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Microprocessor Control of Manufacturing Systems
and
Introduction to Mechatronics
Instructor: Professor Charles Ume
Lecture #9
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Reading Assignments
Reading assignments for the next 4 weeks:
HCS12 Microcontroller: S12CPUV2 Reference Manual
CPU12RG/D: Rev. 2, 11/2001
Reference Book:
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 Assembly Language?
HCS12 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 are
representatives of instructions they represent (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, and put it in S-Record
Format.
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Assembly Language Notations
Address: Common term for memory location. Always written in hexadecimal.
$4000 is the 4000th16 Memory Location (Note: “$” signifies
hexadecimal)
LDAB $4000
Literal Value: A number used as data in program is indicated by “#”. Can be
represented in the following ways:
#$FF = hexadecimal number FF
#%1011 = binary number 1011 (Note: “%” signifies binary)
#123 = decimal number 123
LDAB #$FF; LDAB #$D01A
A Literal Value can be stored in an address
Example: Store Literal Value #$FF in address $4000
LDAB #$FF
STAB $4000
Example 2: Store Literal Value #$FE0A in address $2000
LDD
#$FE0A
STD
$2000
Note: #$FE is stored in address $2000 and #$0A is stored in address $2001
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Assembly Language Directives
Directives: Instructions from the programmer to Assembler NOT to
microcontroller (Page 134 in Ron Bishop)
Example 1: ORG <address>
Store translated/assembled machine language instructions in
sequence starting at given address for any mnemonic instructions that
follow
ORG
$1000
Example 2: END
Stop translating/assembling mnemonics instructions until another
ORG is encountered
(Note: More will be covered in later lectures)
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Left margin of assembly program
Assembly Language Format
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
(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
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Front
ORG
LDAA
DECA
BNE
LDAB
STAB
INCX
SWI
END
$0800
#$100A
Front
$2F,Y
$110C
-
-
86
43
27
3F
100A
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: (See next slide also)
•Inherent
•Immediate
•Extended
•Indexed Indirect
•Direct
•Indexed
•Relative
(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)
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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 [ 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 resumes program execution at this location. If no
address is stored in the SWI vector, the main program will stop execution at
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this point. Used
in this W.
course
to return
control
to Mon12Engineering
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 #$5BEE (Explain what happens)
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 Problem 1
Example : Write a program to add the numbers 1010 and 1110.
Solution
ORG
LDAA
LDAB
ABA
STAA
SWI
END
$1000
#$0A
#$0B
$00
*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 [ LoaD 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
.
.
.
STAA $3000
SWI
END
With Indexed Addressing Mode
LOOP
ORG $1000
LDAA #$20
LDX #$2000
STAA $00,X
INX
CPX #$3001
BNE LOOP
SWI
END
Program on the Left is much longer than the program on Right
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LOOP
Why is Indexed Addressing Mode needed? (Continued)
ORG $1000
LDAA #$20
LDX #$2000
STAA $00,X
INX
CPX #$3001
BNE LOOP
SWI
END
Note:
LDAA[LoaD accumulator A]
LDX [ LoaD Index Register 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
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Etc…..
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Why is Indexed Addressing Mode needed?
Example: Store Data Value #$20 into memory locations $2000 to $3000
ORG $1000
LDY #$1001
LDAA #$20
LDX #$2000
LOOP STAA $00,X
INX
DECY
BNE LOOP
SWI
END
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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.
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Types of Indexed modes of Addressing
Auto Pre-/Post-Increment/Decrement:
 Base index register (X, Y and SP) 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+ EE B1 *Index register X is loaded with
contents of memory location in
stack pointer (same as PULX).
Then stack pointer is
incremented twice.
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Types of Indexed modes of Addressing
 Pre-Increment (in ranges from 1 through 8):
LDX 2,+SP EE A1 *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 6A 2F *Index register X is decremented.
Then content of Accumulator A is
stored in memory location in Index
Register X
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Types of Indexed modes of Addressing
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 Learning Objectives:
 Learn how to use following in index mode of addressing
instructions:
 5-bit signed constant offset: -16 to +15
 9-bit signed constant offset: -256 to +255
 16-bit signed constant offset: -32768 to +32767
 Accumulator offset
16-bit constant indirect indexed
 Learn how to use following base index registers for instructions:
 X, Y, SP and PC
 Note that when:
 Offset is added to base index register to form effective address.
 Content of base register remains unchanged.
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Types of Indexed modes of Addressing
5-Bit Constant Offset Indexed Addressing: -16 to +15
 Index mode uses:
 5-bit signed constant offset 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.
LDAA $00, X
A6 00 *load A with (X + $00)
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The following three statements are equivalent:
STAA -8,X
6A 18 *Offset given in decimal
STAA -$08,X
6A 18 *Offset given in hex
STAA $FFF8,X 6A 18 *Offset given as 16-bit number
Let X contain #$3000. After program executed, content of A
will be stored at address (#$3000 - #$08) = $2FF8
$FFF8 = 1111 1111 1111 1000
0000 0000 0000 0111
1
--------------------------------------------- 0000 0000 0000 1000 = - #$0008
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(Page 21)
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Base
index
register
rr
X
00
Y
01
SP
10
PC
11
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5-bit Constant offset : -16 to +15
LDDA -16, X
A6 10:
-16 = -$10
$10 = 00010000
1’s Comp = 11101111
2’s Comp = 11110000
rr0nnnnn = 00010000 = $10 (this is the postbyte)
bits 0-4 of offset
LDAA 6, SP
A6 86:
6 = $06 = 00000110
rr0nnnnnn = 10000110 = $86 (this is the postbyte)
bits 0-4 of offset
LDAA –14, SP A6 92
-14 = -$0E
0E = 00001110
1’s Comp = 11110001
2’s Comp = 11110010
rr0nnnnn: 10010010 = $92 (this is the postbyte)
bits 0-4 of offset
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Types of Indexed modes of Addressing
9-Bit Constant Offset Indexed Addressing: -256 through +255
 Uses 9-bit signed constant 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.
 Content of base register is not changed after instruction is executed
 MSB (sign bit) of offset is included in instruction postbyte
 Remaining 8 bits are provided as extension byte after instruction
postbyte in instruction flow.
LDAA $FF, X
A6 E0 FF
*Assume X contains $1000 prior to instruct is executed
LDAB -20, Y
E6 E9 EC
*Assume Y contains $2000 prior to instruct is executed
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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Base
index
register
rr
X
00
Y
01
SP
10
PC
11
9-bit Constant offset : -256 to +255
Postbyte Code (xb) is = 111rr0zs
Z = 0 and s = 1 (when offset is a negative number-MSB of offset)
Z = 0 and s = 0 (when offset is a positive number-MSB of offset)
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9-bit Constant offset : -256 to +255
LDAA -25, X
A6 E1 E7:
-25 = -$19
$19 = 00011001
1’s Comp = 11100110
2’s Comp = 11100111 = $E7 (this is the offset)
111rr0zs = 11100001 = E1 (this is the postbyte)
LDAA 30, SP
A6 F0 1E:
30 = $1E (this is the offset)
= 00011110
111rr0zs = 11110000 = F0 (this is the postbyte)
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Types of Indexed modes of Addressing
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16-bit Constant offset : -32768 to +32767
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Base
index
register
rr
X
00
Y
01
SP
10
PC
11
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16-bit Constant offset : -32768 to +32767
Postbyte Code (xb) is = 111rr0zs
 Z = 1 and s = 0 : For the case when the offset is positive
or negative.
 S is not = 1 during postbyte calculation when the offset is
negative because
 Bit 15 of the 2’s complement offset is already 1 which
 Indicates a negative offset number
Postbyte for positive offset is same as postbyte for negative
offset ie EA
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16-bit Constant offset : -32768 to +32767
Postbyte code (xb) is 111rr0zs
LDAA 280, y
A6 EA 0118
280 = $0118 = 0000000100011000 (this is the offset)
111rr0zs = 11101010 = $EA (this is the postbyte)
LDAA -280, Y
A6 EA FE E8
-280 = -$0118
$0118 = 0000000100011000
1’s Comp. = 1111111011100111
2’s Comp. = 1111111011101000 = $FEE8 (this is the
offset)
111rr0zs = 11101010 = $EA (this is the postbyte)
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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 accumulators.
 Value in base index register is not changed.
 Base indexed register can be X, Y, SP or PC and
accumulator can be either 8-bit (A or B) or 16-bit (D)
 Content of A, B, or D accumulator added to base index
register to form address
LDAA B, X
A6 E5
Instruction adds content of accumulator B to X register to form
address from which accumulator A will be loaded. B and X are not
changed by this instruction.
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(Page 21)
Base
index
register
rr
X
00
Y
01
SP
10
PC
11
Detailed Explanation:
Example: LDAA B, X •This is Indexed Addressing Mode with Accumulator Offset.
•Opcode for LDAA is A6 for this mode.
Ans: A6 E5
•From the table above, the formula for postbyte of this mode is: 111rr1aa
•rr is 00 because Base Index Register is X
•aa is 01 because Accumulator used for offset is B
•11100101
= E5 in School
hex of Mechanical Engineering
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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
LOOP
ORG $1000
LDAA #$20
LDX #$2000
STAA $00,X
INX
CPX #$3001
BNE LOOP
SWI
END
With Post-Increment
LOOP
ORG $1000
LDAA #$20
LDX #$2000
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
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 16-Bit constant indirect indexed addressing
 Indexed addressing mode adds 16-bit instruction-supplied offset to
 Base indexed register to form address of memory location that
contains pointer to memory location affected by instruction.
 Instruction itself does not point to address of memory
location to be acted on.
 Square brackets distinguished this addressing mode from
16-bit constant offset indexing.
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16-Bit constant indirect indexed addressing
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LDAA [10, X] or
LDAA [$0A, X]
 Assume:
X holds base address of table of pointers
X has initial value of $1000
#$2000 is stored at addresses $100A and $100B
 When instruction is executed:
#$0A is added to value in X to form address $100A
Address pointer ($2000) is fetched from memory $100A
Value stored in $2000 is read and loaded into A accumulator
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Types of Indexed modes of Addressing - Continued
• 16-Bit Constant Indirect Indexed Addressing
ORG
$1000
LDAB
#$EE
STAB
$0400
*#$EE stored at $0400
LDD
#$0400
STD
$5DBC
*#$0400 stored at $5DBC & $5DBD
LDX
#$5D00
*X contains #$5D00
LDAA
[$BC, X]
*#$BC added to value in X to form
……
……
address $5DBC. Address pointer $0400
......
……
fetched from memory at $5DBC. Value (EE)
store in $0400 read & loaded in Acc. A
……
……
SWI
END
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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
Operand
Address
where
instruction
is located
Opcode
Postbyte
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
Opcode
Postbyte
LDAA 0,X
A6
00
LDAA $02,SP+
A6
B1
LDAA B,Y
A6
ED
SUBB $1040,X
E0
E2
SUBB D,Y
E0
EE
SUBB -14,X
E0
12
Table 1 (Excerpt)
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(Programming
Guide) Engineering
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Hand Assembling
Example: Assemble the following Program
ORG
LDAA
LDAB
ABA
STAA
SWI
END
$1000
#$0A
#$0B
$00
Address
Opcode
Postbyte
Operand
$1000
86
0A
$1002
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
$1000
#$0A
$1100
STAB
SWI
END
$1090
$1100
*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
LDAB
STAB
INCB
ADDB
STAB
SWI
END
$1000
#$0A
$1100
$1100
$1090
Address
1000
1002
1005
1006
1009
100B
Opcode
Postbyte
C6
7B
52
FB
7B
3F
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Operand
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 = 44,44410 = $AD9C
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Example 2 Solution
Solution
*Remember that Port T is input upon reset
ORG $1000
LDY #$AD9C
*Load Y with 44,44410
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 $3000
*Store content of acc. B in $3000
SWI
*Software Interrupt
END
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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.
The George W. Woodruff School of Mechanical Engineering
ME4447/6405
QUESTIONS???
The George W. Woodruff School of Mechanical Engineering