me4447/6405 me 6405ume.gatech.edu/mechatronics_course/lecture9.pdf · me4447/6405 write an assembly...

George W. Woodruff School of Mechanical Engineering, Georgia Tech

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ME 6405

Introduction to Mechatronics

Fall 2005

Instructor: Professor Charles Ume

LECTURE 9


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1. Write an assembly language program to clear the usable internal RAM in the M68HC11E9.

Solution: Clear Locations $0000 - $002C.

ORG $1040LDX #$0000

LOOP CLR $00,XINXCPX #$002DBNE LOOPSWIEND

(NOTE: 1. You could have also written this program to clear $0100 to $01FF2. You should hand-assemble this program.)

Homework 1 Solution


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2. Write a program to add even/odd numbers located in addresses $0000 through $00FE.Solution: (for adding odd numbers)

ORG $1040LDY #$0000 1040 18 CE 0000LDX #$0000 1044 CE 0000

IND LDAA #$FF 1047 86 FFLDAB $00,X 1049 E6 00

0000 00110000 0001 BITB #$01 104B 85 010000 0001

BEQ EVEN 104D 27 02ABY 104F 18 3A

EVEN INX 1051 08STX $2000 1052 FF 2000

1ST time:#$2001: 0000 0001 EORA $2001 1055 B8 2001acc. A: 1111 1111

1111 1110 BNE IND 1058 26 EDSWI 105A 3FEND

ORDECA 4AIf you usethis, then IND shouldbe moveddown to thenext line


ME4447/6405ME4447/6405Write an assembly language program to find the largest signed number in a list of numbersstored in address $0020 thru $0029. Repeat for unsigned number.

FIRST EQU $0020LAST EQU $002AANSWR EQU $0010

ORG $1040 LDX #FIRSTLDAA FIRSTSTAA ANSWRINX Goes with DECB. If DECB is removed, then LDAB #$0A LDAB must be removed

NUB LDAA $00,X CMPA ANSWR (A - ANSWR)

Branch if accumulator A BLE NEXT; (signed)* *(Branch if ≤ zero.is less than ANSWR Branches if 2's complement

OR number in register A was less or equal to the 2's complement number represented by ANSWR)

BLS NEXT; (unsigned)** **(Branch if lower or same, i.e, Branch if register A was lower or same as ANSWR)

STAA ANSWRNEXT INX

CPX #LAST OR DECB BNE NUBSWIEND

Homework 2 Solution


ME4447/6405ME4447/6405 Homework 2 Solution Continued

Signed Example:

#$3F (positive)#$F4 (negative)

Convert Numbers to Decimal:

#$F4 = %11110100

1's Comp #$0B = %000010112's Comp #$0C = %00001100

#$F4 = -1210

#$3F = %00111111 #$3F = 6310

#$3F is larger than #$F4 since 6310 is larger than -1210

Un-Signed Example:

#$3F (positive)#$F4 (positive)

Convert Numbers to Decimal:

#$F4 = %11110100 #$F4 = 24410

#$3F = %00111111 #$3F = 6310

#$F4 is larger than #$3F since 24410 is larger than 6310


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Homework

Homework 3• Write a sample program to output a string of characters to the

screen.

• Write a subroutine to save the first 5 odd (8-bit) numbers pointed to by the x-register (passed in) onto the stack.

Homework 4• Write a program to output a square wave thru port D pin 2. The

output can be observed on the scope, and the period T of the wave should be measured. More than one period wave should be generated. The machine cycle time of the M68HC11E9 should be estimated. Draw the square wave.


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ME4447/6405ME4447/6405 ASCII Character Codes

ASCII – American Standard Code for Information InterchangeASCII assigns a hexadecimal ‘code’ to individual characters

Examples:Character ASCII‘A $41‘E $45‘e $65 (Note: Different codes for lower

and upper case)‘1 $31BS $08 (Note: BS is Backspace. )

A microcontroller must send these codes to a display terminal inorder for the terminal to display them


ME4447/6405ME4447/6405 ASCII Character Codes (continued)

Hex to ASCII Conversion table from Programming Reference Guide Page 58


ME4447/6405ME4447/6405 ASCII Character Codes

An array of characters is called a stringExample:

character array String ASCII Representation‘H ‘e ‘l ‘l ‘o “Hello” $48 $65 $6C $6C $6F

(Note:Character ASCII‘H $48‘e $65‘l $6C‘o $6F

)


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Assembly Directive Types

Assembly Control• ORG, END

Symbol Definition• EQU

Data Definition/Storage Allocation• FCC, FCB, FDB, RMB, ZMB, BSZ, FILL

Listing Control• PAGE, OPT


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Assembly Control


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ORG and END

ORG : Store translated machine language instructions in sequencestarting at given address for any mnemonic instructions that follow

END: Stop translating mnemonics instructions until another ORG is encountered

(Note: These were already discussed in Lecture 7)


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Symbol Definition


ME4447/6405ME4447/6405 EQUEQU lets you refer to a number or an address as a variable name.

Example:VALA EQU $10 *LABEL VALA USED TO REFER TO $10

ORG $1040LDAA #VALA *LOAD HEX NUMBER $10 IN ACCUMULATOR A LDAB VALA *LOAD CONTENT OF MEMORY LOCATION $10

*IN ACC. BSWIEND

Same As:

VALA EQU $10 *LABEL VALA USED TO REFER TO $10ORG $1040LDAA #$10 *LOAD HEX NUMBER $10 IN ACCUMULATOR A LDAB $10 *LOAD CONTENT OF MEMORY

*LOCATION $10 IN ACC. BSWIEND


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Data Definition/Storage Allocation


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FCCFCC – Form Constant Character string• FCC stores ASCII characters into consecutive bytes of memory. • Any printable ASCII characters can be contained in the string. • String is specified between two identical delimiters, which can be

any printable ASCII character. • First non-blank character after the string is used as a delimiter.


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FCC Cont’d

$00

$01

$02

$03

$04

Address Prebyte Opcode Operand

31

32

33

34

35

Example:

ORG $00FCC “12345”

Result


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FCB

FCB – Form Constant Byte• FCB has one or more operands. • Value of each operand is truncated to eight bits, and is

stored in single byte of object program. • Operand may be a numeric constant, character

constant, a symbol or an expression. • Multiple operands are separated by commas, and are

stored in successive memorybytes.


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FCB Cont’d

Example: VALA EQU $10ORG $00FCB $34,’A, $28AC, $0A ,VALA

$00

$01

$02

$03


34

41

AC

0A

Result

$04 10


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FDB

FDB – Form Constant Double Byte• FDB stores a double (two byte) word. • May have one or more operands separated by

commas.• Operand may be a numeric constant, a character

constant, a symbol, or an expression.


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FDB Cont’d

Example:

ORG $00FDB $1234,’&,’G

Note: ASCII value for & is $26ASCII value for G is $47

$00

$01

$02

$03

$04


12

34

00

26

00

47$05

Result


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RMBRMB – Reserve Memory Byte• RMB saves a place in memory for a

number. Example:

ORG $00XVAR RMB 2 *TWO MEMORY

*LOCATIONS $00*and $01 ARE*RESERVED FOR XVAR

ORG $1040LDD #$FFAASTD XVARSWIEND

$00

$01


FF

AA

Result


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Question a student asked:

What happens if you change the previous example to "XVAR RMB 3" instead of "XVAR RMB 2" ? What happens to the 3rd reserved byte when a 2 byte number is stored in XVAR?

Modified Program:

ORG $00XVAR RMB 3

ORG $1040LDD #$FFAASTD XVARSWIEND

Answer: Remains unchanged


ME4447/6405ME4447/6405 ZMB, BSZZMB – Zero Memory Byte and BSZ – Block Storage of Zero• These directives fill a given number of memory locations with zero. • Causes assembler to allocate a block of memory bytes, and each

memory byte is assigned a value of zero. • Both directives do the same thing.• Number of bytes allocated is given in the operand field.

Example:ORG $00ZMB #$02BSZ #$02

$00$01


00

00$02$03

00

00

Result


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FILL

Fill given number of memory locations with anynumber. (Note: Fill uses one byte. If two bytes are specified, then it will truncate it and use LS Byte.)

Example:

ORG $00FILL #$FF, #$02 $00

$01


FF

FF

Result


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Question a student asked:What happens when the previous example is changed to "FILL #$9ABC, #$02" instead of "FILL #$FF,#$02"? What happens if you fill memory with a 2 byte number?

Answer: FILL will just use the LS Byte

Modified Example:

ORG $00FILL #$9ABC,#$02END

(Note: There is no "Go 1040 on the screen since these are just assembly directives and not a program)


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Listing Control


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PAGE

PAGE • The PAGE directive causes a page break in

the list file. • If no source listing is being produced, the

PAGE directive will have no effect. • The directive is not printed on the source

listing.


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OPTOPT• Allows for various options in assembly of a program, including

generating a listing and counting instruction cycles.

Options:nol-no output listing (default)l-do an output listing noc-no cycle number count (default)c-turn on cycle count using zero initial valuecontc-turn cycle count on, begin with last valuecre-create a cross reference table (default anyway) RMBs-create a symbol table (default anyway) EQU

Example:OPT l – Print source listing from this point


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Buffalo Monitor Utility Subroutines


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Buffalo Monitor Utility Subroutines

Buffalo Monitor Utility Subroutines• These subroutines are available for performing I/O

tasks. • A jump table has been set up in ROM

directly beneath the interrupt vectors. • To use these subroutines, execute a jump to

subroutine (JSR) command to the appropriate entry in the jump table.


ME4447/6405ME4447/6405 Subroutine ListingUPCASE If character in accumulator A is lower case alpha, convert to

upper case.WCHEK Test character in accumulator A and return with Z bit set if

character is whitespace (space, comma, tab).

DCHEK Test character in accumulator A and return with Z bit set if

character is delimiter (carriage return or whitespace).

INIT Initialize I/O device.

INPUT Read I/O device.OUTPUT Write I/O device.

OUTLHLF Convert left nibble of accumulator A contents to

ASCII and output to terminal port.

OUTRHLF Convert right nibble of accumulator A contents to

ASCII and output to terminal port.


ME4447/6405ME4447/6405 Subroutine Listing Cont’d

OUTA Output accumulator A ASCII character.OUT1BYT Convert binary byte at address in index register X to two ASCII

characters and output. Return address in index register X pointing tonext byte.

OUT1BSP Convert binary byte at address in index register X to two ASCIIcharacters and output followed by a space. Returns address in indexregister.

OUT2BSP Convert two consecutive binary bytes starting at address in index X tofour ASCII characters and output followed by a space. Returns addressin index register X pointing to next byte.

OUTCCRLF Output ASCII carriage return followed by a line feed.OUTSTRG Output string of ASCII bytes pointed to by address in index register X

until character is an end of transmission ($04).OUTSTRGO Same as OUTSTRG except leading carriage return and line feed is

skipped.INCHAR Waits for you to type an ASCII character from the keyboard, and stores

the corresponding ASCII number in accumulator A, and then outputs the ASCII character to the screen.


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• $FFAO UPCASE Convert character to uppercase• $FFA3 WCHEK Test character for whitespace• $FFA6 DCHEK Check character for delimiter• $FFA9 INIT Initialize I/O device• $FFAC INPUT Read I/O device• $FFAF OUTPUT Write I/O device• $FFB2 OUTLHLF Convert left nibble to ASCII and output• $FFB5 OUTRHLF Convert right nibble to ASCII and output• $FFB8 OUTA Output ASCII character• $FFBB OUTlBYT Convert binary byte to 2 ASCII characters and output• $FFBE OUT1BSP Convert binary byte to 2 ASCII characters and output

followed by space• $FFCl OUT2BSP Convert 2 consecutive binary bytes to 4 ASCII

characters and output followed by space.• $FFC4 OUTCRLF Output ASCII carriage return followed by line feed• $FFC7 OUTSTRG Output ASCII string until end of transmission ($04)• $FFCA OUTSTRGO Same as OUTSTRG except leading carriage return and

line feed is skipped• $FFCD INCHAR Input ASCII character and echo back

To use an I/O subroutine, Jump Sub Routine (JSR) to the specified addresslisted below.

ADDRESS SUBROUTINE FUNCTION


ME4447/6405ME4447/6405 Utility Subroutines Examples

Example: OUTA ORG $1040LDAA #$41 *Load acc. A with ASCII code for the

*character A.JSR $FFB8 *JSR to the subroutine OUTA SWIEND

Resultà A is written to the screen.


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Example: OUTSTRG (Out String)STR1 EQU $2100OUTSTRG EQU $FFC7

ORG STR1FCC “ABCDEFG”FCB #$04

ORG $1040LDX #STR1JSR OUTSTRG *Go to OUTSTRG routine which

*outputs ASCII characters contained in each*address starting from $2100 until*an EOT character is found.

SWIEND

ResultàABCDEFG written to the screen on a new line.

(Note: ASCII end of transmission (EOT) character must be used with theOUTSTRG subroutine to let it know when to stop reading from memory. EOT ASCII = #$04)

Utility Subroutines Examples


ME4447/6405ME4447/6405Difference Between the Utility Subroutines OUTSTRG and

OUTSTRG0

Assume you want to print the following to the screen:MEMORY LOCATION $_____CONTAINS _____ IN DECIMAL.

Use subroutine OUTSTRG to print “MEMORY LOCATION $” to screen on a new line (It always starts printing on a new line, because it outputs carriage return with line feed).Use subroutine OUTSTRG0 to print “CONTAINS” to screen on same line as “MEMORY LOCATION $” (Because it does not output carriage return with line feed)Use subroutine OUTSTRG0 to print “IN DECIMAL.” to screen on the same line

(Note: Refer to Lab website for example program to printdecimal numbers to the screen.)


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Example: OUTLHLF (Out Left Half) and OUTRHLF (Out Right Half)

ORG $1040LDAA #$AF *Puts hex number $AF in acc. AJSR $FFB2 *Goes to subroutine OUTLHLF which converts left

*nibble, #$A, to ASCII number, $41, and then outputs*its ASCII character, ‘A, to the screen.

JSR $FFB5 *Goes to subroutine OUTRHLF which converts right*nibble, #$F, to ASCII number, $46, and then outputs*its ASCII character, ‘F, to the screen.

SWIEND

Result à AF is written to the screen.



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Example: OUT1BYT (Out 1 Byte) and OUT1BSP (Out 1 Byte with Space)Assume that memory contains the following:2100 FA FBNow execute the following code:

ORG $1040LDX #$2100JSR $FFBB *Goes to subroutine OUT1BYT which converts the

*content of the address pointed to by the X register to*two ASCII equivalents and outputs their characters to *the screen, and increments X register by #$1.

JSR $FFBE *Goes to subroutine OUT1BSP which converts the*content of the address pointed to by the X register to*two ASCII equivalents and outputs their characters to the *screen followed by a space.

SWIEND

Resultà FAFB_ sent to screen (Note: last character is a space)



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Example: OUT2BSP (Out 2 Bytes with Space)

Assume that memory contains the following:2100 FA FB

Now execute the following code.ORG $1040LDX #$2100JSR $FFC1 *JSR to subroutine OUT2BSP which converts two

*consecutive binary bytes to their 4 ASCII numbers*and outputs their ASCII characters to the screen followed by a*space.

SWIENDResultàFAFB_ sent to screen (last character is a space).



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Example: OUTCRLF (Output Carriage Return with Line Feed)

ORG $1040JSR $FFC4 *Go to subroutine OUTCRLF

*which outputs ASCII carriage return*followed by a line feed.

SWIEND

Resultà New line on screen (Note:Equivalent to pressingEnter in your computer)



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Example: INCHAR (Input Character)

ORG $1040JSR $FFCD *Goes to subroutine INCHAR. Waits for you to

*type an ASCII character from the keyboard,*and stores the corresponding ASCII number in*accumulator A

SWIEND

Result-> If you type 9 on the keyboard, ASCII number$39 is stored in accumulator A



ME4447/6405ME4447/6405 Homework 3 Solution

Write an assembly language program to output a string of characters to the screen.

Solution: This particular solution will print “Hello” to the screen using the OUTA subroutine

FCC is used to store a string of characters

FCB is used to store End Of Transmission (EOT)(Note: EOT in ASCII is $04)(Note 2: In this particular example any non-printing character can be used)

OUTA EQU $FFB8

ORG $0100FCC “HELLO”FCB #$04

ORG $1040 LDX #$0100

Loop LDAA $00,XCMPA #$04BEQ QuitJSR OUTA; INXBRA Loop

Quit SWIEND


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#$04

1

OUTA EQU $FFB8

ORG $0100FCC “Hello”FCB #$04

ORG $1040 LDX #$0100

Loop LDAA $00,XCMPA #$04BEQ QuitJSR OUTA INXBRA Loop

Quit SWIEND

$0104

$0105

$0103

$0102

$0101

$0100

CPU Registers Used in Program:

Memory Locations Used in Program:

Computer Screen:

Program Initialization

$48

$65

$6C

$6C

$6F

$04

CCR Bit Z:

Index X:

Accum A:

#$0100

#$48

0

CCR Z is 0 since Accum A does not equal $04

Does not branch because CCR Z is 0

H

#$0101

Always Branches

#$65

e

#$0102

Skip Ahead Until Last Loop

llo

#$1005CCR Z is 1 sinceAccum A equals$04

Branch since CCR Z is 1

#$6F

AsciiEquivalent of “Hello”


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Solution 2: This particular solution will print “Hello” to the screen using the OUTSTRG subroutine

OUTSTRG EQU $FFC7

ORG $0100FCC “HELLO”FCB #$04

ORG $1040 LDX #$0100 JSR OUTSTRG SWIEND

(Note: EOT must be used because the OUTSTRG Subroutine uses EOT as the end of the string)


ME4447/6405ME4447/6405 Homework 3 SolutionWrite a subroutine to save the first 5 odd (8-bit) numbers pointed to by the x-register (passed in) onto the stack.

Solution:

SUB1 LDAA #$05PULY

NEXT BRSET $00,X #$01 ODDINXBRA NEXT

ODD LDAB $00,XPSHBINXDECABNE NEXTPSHYRTS

Note: In order for the subroutine to execute, in the main program, a BSR(branch to subroutine) or JSR (jump to subroutine) command must be used inthe main program.


ME4447/6405ME4447/6405 Homework 4 solution

Write a program to output a square wave thru port D pin 2. The outputcan be observed on the scope, and the period T of the wave shouldbe measured. More than one period wave should be generated. Themachine cycle time of the M68HC11E9 should be estimated. Drawthe square wave.


ME4447/6405ME4447/6405 Homework 4 Solution Cont’dSolution:ORG $1040FCB $XXFCB $04LDAA #$04STAA $1009 LDAB #$04

CYCLE LDAA $1041 4STAA $1008 4EORA #$04 2STAA $1041 4LDAA $1040 4

NEXT DECA 2BNE NEXT 3DECB 2BEQ EXIT 3JMP CYCLE 3

EXIT SWIEND

Prebyte Opcode OperandAddress

1040 XX 1041 011042 86 04 1044 B7 10091045 C6 041049 B6 1041104C B7 1008104F 88 041051 B7 10411054 B6 10401057 4A1058 26 FD105A 5A105B 27 03105D 7E 1049105F 3F

Configures PD2as output. Helpsus to output twoperiods.

Outputs signal from PD2

T/2=Machine Cycle Time*[4+4+2+COUNT*(2+3)+4+4+3+2+3] (Note: Count = #$XX)

Switch $1041(toggle)

Delay


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QUESTIONS???

me4447/6405 me 6405ume.gatech.edu/mechatronics_course/lecture9.pdf · me4447/6405 write an assembly...

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