SECT_04_ASSEMBLY - Advanced Microcomputer Systems

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Assembly
Language
Programming
for the
MC68HC11
Assembly language programming
 Recall the 4 design levels for software
development:
– Application
– High Level Language
– Assembly
– Machine code (often called object code)
 Machine code programs are the only ones
that can be directly executed on the
processor -- programs in the higher 3 levels
must be converted to machine code
CONTD
Assembly language programming
– High level language programs must be compiled
» Synthesize an assembly-level intermediate form
» Convert the intermediate form to machine code
– Assembly language programs must be assembled
» Translated into machine code
» The synthesis part of compiling is not required
» One-to-one translation of instructions
Assembly language programming
Assembly level programming
– Advantages (over HLL programs)
» Faster
» Less memory
» Enhanced control over the hardware
– Disadvantages (vs. HLL programming)
» Coding requires more "talent"
» Life cycle support more difficult
» Program tied to a specific HW platform
--- Not portable
Overview of the AS11 assembler
 Assembly language programs consist of
3 types of statements:
» Instructions that will be executed by the
microprocessor
» Comments to document the program operation for
humans
» Assembler directives (pseudo-operations) that tell
the assembler what to do
CONTD
Overview of the AS11 assembler
 Instructions:
– Format is shown below:
label operation operand ;comment
– Labels
» Optional
» Must start in column 1 and begin with an
alphabetic character
» No more than 15 characters long
» Delimit with a space or colon
» May be on a line by itself
Overview of the AS11 assembler
– Operation Operand
» The instruction mnemonic from the
instruction set followed by any needed operands
» The opcode cannot start in column 1 of the
source file
» It a label is present, there must be at least one
space or a colon between it and the opcode
– Comments
» Comments in an instruction line are delimited
by a semicolon
CONTD
Overview of the AS11 assembler
» Everything to the right of a semicolon is treated
as a comment (ignored by the assembler)
» If a line starts with a semicolon or an asterisk in
column 1, the entire line is treated as a comment
» Use enough comments to explain program
function -- not so many as to "flood" reader
» Use meaningful comments; the following is not
very useful
LDDA #$FC ; load $FC into A
Overview of the AS11 assembler
– Directives and pseudo-operations
» These statements cause the assembler to
perform certain actions -- they are converted
directly to executable code
» ORG xxxx
 Place the next byte of the program at
address $xxxx
 The programmer uses this to place logical
program groups into different memory areas
CONTD
Overview of the AS11 assembler
» EQU
label EQU xxxx
 The label is assigned the constant value xxxx
The programmer uses EQU to give meaningful
names to constant numeric data – enhances code
readability
 Examples:
TRUE:
EQU
1
MAXINT
EQU
$FF
Overview of the AS11 assembler
» Memory allocation
 You should always allocate any memory locations
that you use
– Variables (RAM)
– Constants (ROM/EEPROM)
– Strings (ROM/EEPROM)
 FCC ‘ASCII characters’
– Declare a string
(Form Constant Characters)
– Assembler converts characters to
ASCII
– Example:
string1: FCC ‘This is string1.’
Overview of the AS11 assembler
 FDB word,word, . . .
– Declare 16-bit constants
(Form Double Bytes)
– Words may be constants, symbols, or expressions
– If you have more than one word, they must be
separated by commas, with no spaces between them
– Example:
Jump_Table: FDB $E000,$E010,$E020
CONTD
Overview of the AS11 assembler
 FCB byte,byte, . . .
– Declare 8-bit constants
(Form Constant Bytes)
– Bytes may be constants, symbols, or expressions
– Separate bytes with commas, no paces between
them
– Example:
sqr_tbl: FCB $0,$1,$4,$9,$10
Overview of the AS11 assembler
RMB size
– Declare storage for variables
(Reserve Memory Bytes)
– Reserves size number of bytes (bytes are not
initialized)
– Size may be constant, symbol, or Expression
– Example: Declare a 128-byte buffer
at address $100
ORG $100
Buffer: RMB 128
Overview of the AS11 assembler
» INCLUDE “filename”
» INCLUDE <filename>
 Include the specified file into the source code
 Useful for initialization and definition of sections
of code that are common across many programs
 For example, I/O register names
» Output control
 PAGE
 OPT option
– Available options: l, nol, c, noc,
contc, cre, s, crlf, nnf, p50
Overview of the AS11 assembler
» Conditional assembly:
 IFD symbol
 IFND symbol
 ELSE
 ENDIF
– Checks if the symbol has been
defined or not defined
– Each IFD or IFND must have a
matching ENDIF
– ELSE statements are optional
CONTD
Overview of the AS11 assembler
– IF/ELSE/ENDIF blocks may be
nested
– Format:
IFD symbol
; code, more IFs, comments, etc.
ELSE
; code, more IFs, comments, etc.
ENDIF
Overview of the AS11 assembler
 The assembly process
– Most assemblers perform their job by making
2 “passes” over the source code
– Pass 1 identifies all symbolic references to
memory locations, and to the starting locations
of all instructions
» Symbol table is built during Pass 1
CONTD
Overview of the AS11 assembler
Contains name and value for each
symbol
Symbols may be labels or they may
be created with EQU directives
– Pass 2 converts all symbolic references to
absolute memory references and produces the
final object code
» Uses the symbol table information
Overview of the AS11 assembler
Assembler first pass [Sho87]
Assembler second pass [Sho87]
Overview of the AS11 assembler
– Assembler outputs:
» Object code: A specification of the actual bytes
that will be placed in the HC11’s memory to be
executed
 On our systems, this is an ASCII file where
binary code is represented as strings of hex
equivalents
– Motorola S-records format
– Intel hex format
CONTD
Overview of the AS11 assembler
Must be "loaded" into memory, at which time
the ASCII-encoded hex digits are mapped to
actual memory contents
» Listing file: composite file giving both the source
assembly code and the corresponding memory addresses
and their contents
This file is particularly useful for debugging coding
errors
Overview of the AS11 assembler
General Format of Listing File
<address> <code> <source line no.> <source line>
– address = Starting address in memory of the
instruction
– code = Hex digits that are the values put into the
memory locations
– source line no. = line number
CONTD
Overview of the AS11 assembler
– source line = a
instruction
ssembly source code
– Other things that can be seen: Comments,
cycle count, symbol table at the end of file,
etc.
 Example:
018A 8612 203 LDAA #$12
018C 9634 204 LDAA $34
018E B65678 205 LDAA $5678
Overview of the AS11 assembler
 Example: Write a program to find the
minimum value in a list of unsigned integers.
Assume that the starting address of the list is
stored in memory location START, and the
length of the list is stored in location LENGTH.
Store the minimum value in memory location
RESULT.
CONTD
Overview of the AS11 assembler
Pseudocode:
Set min = MAX_INT;
for each item in the list
{
if (list item < min)
{
min = list item
}
}
Store the min in RESULT
Overview of the AS11 assembler
One implementation (in C):
int *START;
int LENGTH;
int RESULT;
int *ptr;
int count;
int min;
min = MAX_INT;
count = LENGTH;
ptr = START;
while (count != 0)
{
if (*ptr < min)
{
min = *ptr;
}
ptr++;
count--;
}
RESULT = min;
Overview of the AS11 assembler
;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;
; Program: FIND_MIN -- Finds the minimum value of
; a list of unsigned integers.
;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;
MAX_INT EQU $FF ; largest unsigned integer
ORG $100
START RMB 2 ; starting address of list
LENGTH RMB 1 ; length of list
RESULT RMB 1 ; minimum value
ORG $E000
FIND_MIN:
CONTD
Overview of the AS11 assembler
FIND_MIN:
LDAA #MAX_INT ; ACCA = min = MAX_INT
LDAB LENGTH ; ACCB = count = LENGTH
LDX START ; IX = ptr = START ;
LOOP:
CMPA 0,X ; compare list item with min
BLS END_LOOP ; branch if min <= list item
LDAA 0,X ; else, update min
END_LOOP:
INX ; increment list pointer
DECB ; decrement counter
BNE LOOP ; and repeat if counter > 0 ;
STAA RESULT ; store minimum value
Overview of the AS11 assembler
Overview of the AS11 assembler
 Disassembly
– Translate from object code/machine code back to source
code
– You usually lose symbols, labels, etc.
– Example: (problem 12, chapt. 2 in text)
E000 7F 50 00 7C 50 00 CE 10 - 00 18 CE 20 00 08 18 09
E010 A6 00 18 A7 00 78 50 00 - 24 F3 01 01 01 01 01 01
CONTD
Overview of the AS11 assembler
CLR $5000
INC $5000
LDX #$1000
LDY #$2000
LABEL:
INX
DEY
LDAA 0,X
STAA 0,Y
LSL $5000
BCC LABEL
– What does this program do?
Overview of the AS11 assembler
 Example: Exercise 13 from Chapt. 2
– Write a program to clear the first 20 bytes of
RAM (set them to zero)
» One solution:
ORG $B600
LDX #$00 ; start with address $00
Loop:
CLR $0,X ;
clear the byte
INX ;
increment addr pointer
CPX #20 ;
addr < 20?
BLO Loop ;
branch if so
Overview of the AS11 assembler
 Example: Exercise 13 from Chapt. 2
– A slightly different solution
ORG $B600
LDX #19 ; s
tart with end of block
Loop:
CLR $0,X ;
clear the byte
DEX ;
decrement addr pointer
BNE Loop ;
branch if not done
CLR $0,X
; why do we need this?
Overview of the AS11 assembler
 Example:
– Write a routine that will delay for a specified number of
milliseconds. (The number of milliseconds will be stored in
memory location DELAY_VALUE). Assume the system
uses an 8 MHz crystal. The delay should be accurate to +/1%.
» Remember that the E-clock, or system clock, is the
internal clock frequency of the processor.
CONTD
Overview of the AS11 assembler
 E-clock = crystal frequency / 4
= 8 MHz / 4
= 2 MHz
 Each clock period is therefore 0.5 µsec
 1 ms = 2000 clock cycles
 So, we need to write a routine that will
take 2000 clock cycles (+/- 20 cycles) to
execute
Overview of the AS11 assembler
 Solution(?):
DELAY_VALUE RMB
1
DELAY:
LDAA #LOOP_CNT ;
2 cycles
LOOP:
DECA ;
2 cycles
BNE LOOP ;
3 cycles
;
DEC DELAY_VALUE ;
6 cycles
BNE DELAY ;
3 cycles
– What value should we use for LOOP_CNT?
Overview of the AS11 assembler
 Corrected solution:
DELAY_VALUE RMB 1
DELAY:
LDAA #LOOP_CNT ;
2 cycles
LOOP:
NOP ;
2 cycles
NOP ;
2 cycles
DECA ;
2 cycles
BNE LOOP ;
3 cycles
;
DEC DELAY_VALUE ;
6 cycles
BNE DELAY ;
3 cycles
– Now what should LOOP_CNT be?
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