Transcript 05_basic_computer_part2

Basic Microcontroller System
Microcontroller-Based System
Memory
CPU
I/O
Interface
To I/O
BUS
CPU: Central Processor Unit
Microcontroller
I/O: Input/Output
e.g. M68HC11
Memory: Program and Data
Bus: Address signals, Control signals, and Data signals
Central Processing Unit
(CPU)
68HC11
68HC11 Register Set
7
Accumulator A
15
15
15
15
15
15
Clock is implied
0 7
Accumulator B
Double Accumulator D
INDEX REGISTER X
INDEX REGISTER Y
STACK POINTER
PROGRAM COUNTER
0
0
0
0
0
00
68HC11 Memory Address
Space
8 bits
$0000
$0001
$0002
$0003
$FFFC
$FFFD
$FFFE
$FFFF
Instruction Execution Cycle
• Fetch – Decode - Execute
– Fetch stage
• Operations code (op-code) is loaded from memory
into the Instruction Register (IR)
– Decode stage
• Instruction is “decoded” into a set of “microinstructions” (e.g. FSM)
– Execution stage
• The micro-instructions are executed. This may
involve loading the “operand” from memory.
Instruction Execution Cycle
• Comments:
– Each instruction requires N clock cycles for
execution. N varies from 2 to 12
– An Op-code is also referred to as
• Machine code which are the binary codes that are
used to represent the instruction.
Simple Example
Reset EQU $FFFE ; Set symbol Reset to FFFE16
Program EQU $E000 ; Set symbol Program to E00016
ORG Program
; Set the assembler’s location to the value
; represented by symbol Program.
Top: LDAA #$23
; Load the A register with 2316
LDAB #$BE
; Load the B register with BE16
ABA
; Add the A and B registers. Result is stored in A.
STAA $1000
; Store the value in the A register in
; memory location $1000. The A reg is
; unchanged.
L0: BRA L0
Org Reset
; Branch to Label L0
; set the assembler’s location to the value
; represented by symbol Reset.
FDB Top
; Form a double byte from symbol TOP. Stored it at the
; current memory location.
Simple Example
• We can also represent this as:
– A  $23 ; The A register is assigned $23
– B  $BE ; The B register is assigned $BE
–AA+B; A=A+B
– ($1000)  A ;The contents at memory location
; $0100 is replaced with $23+$BE
Simple Example (cont)
• The assembler will convert our program to:
Address
$E000
$E002
$E004
$E005
$E008
……..
$FFFE
Opcode
86
c6
1b
b7
20
Operand
23
be
Operand
10
fe
00
e0
00
What do all of these numbers mean?
Simple Example (cont)
• Example: $E000 86 23
• $E000:
– This is the memory location or address where the program or
data is located.
• $86
– This is the Operation Code or opcode
– The opcode is the Hex (i.e. binary) representation of the
instruction.
• $86 means LDAA # or
– Load the A register with a constant value for the 68HC11
• Where is the value to be loaded?
• $23
– This is the operand for the opcode. For this instruction, 23 is the
constant value to be loaded into the A register.
Simple Example (cont)
• Example: $E002 c6 be
• $E002:
– This is the memory location or address where this program or
data is located.
• $C6 (note the change from the previous instruction)
– This is the Operation Code or opcode
– The opcode is the Hex (i.e. binary) representation of the
instruction.
• $c6 means LDAB # or
– Load the B register with a constant value for the 68HC11
• Where is the value to be loaded?
• $be
– This is the operand for the opcode. For this instruction, 23 is the
constant value to be loaded into the A register.
Simple Example (cont)
• $E004 1b
– $1b = ABA
– Note: we don’t need an operand for this one
• $E005 b7 10 00
– $b7 = STAA in an extended memory address
– address is stored at the operand
– $1000 is the storage address
• Note: the contents at location $1000 are
overwritten
Simple Example (cont)
• $E008 20 fe
– $20 = This is the BRAnch opcode.
• The program counter (PC) is set to
• PC= PC + relative address (signed addition)
– $fe is the relative address.
– After fetching the instruction, the PC points to the next
instruction.
– In this case
•
•
•
•
PC = $E00A, after fetching this instruction
PC = $E00A + $FE = $E00A + $FFFE(sign extended)
PC = $E008 (or we branch right back to this instruction)
This is an infinite loop. How do we get out of this?
Simple Example (cont)
• Hit the RESET button!!!!!!
• $FFFE E0 00
– This is data (not program) stored in the
Interrupt Vector Table (IVT) at the Reset
interrupt location.
– This will cause our program to begin
executing the program stored at location
$E000 whenever we hit the Reset button.
Simple Example (cont)
• In memory, our program appears as
$E000: 86 23 c6 be 1b b7 10 00 20 fe
………..
$FFFE: E0 00
When we hit the Reset button, the Program
counter is set equal to the data stored at location
$FFFE.
Instruction Execution Cycle
• How does this all work?
– Recall, Instruction Execution Cycle
– Fetch – Decode – Execute
• Fetch stage
– Operations code (op-code) is loaded from memory into
the Instruction Register (IR)
• Decode stage
– Instruction is “decoded” into a set of “micro-instructions”
(e.g. FSM)
• Execution stage
– The micro-instructions are executed. This may involve
loading the “operand” from memory.
Simple Example (cont)
•
When we hit the Reset button, the
Program counter is set equal to the data
stored at location $FFFE, or
1. PC  ($FFFE) = $E000
– Program counter is loaded with the contents
of memory location $FFFE
Simple Example (cont)
2.
Fetch Opcode
IR  ($E000) = 86
3.
Decode Opcode
CPU decodes as LDAA #
4.
Execute opcode
CPU fetches operand as next byte from program memory.
Instruction is executed.
Program counter is incremented PC = PC + 2
•
Process repeats for next byte
Simple Example (cont)
2.
Fetch Opcode
IR  ($E002) = c6
3.
Decode Opcode
CPU decodes as LDAB #
4.
Execute opcode
CPU fetches operand as next byte from program memory.
Instruction is executed.
Program counter is incremented PC = PC + 2
•
•
Process repeats for next byte
…………..
TPS Quiz
THRSIM11
Design Procedure
1.
2.
3.
4.
5.
6.
7.
Use THRSIM11 to develop program
Assemble program with THRSIM11 assembler
Correct any assembly errors
Simulate program using THRSIM11 simulator
Correct any logical errors
Download program to board (if needed)
Correct any hardware errors
Assembler
• An assembler converts the assembly
language program into machine code.
• The machine code is also known as the
op-code. We will use a Mnemonic (e.g.
LDAA) to represent this op-code.
Assembler – THRSIM 11
• We’ll use the THRSIM11 assembler
• Syntax: () = Optional
– (Label) Opcode (Operand) (Comment)
– Ex:
• Loop:
•
•
LDAA #$2F Load Accumulator A with
Hex $2f
Label Field - Optional
• Must start in the first position of line
• One to fifteen characters
– Upper- and lower-case letters a-z
• Case sensitive
– Digits 0-9, Period (.), Dollar sign ($), or underscore (_)
• First character must be
– Alphabetic, period, or underscore
• Last character should be a colon (:) (BP)
• Label may be by itself
Label Field - Optional
• Labels must be unique
• If an asterisk (*) is the first character, the
rest of the line is considered a Comment.
Op-code or Operation Field
• Required field
• Contains mnemonic for the operation or
assembler directive (or pseudooperation)
• Not case sensitive
• Must not begin in the first position
– Will be treated as label if it begins in the first
position
Operand Field
• Operand of Op-Code (if, needed)
• Can consist of:
– Symbols – assembler replaces the symbol
with its value
• Ex: COUNT EQU $5F
•
LDAA COUNT
• EQU is an assembler directive. It instructions the
assembler to replace the symbol COUNT with the
value of $5F, so the above is equivalent to:
–
LDAA $5F
Operand Field
• Constants - Value does not change
– Formats:
•
•
•
•
•
$ = Hex: Ex: COUNT EQU $5A
% = Binary: Ex: COUNT EQU %01011010
Blank = Decimal: Ex: COUNT EQU 70
‘ = ASCII: COUNT EQU ‘0 (i.e. $30)
Note: assembler will convert each format to HEX
– Constants must be consistent with
• bit width (i.e. 8-bit or 16-bit)
• type (i.e. signed or unsigned)
Operand Field
• Expressions –Assembler can evaluate
simple expressions to compute the value
of the operand
– Operators
•
•
•
•
+ = Addition
- = Subtraction
* = Multiplication
/ = Division
Operand Field
• Expressions –
– Operators (continued)
•
•
•
•
% = Remainder after division
& = Bitwise AND
| = Bitwise OR
^ = Bitwise XOR
– Evaluated by assembler from left to right
– Signed two’s complement arithmetic
– Can use only constants
Operand Field
• Expressions –
– Examples
• COUNT EQU $1F
• EX1: LDAA COUNT+$0A
– A  ($1F+$0A) = ($19)
Assembler Directive: EQU
•
•
•
•
Equates a symbol to a value
Syntax:
Label EQU Expression (comment)
Note: the Label is NOT stored in memory. Label
is just an alias for expression
• Ex:
– TEST EQU $3000
–
LDX #TEST
– This is the same as
LDX #$3000
Assembler Directive: ORG
•
•
•
•
Set assembler’s counter to expression
Syntax:
(Label) ORG Expression (comment)
Ex:
– Code EQU $3000
–
ORG Code
– Assembler assumes Program Counter now
contains the value of $3000
Assembler Directive: RMB
•
•
•
•
Reserve memory bytes
Syntax:
(Label) RMB Expression (comment)
Ex:
– Buffer1 RMB $10
– Buffer2 RMB $10
– The assembler saves $10 bytes of RAM for a data
buffer located at address Buffer1. For example, if
Buffer1 is located at address $2000, assembler will
locate Buffer2 at $2000+$0010 = $2010.
Assembler Directive: FCB
• Form constant byte
• Syntax:
• (Label) FCB Expression, (Exp), (Exp), (comment)
• Stores a constant byte in a memory location. Note:
difference with EQU
• Ex:
–
–
–
–
–
Data
EQU $1000
ORG Data
Buffer1 FCB $10,$FA,$2F
Result is at memory address $1000, we have
$1000: 10 FA 2F
Assembler Directive: FDB
• Stands for “Form Double-Byte constant”
Using square brackets [ ] to denote optional elements
• Syntax:
– [label][:] FDB expression [, expression] … [, expression] [comment]
• Semantics (meaning, function, behavior):
– Stores one or more 16-bit values in subsequent memory
locations.
• Uses “Big Endian” format: stores MSB first, LSB second.
• Note the difference between this and EQU!
– EQU assigns the symbol (label) to the value of given Expression
– FCB, FDB, etc. assign the symbol to the current address, and
specify the contents of memory starting at that address
• Example:
Buffer1 FDB $10FA,$2FFF
– Contents of memory bytes starting at address Buffer1:
• $10, $FA, $2F, $FF
Assembler Directive: FCC
• Stands for “Form Constant Character string”
• Syntax:
– [label][:] FCC “string“ [comment]
• Converts the given string into its ASCII equivalent.
– ASCII = American Standard Code for Information
Interchange
• See http://www.asciitable.com for a list of codes.
• Example:
String:
FCC “Hello“
– The data bytes stored at address String are
• $48, $65, $6c, $6c, $6f
A few other useful directives
• BSZ – “Block Storage of Zeros”
– A.k.a. ZMB “Zero Memory Bytes”
– Like RMB, but clears each memory byte to 0.
• FILL – “Fill memory with constant”
– Syntax: FILL value, howMany
– Stores value at next howMany bytes.
• OPT – “Assembler output options”
– Syntax: OPT opt1 [, opt2 …]
• Each opti is one of these tags: c, cre, l, noc, nol, or s.
– See textbook, pp.29 & 31 for details.
• PAGE – “Output a page break in program listing”
Comment Field
• Available with all assembler directives and
instructions.
• Placed after operand field
– Use asterisk to indicate comment in first
character position
• Need comments at the beginning of
program (BP)
• Need comment on every line. (BP)
EEL-4746 Best Practices
EEL-4746 Best Practices
1. All programs submitted for homework
assignments must begin with the following
comment:
***************************
*
EEL-4746 Spring 2004 Semester
*
Homework #N – Due (Due Date)
*
Problem #M
*
Name of Partner A
*
Name of Partner B
****************************
*
Description of Program
EEL-4746 Best Practices
2. All subroutines must begin with the
following comment
***************************
*
Subroutine Name: Mysub
*
Input parameter list:
*
Output parameter list:
*
Registers changed list:
****************************
*
Description of Subroutine
EEL-4746 Best Practices
3. All lines must contain a comment
4. All labels must end with a colon (“:”)
5. Must declare and use a symbol for all
constants, using EQU.
EEL-4746 Best Practices
6. Program must have a well-organized overall format,
such as:
Header Comment, from previously
***********************
*
Standard Symbols
Data
EQU $0200 ; Start of RAM in U5, to $FFF
Program EQU $1040 ; Just above config registers
Stack
EQU $7FFF ; Top of RAM in U5
Reset
EQU $FFFE ; Reset vector location
************************
(Define your own symbols here)
***********************
*
Program area
Start:
ORG Program
1st program line
; comment
***********************
*
Data area
Symbol
ORG Data
Directive …
Example
• Write an 68HC11 assembly language
program to monitor the temperature of a
power plant. If the temp >= 50C, sound the
alarm.
– Given the following memory addresses
• Analog to digital converter = temperature (in Hex)
• Port B – bit 0 = Output alarm
– 0 – no alarm
– 1 – alarm will sound
Pseudo-Code
MC68HC11 Instruction Set
M68HC11 Instruction Set
• Categories
– Load and Store
– Stack
– Transfer
– Decrement and Increment
– Clear and Set
– Shift and Rotate
– Arithmetic Instructions
Arithmetic Instructions
•
•
•
•
•
Add and Subtract
Decimal Arithmetic
Negating Instruction
Multiplication
Division
Other Instructions
•
•
•
•
•
•
•
Logic
Data Test
Conditional Branch
Unconditional Jump and Branch
Subroutines
Interrupt
Miscellaneous
Instruction Cycle
1. IR  (PC)
 Opcode is fetched into Instruction Register
2. Instruction is decoded
3. Data or address is loaded from memory
(if needed)
 Program Counter is updated
4. Instruction is executed (if needed)
Load Instructions
• Mnemonic Operation
– LDAA – Load Accumulator A
– LDAB – Load Accumulator B
– LDD - Load Accumulator D
– LDS - Load Stack Pointer
– LDX - Load Index X Register
– LDY - Load Index Y Register
• Addressing modes:
• Condition codes:
All except INH
Set N,Z and V=0
Mnemonic  Machine Code
• Immediate Mode
– LDAA #$02

86 02
• Direct Addressing Mode
– LDAA $02

96 02
• Extended Addressing Mode
– LDAA $1002 
B6 10 02
• Register Indexed (X = $1000)
– LDAA $02, X 
A6 02
• Register Indirect (X = $1002)
– LDAA 0,X

A6 00
Store Instructions
• Mnemonic Operation
–
–
–
–
–
–
STAA – Store Accumulator A
STAB – Store Accumulator B
STD – Store Accumulator D
STS – Store Stack Pointer
STX – Store Index X Register
STY – Store Index Y Register
• Addressing modes: All except IMM and INH
• Condition codes: N,Z and V=0
Mnemonic  Machine Code
• Immediate Mode
– STAA #$02

(Illegal operation)
• Direct Addressing Mode
– STAA $02

5A 02
• Extended Addressing Mode
– STAA $1002

7A 10 02
• Register Index (X = $1000)
– STAA $02, X 
6A 02
• Register Indirect (X = $1002)
– STAA 0,X

6A 00
Notation: Memory Locations
• $C000: 12 34 56 78 9A BC DE F0
– This means the hex bytes shown are
contained in consecutive memory locations
starting from address $C000.
• In other words:
– [$C000]=$12, [$C001]=$34, [$C002]=$56,
[$C003]=$78, [$C004]=$9A, [$C005]=$BC,
[$C006]=$DE, [$C007]=$F0
Notation: Memory Locations
• You can visualize the
memory bytes as
arranged in a table…
– $C000: 12 34 56 78
9A BC DE F0
– Gives this table:
Address
$C000
$C001
$C002
$C003
$C004
$C005
$C006
$C007
Data
$12
$34
$56
$78
$9A
$BC
$DE
$F0
16-bit Load and Store
• For a 16-bit (2 byte) load and store, the high byte is stored at the
lower address and the low byte is stored at the higher address.
– This is called “big endian” byte ordering
• The “big end” of the word comes first.
•
•
•
•
•
Ex: $C000: 12 34 56 78 9A BC DE
LDAA $C001 A  ($C001) = $34
LDX $C002
X  $5678
STX $C004
($C004)  $5678
$C000: 12 34 56 78 56 78
Transfer Register Instructions
• Mnemonic Operation
–
–
–
–
–
–
–
–
TAB : Transfer A to B:
TBA : Transfer B to A:
TSX : Transfer SP to X:
TSY : Transfer SP to Y:
TXS : Transfer X to SP:
TYS : Transfer X to SP:
XGDX: Exchange D and X:
XGDY: Exchange D and Y:
B  (A)
A  (B)
X  (SP) + 1
Y  (SP) +1
SP  (IX) -1
SP  (IY) – 1
X  D
Y  D
• Addressing modes: INH Only
• Condition codes: N,Z and V=0 (TAB and TBA
only)
Decrement Instructions
• Mnemonic Operation
–
–
–
–
–
DECA
DECB
DES
DEX
DEY
: Decrement A: A <- (A) – 1
: Decrement B: B <- (B) – 1
: Decrement SP: SP <- (SP) -1
: Decrement X : X <- (X) - 1
: Decrement Y : Y <- (Y) – 1
• Addressing modes: INH Only
• Condition codes:
– DECA,DECB: N,Z and V
– DEX, DEY: Z
– DES: none
Decrement Instructions
• Mnemonic Operation
– DEC : Decrement Mem: (M) <- (M) -1
• Addressing modes: All but INH
• Condition codes: N,Z and V
Increment Instructions
• Mnemonic Operation
–
–
–
–
–
INCA
INCB
INS
INX
INY
: Increment A: A <- (A) + 1
: Increment B: B <- (B) + 1
: Increment SP: SP <- (SP) + 1
: Increment X : X <- (X) + 1
: Increment Y : Y <- (Y) + 1
• Addressing modes: INH Only
• Condition codes:
– INCA,INCB: N,Z and V
– INX, INY: Z
– INS: none
Increment Instructions
• Mnemonic Operation
– INC : Increment Mem: (M) <- (M) +1
• Addressing modes: All but INH
• Condition codes: N,Z and V
Clear Instructions
• Mnemonic Operation
– CLR : Clear Memory: (M) <- 0
– CLRA : Clear A: A <- 0
– CLRB : Clear B: B <- 0
• Addressing modes:
– CLR: EXT, IX, IY
– CLRA, CLRB: INH only
• Condition codes: N=0,Z=1,V=0,C=0
Clear and Set Bit Instructions
• Mnemonic Operation
– BCLR : Clear Bits
– BSET : Set Bits
• Addressing modes: Direct, Index
• Condition codes: N,Z,V=0
• Example:
– BCLR $33 $AA
• Clear bits 2,4,6, and 8 at memory add $33
Arithmetic Instructions - ADD
• Mnemonic Operation
–
–
–
–
–
–
–
–
ABA: Add B to A: A <- (A) + (B)
ABX: Add B to X: X <- (X) + (B)
ABY: Add B to Y: Y <- (Y) + (B)
ADDA: Add memory to A: A <- (A) + (M)
ADDB: Add memory to B: B <- (B) + (M)
ADDD: Add memory to D: D <- (D) + (M:M+1)
ADCA: Add memory to A and C: A<- (A)+ (M) +C
ADCB: Add memory to B and C: B<- (B)+ (M) +C
• Addressing modes: INH or All except IHN
• Condition codes: N,Z,V,C (except X and Y)
Arithmetic Instructions - Sub
• Mnemonic Operation
–
–
–
–
–
–
SBA: Sub B from A: A <- (A) + (B)
SUBA: Sub memory from A: A <- (A) - (M)
SUBB: Sub memory from B: B <- (B) - (M)
SUBD: Sub memory from D: D <- (D) - (M:M+1)
SBCA: Sub memory and C from A: A<- (A)-(M)-C
SBCB: Sub memory and C from B: B<- (B)-(M)-C
• Addressing modes: INH or All except IHN
• Condition codes: N,Z,V,C
Arithmetic Instructions - Neg
• Mnemonic Operation
– NEG: 2’s comp memory: (m)  -1*(M)
– NEGA: 2’s comp A : A  -1*(A)
– NEGB: 2’s comp B : B  -1*(B)
• Addressing modes: INH or Ext, Ix, Iy
• Condition codes: N,Z,V,C
Condition Code Register
• Special Register used for control and
provide arithmetic information to
programmer and CPU.
Condition Code Register
S
X
H
I
N
Z
V
C
Condition Code Register
7
6
5
S
X
H
4
3
2
1
0
I
N
Z
V
C
S = Stop
X = X Interrupt Bit
I = Interrupt Mask
Control Bits
N = Negative
Z = Zero
V = Overflow
C = Carry
H = Half Carry
Arithmetic Bits
Compare Instructions - CMP
• Mnemonic Operation
–
–
–
–
–
–
CBA: Compare B to A:
CMPA: Compare memory to A:
CMPB: Compare memory to B:
CMPD: Compare memory to D:
CMPX: Compare memory to X:
CMPY: Compare memory to Y:
A-B
A - (M)
B - (M)
D - (M:M+1)
X - (M:M+1)
Y – (M:M+1)
• Addressing modes: INH or All except IHN
• Condition codes: N,Z,V,C
– Note: Only Condition Code Register (CCR) is
changed by these instructions. All other registers
unaffected.
Examples of SBA (A−B) results
Minuend (A)
7F
80
81
FF
A−B
00
01
(0)
(1)
(127)
(128, −128)
(129, −127)
(255, −1)
00
00
NZVC
01
NZVC
7F
NZVC
80
NZVC
81
NZVC
FF
NZVC
FF
NZVC
00
NZVC
7E
NZVC
7F
NZVC
80
NZVC
FE
NZVC
81
NZVC
82
NZVC
00
NZVC
01
NZVC
02
NZVC
80
NZVC
80
NZVC
81
NZVC
FF
NZVC
00
NZVC
01
NZVC
7F
NZVC
7F
NZVC
80
NZVC
FE
NZVC
FF
NZVC
00
NZVC
7E
NZVC
01
NZVC
02
NZVC
80
NZVC
81
NZVC
82
NZVC
00
NZVC
(0)
Subtrahend (B)
01
(1)
7F
(127)
80
(128, −128)
81
(129, −127)
FF
(255, −1)
Branch Instructions
• A Branch Instruction typically follows a
Compare instruction. The Branch
instruction will branch if certain CCR bits
are set or clear
Branch Instructions
• BRA – Branch Always
– Unconditional “Branch” (i.e. GoTo)
• BEQ - Branch if equal
– Z bit = 1
• BNE – Branch if not equal
– Z bit = 0
• BCC – Branch if Carry Clear
– C bit = 0
• BCS – Branch if Carry Set
– C bit = 1
Branch Instructions
• BMI - Branch if Minus
– Branch if N = 1
• BPL – Branch if “Plus” (Positive or Zero)
– Branch if N = 0
• BVS – Branch if Overflow Set
– Branch if V=1
• BVC – Branch if Overflow Clear
– Branch if V=0
Branch Instructions
Unsigned
• BHI - Branch if High
– Unsigned, Branch if C OR Z = 0
• BHS – Branch if Higher or Same
– Unsigned, Branch if C=0
• BLO – Branch if Low
– Unsigned, Branch if C = 1
• BLS – Branch if Lower or Same
– Unsigned, Branch if C OR Z = 1
Branch Instructions
Signed
• BGE - Branch if Greater Than or Equal
– Signed, Branch if N XOR V = 0
• BGT – Branch if Greater Than
– Signed, Branch if Z OR (N XOR V) = 0
• BLE – Branch if Less Than or Equal
– Signed, Branch if Z OR (N XOR V) = 1
• BLT – Branch if Less Than
– Signed, Branch if (N XOR V) = 1
Signed vs. Unsigned Inequalities
Branch
Signed
Test
Rationale
BLT
NV = 1
A<B if A−B is really negative;
but if V=1 then the sign bit is wrong.
BGE
NV = 0
A≥B if it’s not true that A<B.
BLE
Z+(NV) = 1
A≤B if either A<B,
or if A=B (that is if A−B=0, so Z=1).
BGT
Z+(NV) = 0
A>B if it’s not true that A≤B.
Unsigned
BLO,BCS C = 1
BHS,BCC C = 0
BLS
C+Z = 1
BHI
C+Z = 0
A<B if A−B produced a carry
(borrow) out of its high-order bit.
A≥B if it’s not true that A<B.
A≤B if either A<B or A=B (Z=1).
A>B if it’s not true that A≤B.
The Seven Possible Subtraction/
Comparison Outcomes
• They are: (1) Normal, (2) Carry, (3) Overflow,
(4) Zero, (5) Negative, (6) Negative-Carry, and
(7) Negative-Overflow-Carry.
Examples
Subtract
B from A (SBA)
A−B=?
Condition
Code Flags
Universal
Signed Only
Unsigned
B B B B B B B B B B B B B B
N E M P G L G L V V H H L L
N Z V C E Q I L E E T T S C I S O S
FF − 80 = 7F
N Z V C 
 

  
01 − FF = 02
N Z V C 
 


81 − 7F = 02
N Z V C 

80 − 80 = 00
N Z V C
FF − 7F = 80
N Z V C 



  
80 − 81 = FF
N Z V C 




7F − 81 = FE
N Z V C 



 
  

 



 


 
 
Proof this list is exhaustive
• If Z=1, then clearly N=V=C=0.
– Thus our list only includes one line where Z=1.
• If V=1, then C=N, because:
– If V=1, then either A−B>127, or A−B < −128.
• If A−B>127, then A>0, and B<0 (so that −B>0).
– Thus, A<$80 and B≥$80, so A−B will do a carry (C=1).
– Also, the result of A−B will be <0 (≥$80), so N=1.
• If A−B<−128, then A<0 and B>0 (so that −B<0).
– Thus, A≥$80 and B<$80, so A−B causes no carry (C=0).
– Also, the result of A−B will be >0 (<$80), so N=0.
– In either case, we note that C=N.
• For V=0 we list all 4 combinations of C and N,
but for V=1 we need only C=N=0 and C=N=1.
Branch Examples
• Example 1
–
–
–
LDAA #$F2 (-14, signed) (242 unsigned)
LDAB #$F0 (-16 signed) (240 unsigned)
CBA (Compute A-B)
•
Result is $F2-$F0 = $F2 + $10 = $102 => $02
–
–
–
–
N bit = 0
Z bit = 0
V bit = 0
C bit = 0
(result is not negative)
(result is not zero)
(2’s comp overflow did not occur)
(this is really, not carry out)
– Note: A > B for both signed and unsigned numbers.
– The following instructions will branch
• BNE,BGE,BGT,BHI,BHS
Branch Example
• Example 2
–
–
–
LDAA #$F2 (-14, signed) (242 unsigned)
LDAB #$F2 (-14, signed) (242 unsigned)
CBA (Compute A-B)
•
Result is $F2-$F2 = $F2+$0E=$100 => $00
–
–
–
–
N bit = 0
Z bit = 1 (result is zero)
V bit = 0
C bit = 0 (not carry out)
– Note: A = B for both signed and unsigned numbers.
– The following instructions will branch
• BEQ,BGE,BLE,BHS,BLS
Branch Example
• Example 3
–
–
–
LDAA #$F2 (-14, signed) (242 unsigned)
LDAB #$FF (-1, signed) (255 unsigned)
CBA (Compute A-B)
•
Result is $F2-$FF = $F2 + $01 = $0F3
–
–
–
–
N bit = 1 (result is negative: -13)
Z bit = 0
V bit = 0
C bit = 1 (not carry out)
Note: A < B for both signed and unsigned numbers.
The following instructions will branch
BNE,BLE,BLT,BLO,BLS
Branch Example
• Example 4
–
–
–
LDAA #$F2 (-14, signed) (242 unsigned)
LDAB #$02 (2, signed) (2 unsigned)
CBA (Compute A-B)
•
Result is $F2-$02 = $F2+$FE=$1F0
–
–
–
–
N bit = 1 (result is negative: signed- positive unsigned)
Z bit = 0
V bit = 0
C bit = 0 (not carry out)
– Note: A < B for signed and A>B unsigned numbers.
– The following instructions will branch
• BNE,BLE,BLT,BHI,BHS
Branch Example
• Example 5
–
–
–
LDAA #$02 (2, signed) (2 unsigned)
LDAB #$F2 (-14, signed) (242 unsigned)
CBA (Compute A-B)
•
Result is $02-$F2 = $02+$0E=$10
–
–
–
–
N bit = 0 (result is positive signed: negative unsigned)
Z bit = 0
V bit = 0
C bit = 1 (not carry out)
– Note: A > B for signed and A<B unsigned numbers.
– The following instructions will branch
• BNE,BGE,BGT,BLO,BLS
Branch Example
• Example 6
–
–
–
LDAA #$80 (-128 signed, +128 unsigned)
LDAB #$7F (+127 signed, +127 unsigned)
CBA (Compute A-B)
•
Result is $80-$7F = $80+$81=$101
–
–
–
–
N bit = 0 (result is negative unsigned, positive signed)
Z bit = 0
V bit = 1 (Two’s comp overflow)
C bit = 0 (not carry out)
• Two’s comp overflow occurs when the carry into the
MSB is not equal to the carry out of the MSB. In this
case, carry in =0 and carry out = 1
Branch Example
• Example 6
–
–
–
LDAA #$80 (-128 signed, +128 unsigned)
LDAB #$7F (+127 signed, +127 unsigned)
CBA (Compute A-B)
•
Result is $80-$7F = $80+$81=$101
–
–
–
–
N bit = 0 (result is negative unsigned, positive signed)
Z bit = 0
V bit = 1 (Two’s comp overflow)
C bit = 0 (not carry out)
– Note: A < B for signed and A>B unsigned numbers.
– The following instructions will branch
• BNE,BLE,BLT,BHS,BHI
Example 4-43
• Let A=$FF and ($1000) = $00
• Given Code Fragment:
– CMPA $1000
• Indicate whether each of the following
branches will be taken
• BGE, BLE, BGT, BLT, BEQ, BNE, BHS,
BLS, BHI, BLO
Example 4-43: Solution
• Let A=$FF and ($1000) = $00
• BGE, BLE, BGT, BLT, BEQ, BNE, BHS,
BLS, BHI, BLO
• CMPA $1000
– As an unsigned number A = 255
– So, for unsigned numbers A>$00, A <> $00
• Unsigned Branches
– BHS yes, BLS no, BHI yes, BLO no
Example 4-43: Solution
• Let A=$FF and ($1000) = $00
• BGE, BLE, BGT, BLT, BEQ, BNE, BHS,
BLS, BHI, BLO
• CMPA $1000
– As a signed number A = -1
– So, for signed numbers A<$00, A <> $00
• Signed Branches
– BGE no, BLE yes, BGT no, BLT yes
Example 4-43:
Solution Summary
• Signed Branches
– BGE no, BLE yes, BGT no, BLT yes
• Unsigned Branches
– BHS yes, BLS no, BHI yes, BLO no
• Other Branches
– BEQ no, BNE yes
TPS Quiz
Test Instructions - TST
• Test if Memory = 0
• Mnemonic Operation
– TST: Test memory = 0: (m) - 0
– TSTA: Test A = 0 : A - 0
– TSTB: Test B = 0: B - 0
• Addressing modes: INH or Ext, X, Y
• Condition codes: N,Z,V=0,C=0
Test Example
PULA ; A  (SP)
* Pull instruction does not affect CCR
TSTA ; Tests if A=0
* This instruction will set N and Z bits
BPL Label ; Branch if A>0
•
Test Bits Instructions - BIT
• Mnemonic Operation
– BITA: AND Register A with memory: A AND (MEM)
– BITB: AND Register B with memory: B AND (MEM)
• Addressing modes: IMM,DIR, Ext, X, Y
• Condition codes: N,Z,C=V=0
• Note: Memory does NOT change, Only CCR
Bit Example
Memory EQU $1000
MASK EQU %10000000
………
LDAA #MASK ; A  ($80)
BITA Memory ; Result = $80 AND $FF = $80
; N=1 and Z=0
ORG Data
Memory FCB $FF
*
Shifting Instructions
Shift Bit Instructions - ASL
• Mnemonic Operation
–
–
–
–
ASL Arithmetic Shift Left Memory
ASLA: Arithmetic Shift Left A
ASLB: Arithmetic Shift Left B
ASLD: Arithmetic Shift Left D (16 bits)
• Addressing modes: INH or Direct, Index
• Condition codes: N,Z,V,C
ASL
C
7
6
5
4
3
Multiplies by 2
2
1
0
0
Shift Bit Instructions - ASR
• Mnemonic Operation
–
–
–
–
ASR Arithmetic Shift Right Memory
ASRA: Arithmetic Shift Right A
ASRB: Arithmetic Shift Right B
ASRD: Arithmetic Shift Right D (16 bits)
• Addressing modes: INH or Direct, Index
• Condition codes: N,Z,V,C
ASR
7
6
5
Sign bit preserved
4
3
Divides by 2
2
1
0
C
Shift Bit Instructions - LSL
• Mnemonic Operation
–
–
–
–
LSL Logical Shift Left Memory
LSLA: Logical Shift Left A
LSLB: Logical Shift Left B
LSLD: Logical Shift Left D (16 bits)
• Addressing modes: INH or Direct, Index
• Condition codes: N,Z,V,C
LSL
C
7
6
5
4
3
Multiplies by 2
2
1
0
0
Shift Bit Instructions - LSR
• Mnemonic Operation
–
–
–
–
LSR Logical Shift Right Memory
LSRA: Logical Right Left A
LSRB: Logical Right Left B
LSRD: Logical Right Left D (16 bits)
• Addressing modes: INH or Direct, Index
• Condition codes: N=0,Z,V,C
LSR
0
7
6
5
4
3
Divides by 2
2
1
0
C
Shift Bit Instructions - ROR
• Mnemonic Operation
– ROR Rotate Right Memory
– RORA: Rotate Right A
– RORB: Rotate Right B
• Addressing modes: INH or Direct, Index
• Condition codes: N,Z,V,C
ROR
7
6
5
4
3
2
1
0
C
Shift Bit Instructions - ROL
• Mnemonic Operation
– ROL Rotate Left Memory
– ROLA: Rotate Left A
– ROLB: Rotate Left B
• Addressing modes: INH or Direct, Index
• Condition codes: N,Z,V,C
ROL
7
6
5
4
3
2
1
0
C
Logic Instructions
• Mnemonic Operation
–
–
–
–
–
–
ANDA: Logical A and mem: A  A AND (M)
ANDB: Logical B and mem: B  B AND (M)
EORA: Exclusive OR A and mem: AA XOR (M)
EORB: Exclusive OR B and mem: BB XOR (M)
ORAA: OR A and mem: A A OR (M)
ORAB: OR B and mem: B B OR (M)
• Addressing modes: All except IHN
• Condition codes: N,Z,V,C
Complement Instructions
• Mnemonic Operation
– COM: 1’s Complement of mem: (M)  (M)
– COMA: 1’s Comp of A: A  (A)
– COMB: 1’s Comp of B: B(B)
• Addressing modes: INH or EXT, X, Y (mem)
• Condition codes: N,Z,V=0,C=1
Subroutine Instructions
JMP/JSR/BSR/RTS
JMP Instructions
• Jump to Address
– Similar to BRA but uses absolute address
• Mnemonic Operation
– JMP Address
• Addressing modes: EXT, X, Y
– Condition codes: none
– PC  Address
JMP Example
Program EQU $E000
Stack
EQU $00FF
ORG Program
E000: 8E 00 FF Top:
E003: 7E E0 03 Done:
********** Compare with
E000: 8E 00 FF Top:
E003: 20 FE
Done:
LDS #Stack
JMP Done
LDS #Stack
BRA Done
Why use JMP instead of BRA?
Why use JMP instead of BRA?
We can only Branch up to -128 to +127 bytes from the
current address. If we need to Branch to an address
outside of this range, we’ll need to use JMP instruction.
* EXAMPLE. Assume FAR_LABEL is +$1000 bytes away
from this address
TSTA
BEQ FAR_LABEL
LDAA #NUMA ; This is the A<>0 code
* This code will give us an error (out of range)
Why use JMP instead of BRA?
* We need to use a JMP instead
TSTA
BNE L1
; Use a BNE because the label
; is local.
JMP FAR_LABEL
L1: LDAA #NUMA ; This is the A<>0 code
JSR Instructions
• Jump to Subroutine
– Similar to JMP except we have the ability to return to
the calling program. Same as HLL.
• Mnemonic Operation
– JSR Address
• Addressing modes: DIR, EXT, X, Y
– Condition codes: none
• PSH PC ; Program Counter pushed onto stack
• PC  Address
JSR Instruction
• JSR
– Pushes Program Counter onto Stack
• Note: PC is pointing to the next instruction
– Loads PC with Addressing of Subroutine
• Starts executing program beginning at EA
– How do we return from the subroutine?
BSR Instructions
• Branch to Subroutine
• Mnemonic Operation
– BSR Relative_Address
• Addressing modes: REL
– Condition codes: none
• PSH PC ; Program Counter pushed onto stack
• PC  PC + Relative_Address
• Note: PC is pointing to the NEXT instruction.
JMP/JSR/BSR Instructions
• JMP – Jump to Address
– Unconditional jump to address
• JSR- Jump to Subroutine
– Unconditional jump to subroutine
• BSR – Branch to Subroutine
– Unconditional branch to a subroutine
RTS - Instruction
• RTS – Return from Subroutine
– PULL PC from Stack
• Recall this contains the EA of the instruction
following the original JSR/BSR
– PC  Value pulled from stack
• Execute program
– Note: You must POP anything you have
PUSHed onto the stack or the RTS will start
executing the wrong program
Special Arithmetic Operations
Hexadecimal Number System
•
•
•
•
•
Base 16
Sixteen Digits: 0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F
Example: EF5616
Positional Number System
16
n1
4
3
2
1
16 16 16 16 16
0
0000
0001
0
1
0100
0101
4
5
1000
1001
8
9
1100
1101
C
D
0010
0011
2
3
0110
0111
6
7
1010
1011
A
B
1110
1111
E
F
Packed Binary Coded Decimals
(Packed BCDs)
• Use 4-bits 0-9 to represent a decimal number
• Example: 123410
– As hex = $4D2
– As BCD =$0102, $0304
– As Packed BCD = $1234
• Example: 5137
– As Hex = $1411
– As BCD = $0501,$0307
– As Packed BCD = $5137
Subroutine Example
*************************
*
Subroutine Name: Bcd2Asc
*
Input parameter list:
A
*
Output parameter list: A,B
****************************
* This routine converts a Packed BCD byte in A
* into a pair of ASCII characters in A,B.
**************************************************
LOWER
EQU
$0F
; Mask for low nybble
UPPER
EQU
$F0
; Mask for high nybble
ASCII
EQU
$30
; ASCII code for numeral 0
Bcd2Asc: TAB
; Copy A register to B
ANDB #LOWER
; Mask-select lower nybble of B
ADDB #ASCII
; Convert B to ASCII code
ANDA #UPPER
; Mask-select upper nybble of A
LSRA
; Logical shift right 4 positions
LSRA
LSRA
LSRA
ADDA #ASCII
; Convert A to ASCII
RTS
; Return result in A,B
Using Subroutine
* Standard equates
CODE
EQU $E000
DATA
EQU $0000
STACK
EQU $7FFF
EOS
EQU $FF
array
result
ORG
FCB
RMB
ORG
Main:
LDS
LDX
LDY
Loop:
LDAA
CMPA
BEQ
JSR
STD
INX
INY
INY
BRA
Done:
WAI
Infloop: BRA
;
;
;
;
Start
Start
Start
We’ll
code in EEPROM
data in page 0 internal RAM
stack at top of external RAM.
use hex $FF to mark end-of-string of BCD bytes.
DATA
$01,$23,$45,$67,$89,EOS
10
; Begin data segment.
; Array of packed BCD bytes
; We'll put the resulting chars here
CODE
#STACK
#array
#result
0,X
#EOS
Done
BCD2ASC
0,Y
;
;
;
;
;
;
;
;
;
;
;
Loop
; Do it all again
; End of main program, wait for interrupts
; If we ever get here, stay here
Infloop
Begin code segment.
Load stack pointer
Load address of data array
Ditto for result array
Get byte to convert
Compare byte with end of string
If end of string, then stop
Call subroutine to do conversion
Store D in result array
Go to the next input byte
Go to next output word
Decimal Arithmetic Instruction
• DAA – Decimal Adjust A Instruction
– Used to adjust the result of the addition of two
packed BCDs
– Use immediately after the ADDA instruction.
• Affects the A register
– Example:
•
•
•
•
34 = $22 (hex) or $34 as packed BCD
29 = $1D (hex) or $29 as packed BCD
$34+$29 = $5D (not a valid packed BCD)
DAA: A  (A) +$06 = $63 (correct packed BCD)
Decimal Arithmetic Instruction
• DAA – Decimal Adjust Instruction
– Why do we add $06?
– Note:
• $01+$09 = $0A + $06 = $10
– What about $01+$02 = $03?
– DAA checks CCR to determine if adjust is
needed, so DAA: $03+$00 = $03
– See reference manual (p.536) for a
complete list of cases
MUL Instruction
• MUL – Multiply Instruction
– Multiplies 8-bit unsigned numbers loaded into the A
and B registers
– D  AB
– Max result = $FF$FF = $FE01 = 65,020 =2552
– For signed multiplication:
• Determine sign of A and B
– (−)(−)=(+), (+)(+)=(+),
– NEG negative numbers
– MUL, NEG (if needed)
(−)(+)=(−),
• Range: -1282 = -16,384 = $E100 to
+1272 = 16,129 = $3F01
(+)(−)=(−)
IDIV Instruction
• IDIV – Integer Divide Instruction
– Divides 16-bit unsigned D register by the 16-bit
unsigned X register
– Quotient is stored in the X register
• X  Integer portion of D/X
– Remainder is stored in the D register
• D  D – Integer portion of D/X
– If X is $0000, C 1, and D  $FFFF
– For signed division
• Determine sign of A and B
– (-)/(-)=+ (+)/(+)=+ (-)/(+)=(-) (+)/(-)=– NEG negative numbers
– IDIV, NEG (if, needed)
Fractional Numbers
• Recall 20=1, 21=2, 22=4, ….
• But, we can also go the other way
– 2-1=0.5, 2-2=0.25, 2-3=0.125, 2-4=0.0625,…
• In general, we have 2-n=1/2n
• So, for example, to represent
– 2.5 = %10.10 and 5.25 = %101.01
• Check: 2+0+0.5+0=2.5
• Check: 4+0+1+0+0.25=5.25
Fractional Numbers
• We have
– 2.5 = %10.10 and 5.25 = %101.01
• As 8 bit numbers, these become
– %000010.10 and %000101.01
• Now, we really DON’T have a way to represent the
decimal point in our numbers, so we have to
remember where it is located.
• So really, we have
• 2.5 = %00001010 = $0A
• 5.25 = %00010101 = $15
• And we remember that we are using 2 bits for the
fractional part.
Fractional Numbers
• Let’s use 16-bit numbers where 4-bits (or one
nybble) are reserved for the fractional part
• So, Y = $NNN.N
• Example, What is 375.875 in Hex?
• 375  $177
• 0.875 = 1*0.5+1*0.25+1*0.125+0*0.0625
= %1110 = $E
• Or, 375.875 = $177E
Fractional Numbers
• Let’s use 16-bit numbers where 4-bits (or one
nybble) are reserved for the fractional part
• So, Y = $NNN.N
• Example, What is $24E3 as a fractional decimal?
• $24E  590
• $3 = 0*0.5+0*0.25+1*0.125+1*0.0625=0.1875
• Or, $24E3 = 590.1875
FDIV Instruction
• FDIV – Fractional Divide Instruction
– Divides 16-bit unsigned D register by the
16-bit unsigned X register
– D register is assumed less than X register.
(Answer is less than one)
– Radix point is assumed to be the same
– Fractional quotient is stored in the X register.
Ex: %0.1010….
– Remainder is stored in the D register
– If X is $0000, C 1, and D  $FFFF
Miscellaneous Instructions
NOP Instruction
• NOP – No operation
– Performs “no operation”
– Can be used for “crude” timing routines
• 2 Clock Cycles for each instruction
– Also used for debugging
• Insert NOPs in place of SWI
– Software Interrupts
STOP Instruction
• STOP– STOP operation
– Puts the CPU “to sleep”
• Can be “awakened” by using an interrupt
– Can be disabled by setting the STOP bit
Interrupt
Instructions
Interrupt Instructions
• CLI– Clear Interrupt Mask
– Clears I bit to 0
• SEI – Set Interrupt Mask
– Sets I bit to 0
• RTI – Return from Interrupt
• SWI – Software Interrupt
• WAI – Wait for Interrupt
End of Section
TPS Quiz
Example 4-43
• Let A=$FF and ($1000) = $00
• A = -1 (signed), A=255 (unsigned)
• Given Code Fragment:
– CMPA $1000
– $FF - $00 = $FF+$00 = $FF
– Condition Code Register
•
•
•
•
N=1
Z=0
V=0
C=0
Example 4-43: Solution
• We have: A=$FF, Mem=$00 (CMPA Mem)
• CCR: N=1, Z=0, V=0, C=0
• Mnemonic:Condition checked|Our example| Yes or No
• BEQ: Z = 1| Z = 0 | No
• BNE: Z = 0| Z = 0 | Yes
Example 4-43: Solution
Unsigned Branches
• We have: A=$FF, Mem=$00 (CMPA MEM)
• CCR: N=1, Z=0, V=0, C=0
•
•
•
•
•
Condition | Example | Yes or No
BHS: C =0| C=0| Yes
BLS: C OR Z = 1| 0 or 0 = 0| No
BHI: C OR Z = 0 | 0 or 0 = 0| Yes
BLO: C = 1| C=0| No
Example 4-43: Solution
Signed Branches
• We have: A=$FF, Mem=$00 (CMPA Mem)
• CCR: N=1, Z=0, V=0, C=0
•
•
•
•
•
Branch : Condition | Example | Yes or No
BGE: (N XOR Z) = 0 | (1 XOR 0) = 1 | No
BLE: Z OR (N XOR Z) = 1| (0 OR (1 XOR 0)) =1|Yes
BGT: Z OR (N XOR Z) = 0| (0 OR ( 1 XOR 0)) =1| No
BLT: (N XOR Z) = 1 | (1 XOR 0) = 1 | YES