAA $00,X INX CPX #$3001 BNE LOOP SWI END Why is Indexed Addressing Mode needed? (Continued) Note: LDX [ LoaD accumulator X] STAA [ STore Accumulator A] INX [ INcrement X] CPX [ ComPare X] BNE [Branch if Not Equal] ( is using relative addressing in conjunction with label LOOP ) LOOP, BNE LOOP, INX, and CPX #$3001 creates a loop. Loop1: Data in accumulator A (#$20) is stored at $2000 + $00 Data in X is incremented #$2000 + #$0001 = #$2001 Data in X is compared to #$3001 Not equal so do another loop Loop2: Data in Accumulator A (#$20) is stored at $2001 Data in X is incremented #$2001 + #$0001 = #$2002 Data in X is compared to #$3001 Not equal so do another loop Etc..
Types of Indexed modes of Addressing Indexed addressing may be implemented in multiple ways. The HCS12 CPU uses the X or Y index registers, Stack Pointer or Program Counter as the base index register for the instruction. The offset is then added to the base index register to form an effective address. The different indexed addressing modes are: Constant offset 5-, 9- or 16-bit signed offset The assembler will interpret all hex values < $8000 as positive numbers Hex values with a 1 in the 15 th bit are negative (recall 2 s compliment) The following three statements are equivalent: STAA -8,X Note: offset given in decimal STAA -$08,X Note: offset given in hex STAA $FFF8,X Note: offset given as 16-bit number Accumulator Offset A, B, or D accumulator added to base index register to form address Contents of accumulator is unsigned offset LDAB #$FF STAA B,X Note: Accumulator B is unsigned offset (=255 10 ) Value contained in accumulator A is stored in effective address formed by adding 255 The to George contents W. Woodruff of X School of Mechanical Engineering
Types of Indexed modes of Addressing - Continued Auto Pre-/Post-Increment/Decrement The base index register may be automatically incremented/decremented before or after instruction (Program Counter may not be used as the base register) No offset is available May be incremented/decremented 1 to 8 times Post-Increment: LDX 2,SP+ Note: Index register X is loaded with the contents of the memory location in the stack pointer, then the stack pointer is incremented twice Pre-Increment: LDX 2,+SP Note: the stack pointer is incremented twice, then Index register X is loaded with the contents of the memory location in the stack pointer Pre-Decrement: STAA 1,-X Note: Index register X is decremented, then the contents of Accumulator A are stored in the memory location contained in Index Register X
Why is Pre-/Post-Increment/Decrement Useful? Example: Store Data Value #$20 into memory locations $2000 to $3000 Without Post-Increment ORG $1000 LDAA #$20 LDX #$2000 LOOP STAA $00,X INX CPX #$3001 BNE LOOP SWI END With Post-Increment ORG $1000 LDAA #$20 LDX #$2000 LOOP STAA 1,X+ CPX #$3001 BNE LOOP SWI END Note: 1 refers to the number of post increments, not an offset! Program on the Left requires 1 more byte of program memory and takes 1 more cycle to execute per run through the loop then the program on the right. This may make a large difference when the program is large and complex or when dealing with values larger than 16-bits.
Types of Indexed modes of Addressing - Continued Indexed-Indirect Like other indexed addressing modes, an offset and base index register are added to form an effective address However, the effective address is understood to contain the address of the memory location containing the address to be acted upon. Example: Clear the contents of the memory location pointed to by a pointer located in memory location $2000 ORG $2000 PTR RMB 2 *Note: The user places the address of the memory location to be cleared here ORG $1000 LDX $2000 CLR [$00,X] SWI END (Note: CLR [ CLeaR Memory Location]) Useful for efficiently forming switch case statements and other program flow operations The George in high W. Woodruff level programming School of Mechanical languages Engineering
As stated before the Assembler translates an assembly language program into a machine language program Format of machine language program Address where instruction is located Opcode Instruction Postbyte Operand Or Blank if instruction does not use Operands (Note: This format is for Lecture and Tests only!! The real format the assembler outputs is S19 and will be shown to you in Lab)
Postbyte and Opcode Reference All Mnemonics and associated Op-codes can be found in Programming Reference Guide pages 6-19 Example: Programming Reference Guide Page 12 (Note: LDAA outlined in red)
Postbyte Postbyte allows an op-code to be used for more than one instruction. Determined from Tables 1, 3 or 4 in the programming reference guide Instruction Opcode Postbyte LDAA 0,X A6 00 LDAA $02,SP+ A6 B1 LDAA B,Y A6 ED SUBB $1040,X E0 E0 SUBB D,Y E0 EE SUBB -14,X E0 12 Table 1 (Excerpt)
The George W. Woodruff (Programming School Reference of Mechanical Guide) Engineering
Hand Assembling Example: Assemble the following Program ORG $1000 LDAA #$0A LDAB #$0B ABA STAA $00 SWI END Address Opcode Postbyte Operand $1000 86 0A $1002 C6 0B $1004 18 06 $1006 5A 00 $1008 3F (Note: $1002 since $86 is now at $1000 and $0A is at $1001)
Example Problem 1 (revisited) ORG $1000 LDAB #$0A *Load acc. B with number 0A STAB $1100 *Store acc. B in address $1100 INCB *Increment acc. B by 1 ADDB $1100 *Add memory location $1100 *to acc. B STAB $1090 *Store acc. B in address $1090 SWI *Software interrupt END
Hand Assemble Example Problem 1 (Revisited) ORG $1000 LDAB #$0A STAB $1100 INCB ADDB $1100 STAB $1090 SWI END Address Opcode Postbyte Operand 1000 C6 0A 1002 7B 1100 1005 52 1006 FB 1100 1009 7B 1090 100B 3F
Example Problem 2 Write a short assembly language program that stores the content of Port T in memory location $3000 after waiting for 0.05 seconds for the input data. Solution Recall: One machine cycle = 0.125 x 10-6 s (8 MHz Bus Clock) We want the HCS12 to wait 0.05 s/0.125 x 10-6 s = 400,000 cycles One good way to make the HCS12 wait is to create a loop.
Wait Loop LDY #$C34F 2 cycles LDD #$0000 2 cycles LOOP ABA 2 cycles CPX $2000 3 cycles DEY 1 cycle BNE LOOP 3 cycles Note: These instructions are included to increase the operation time Assume the number of loops needed to wait is 2 bytes (2 + 3 + 1 + 3)*X + 4 = 400,000 cycles X = 44,444 10 = $AD9C
Example 2 Solution Solution *Remember that Port T is input upon reset ORG $1000 LDY #$C34F *Load Y with 44,444 10 LDD #$0000 *Clear Accumulator D LOOP ABA *Add the contents of B to A CPX $2000 *Compare X with the contents of *$2000 DEY *Decrement Y BNE LOOP *Branch to LOOP if Y is not equal *to zero LDAB $0240 *Load acc. B with content of $0240 STAB $0020 $3000 *Store content of acc. B in SWI *Software Interrupt END
Homework Homework 1 Write an assembly language program to clear the internal RAM in the MC9S12C32. Write a program to add even/odd numbers located in addresses $0800 through $0900. Homework 2 Write a program to find the largest signed number in a list of numbers stored in address $0A00 through $0BFF. Repeat for an unsigned number.
QUESTIONS???