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Reading assignments for this week and next week
Read Chapters 5-8 in Basic Microprocessors and the 6800, by Ron
Bishop.
Chapter 5
Chapter 6
Chapter 7
Chapter 8
Microcomputers-What Are They?
Programming Concepts
Addressing Modes
M6800 Software
There will be questions and answers the rest of this week and next week based on your reading assignment.
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Why use
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 [ L oa D
A ccumulator 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, and put it in S-Record
Format.
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Language Notations
Address: Common term for memory location. Always written in hexadecimal.
$4000 is the 4000 hexadecimal) th
16
Memory Location (Note: “$” signifies
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 can be 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)
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Assembly
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)
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Assembly Language Format
(Note: Last three options are called Operands )
A Tab (8 white spaces) or Label
(Note: A Label is another
Assembly
Directive and will be covered in later lectures)
Assembly Directive
Or
Mnemonic
Instruction
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
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ORG $0800
LDAA #$100A
-
BNE Front
LDAB $2F,Y
STAB $110C
-
Front INCX
-
SWI
END
27 0E
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Addressing Modes
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
•Indexed Indirect
•Direct
•Indexed
•Relative
(Note: All instructions are not capable of all addressing modes.
Example: BLE [ B ranch if L ess than or E qual to Zero] is only capable of Relative addressing mode)
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Example: Programming Reference Guide Page 6
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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 [ CL ea R accumulator A ] is an instruction using Inherent Addressing
(NOTE: SWI [ S oft W are I nterrupt] 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 in this course to return control to Mon12 Program )

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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
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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
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Example : Write a program to add the numbers 10
10 and 11
10
.
Solution
ORG $1000
LDAA #$0A
LDAB #$0B
ABA
STAA $00
SWI
END
*Puts number $0A in acc. A
*Puts number $0B in acc. B
*Adds acc. B to acc. A
*Stores results in address $00
*Software interrupt
LDAB and LDAA use immediate addressing mode
STAA uses direct addressing mode
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Address were Data that Mnemonic instruction uses is stored (Continued)
Indexed:
• Data is located within Memory locations $00 to $FFFF
Example: Store content of $2003 in Register A
LDX #$2000
LDAA $03,X
Loads accumulator A with Data value stored at memory location
$2003
X + $03 = $2000 + $03 = $2003
(Note: LDX [ L oa D index register X ])
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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
END
.
.
.
STAA $3000
SWI
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
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Why is Indexed Addressing Mode needed?
Example: Store Data Value #$20 into memory locations $2000 to $3000
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ORG $1000
LDAA #$20
LDX #$2000
LOOP STAA $00,X
INX
CPX #$3001
BNE LOOP
SWI
END
Note: LDX [ L oa D accumulator A ]
STAA [ ST ore A ccumulator A ]
INX [ IN crement X ]
CPX [ C om P are X ]
BNE [ B ranch if N ot E qual] ( 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

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Types of Indexed modes of Addressing
Indexed addressing may be implemented in multiple ways. HCS12 CPU uses X,
Y, SP or PC as base index register for instruction. O ffset is then added to base index register to form effective address .
Constant offset
5-, 9- or 16-bit signed offset (-16 to +15, -256 to +255, -32768 to +32767)
5-Bit Constant Offset Indexed Addressing
Index mode uses 5-bit signed offset which is added to base index register
(X, Y, SP or PC) to form effective address of memory location that will be affected by instruction. Offset ranges from -16 through +15. Majority of indexed instructions in real programming use offsets that fit in shortest 5-bit form of indexed addressing. Contents of base registers remain unchanged.
LDAA $00, X *load A with (X + $00)
The following three statements are equivalent:
STAA -8,X Note: offset given in decimal
STAA -$08,X
STAA $FFF8,X
Note: offset given in hex
Note: offset given as 16-bit number
Let X contain #$3000. After program executed, content of A will be stored at address (#$3000 - #$08) = $2FF8

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Types of Indexed modes of Addressing
9-Bit Constant Offset Indexed Addressing.
Uses 9-bit signed offset which is added to base index register (X, Y, SP or PC) to form effective address of memory location affected by instruction. Offset ranges from -256 through +255. Contents of base registers are not changed after instruction is executed. MSB (sign bit) of offset is included in instruction postbyte and remaining 8 bits are provided as extension byte after instruction postbyte in instruction flow.
LDAA $FF, X *Assume X contains $1000 prior to instruct. Execut.
A6 E0 FF
LDAB -20, Y *Assume Y contains $2000 prior to instruct. Execut.
E6 E9 EC
First instruction will load A with value from ($1000 + $FF) = $10FF
Second instruction will load B with value from ($2000 – 20) = $IFEC
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Types of Indexed modes of Addressing
Accumulator Offset Indexed Addressing
In this indexed addressing mode, effective address is sum of values in base index register and unsigned offset in one of accumulator. Value in base index register is not changed. Indexed register can be X, Y, SP or PC and accumulator can be either 8-bit (A or B) or 16-bit (D)
– A, B, or D accumulator added to base index register to form address
– Content of accumulator is unsigned offset
LDAA B, X
Instruction adds B to X to form address from which A will be loaded. B and X are not changed by this instruction.
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Types of Indexed modes of Addressing - Continued
•Auto Pre-/Post-Increment/Decrement
•Base index register may be automatically incremented/decremented before or after instruction ( Program Counter may not be used as base register )
•No offset is available
•May be incremented/decremented 1 to 8 times
Post-Increment (in ranges from 1 through 8):
LDX 2,SP+ *Index register X is loaded with contents of
(Same as PULX) memory location in stack pointer, then stack pointer is incremented twice
Pre-Increment (in ranges from 1 through 8):
LDX 2,+SP *Stack pointer is incremented twice, then Index register X is loaded with contents of memory location in stack pointer
Pre-Decrement (in ranges from -8 through -1):
STAA 1,-X *Index register X is decremented, then
Accumulator A is stored in memory location stored in Index Register X
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Indexed Addressing Mode Postbyte Encoding (xb)
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Indexed Addressing Mode Postbyte Encoding (xb) - Continued
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Why is Pre-/Post-Increment/Decrement Useful?
Example: Store Data Value #$20 into memory locations $2000 to $3000
Without Post-Increment With Post-Increment
ORG $1000
LDAA #$20
LDX #$2000
LOOP STAA $00,X
INX
CPX #$3001
BNE LOOP
SWI
END
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 than 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.
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Types of Indexed modes of Addressing - Continued
•Indexed-Indirect [Differ Discussion of this for Now]
•Like other indexed addressing modes, offset and base index register are added to form an effective address
• Offset is either Accumulator D or a 16 bit constant offset
•However, effective address is understood to contain address of memory location containing address to be acted upon.
Example: Clear the contents of the memory location pointed to by a pointer located in memory location $2000
PTR
ORG $2000
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 [ CL ea R Memory Location])
•Useful for efficiently forming switch case statements and other program
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Indexed
Example: Hex number is stored in $2000
LDD #$2000 *Loads accumulator D with #$2000
STD $100A *Stores content of Register D in Addresses $100A and $100B
LDX #$1000 *Loads Register X with #$1000
LDAA [10,X] *#$0A is added to #$1000 in X-Register to form $100A. Next, CPU
*goes to memory (address) locations $100A and $100B and fetch
*address pointer $2000. Value stored in address $2000 is read and
*loaded into accumulator A.
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As stated before the Assembler translates an assembly language program into a machine language program
Format of machine language program
Instruction
Address where instruction is located
Opcode 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)
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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 )
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Indexed Addressing Mode Postbyte Encoding (xb)
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Indexed Addressing Mode Postbyte Encoding (xb) - Continued
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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
LDAA 0,X
LDAA $02,SP+
LDAA B,Y
SUBB $1040,X
SUBB D,Y
SUBB -14,X
Opcode Postbyte
A6
A6
A6
E0
E0
E0
00
B1
ED
E2
EE
12
Table 1 (Excerpt)
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Hand Assembling
Example: Assemble the following Program
ORG
LDAA #$0A
LDAB
$1000
ABA
STAA $00
SWI
END
#$0B
Address Opcode Postbyte Operand
$1000
$1002
86 0A
C6 0B
$1004 18 06
$1006 5A
$1008 3F
00
(Note: $1002 since $86 is now at
$1000 and $0A is at $1001)
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Example Problem 1 (revisited)
ORG
LDAB
STAB
INCB
ADDB
STAB
SWI
END
$1000
#$0A
$1100
$1100
$1090
*Load acc. B with number 0A
*Store acc. B in address $1100
*Increment acc. B by 1
*Add memory location $1100
*to acc. B
*Store acc. B in address $1090
*Software interrupt
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Hand Assemble Example Problem 1 (Revisited)
ORG
STAB $1090
SWI
END
$1000
LDAB #$0A
STAB $1100
INCB
ADDB $1100
Address Opcode Postbyte Operand
1000 86
1002 7B
1005 52
1006 FB
1009 7B
100B 3F
0A
1100
1100
1090
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Example Problem 2
Write a short assembly language program that stores the content of Port T in memory location $3000 after waiting for 0.05 seconds for the input data.
Solution
Recall: One machine cycle = 0.125 x 10 -6 s (8 MHz Bus Clock)
We want the HCS12 to wait 0.05 s/0.125 x 10 -6 s = 400,000 cycles
One good way to make the HCS12 wait is to create a loop.
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Wait Loop
LDY #$AD9C
LDD #$0000
LOOP ABA
CPX $2000
DEY
BNE LOOP
2 cycles
2 cycles
2 cycles
3 cycles
1 cycle
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
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Example 2 Solution
Solution
*Remember that Port T is input upon reset
ORG $1000
LDY #$AD9C
LOOP
LDD #$0000
ABA
*Load Y with 44,444
10
*Clear Accumulator D
*Add the contents of B to A
CPX $2000 *Compare X with the contents of
*$2000
*Decrement Y DEY
BNE LOOP *Branch to LOOP if Y is not equal
*to zero
LDAB $0240 *Load acc. B with content of $0240
STAB $3000 *Store content of acc. B in $3000
SWI *Software Interrupt
END
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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.
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QUESTIONS???
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