- 341 Views
- Uploaded on
- Presentation posted in: General

Chapter 2 HCS12 Assembly Programming

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Chapter 2HCS12 Assembly Programming

- Assembler directives
- Defines data and symbol
- Reserves and initializes memory locations
- Sets assembler and linking condition
- Specifies output format
- Specifies the end of a program

- Assembly language instructions
- HCS12/MC9S12 instructions

- Comments
- Explains the function of a single or a group of instructions

Label field

Optional

Starts with a letter and followed by letters, digits, or special symbols (_ or .)

Can start from any column if ended with “:”

Must start from column 1 if not ended with “:”

Operation field

Contains the mnemonic of a machine instruction or an assembler directive

Separated from the label by at least one space

Operand field

Follows the operation field and is separated from the operation field by at least one space

Contains operands for instructions or arguments for assembler directives

Comment field

Any line starts with an * or ; is a comment

Separated from the operand and operation field for at least one space

Optional

Example

loopADDA #$40; add 40 to accumulator A

(1) “loop” is a label

(2) “ADDA” is an instruction mnemonic

(3) “#$40” is the operand

(4) “add #$40 to accumulator A” is a comment

movb0,X,0,Y; memory to memory copy

(1) no label field

(b) “movb” is an instruction mnemonic

(c) “0,X,0,Y” is the operand field

(d) “; memory to memory copy” is a comment

- END
- Ends a program to be processed by an assembler
- Any statement following the END directive is ignored.

- ORG
- The assembler uses a location counter to keep track of the memory location where the next machine code byte should be placed.
- This directive sets a new value for the location counter of the assembler.
- The sequence
ORG $1000

LDAB #$FF

places the opcode byte for the instruction LDAB #$FF at location $1000.

dc.b (define constant byte)

db (define byte)

fcb (form constant byte)

-These three directives define the value of a byte or bytes that will be placed at a given

location.

-These directives are often preceded by the org directive.

-For example,

org $800

arraydc.b $11,$22,$33,$44

dc.w (define constant word)

dw (define word)

fdb (form double bytes)

-Define the value of a word or words that will be placed at a given location.

-The value can be specified by an expression.

-For example,

vec_tabdc.w$1234, abc-20

- Used to define a string of characters (a message)
- The first character (and the last character) is used as the delimiter.
- The last character must be the same as the first character.
- The delimiter must not appear in the string.
- The space character cannot be used as the delimiter.
- Each character is represented by its ASCII code.
- Example
msgfcc “Please enter 1, 2 or 3:”

fill (fill memory)

-This directive allows the user to fill a certain number of memory locations with a

given value.

-The syntax is fill value,count

-Example

space_line fill $20,40

ds (define storage)

rmb (reserve memory byte)

ds.b (define storage bytes)

-Each of these directives reserves a number of bytes given as the arguments to the

directive.

-Example

bufferds 100

reserves 100 bytes

ds.w (define storage word)

rmw (reserve memory word)

-Each of these directives increments the location counter by the value indicated in

the number-of-words argument multiplied by two.

-Example

dbufds.w 20

reserves 40 bytes starting from the current location counter

equ (equate)

-This directive assigns a value to a label.

-Using this directive makes one’s program more readable.

-Examples

arr_cnt equ 100

oc_cnt equ 50

- loc
- This directive increments and produces an internal counter used in conjunction with
- the backward tick mark (`).
- By using the loc directive and the ` mark, one can write program segments like the
- following example, without thinking up new labels:
- locloc
- ldaa #2ldaa #2
- loop`decasame asloop001deca
- bne loop`bne loop001
- locloc
- loop`brclr 0,x,$55,loop`loop002brclr 0,x,$55,loop002

- Macro
- A name assigned to a group of instructions
- Use macro and endm to define a macro.
- Example of macro
- sumOf3macroarg1,arg2,arg3
- ldaaarg1
- addaarg2
- addaarg3
- endm
- Invoke a defined macro: write down the name and the arguments of the macro
- sumOf3$1000,$1001,$1002
- is replaced by
- ldaa$1000
- adda$1001
- adda$1002

- Problem definition: Identify what should be done.
- Develop the algorithm.
- Algorithm is the overall plan for solving the problem at hand.
- An algorithm is often expressed in the following format:
- Step 1
- …
- Step 2
- …

- Another way to express overall plan is to use flowchart.

- Develop the algorithm.
- Programming. Convert the algorithm or flowchart into programs.
- Program testing
- Program maintenance

Example 2.4 Write a program to add the values of memory locations at $1000, $1001, and

$1002, and save the result at $1100.

Solution:

Step 1

A m[$1000]

Step 2

A A + m[$1001]

Step 3

A A + m[$1002]

Step 4

$802 A

org$1500

ldaa$1000

adda$1501

adda$1002

staa$1100

end

Programs to Do Simple Arithmetic (2 of 5)

Example 2.4 Write a program to subtract the contents of the memory location at $1005 from the sum of the memory locations at $1000 and $1002, and store the difference at $1100.

Solution:

org $1500

ldaa $1000

adda $1002

suba $1005

staa $1000

end

Programs to Do Simple Arithmetic (3 of 5)

Example 2.6 Write a program to add two 16-bit numbers that are stored at $1000-$1001 and $1002-$1003 and store the sum at $1100-$1101.

Solution:

org $1500

ldd $1000

addd$1002

std $1100

end

The Carry Flag

- bit 0 of the CCR register

- set to 1 when the addition operation produces a carry 1

- set to 1 when the subtraction operation produces a borrow 1

- enables the user to implement multi-precision arithmetic

Programs to Do Simple Arithmetic (4 of 5)

Example 2.7 Write a program to add two 4-byte numbers that are stored at $1000-$1003 and $1004-$1007, and store the sum at $1010-$1013.

Solution: Addition starts from the LSB and proceeds toward MSB.

org$1500

ldd$1002; add and save the least significant two bytes

addd$1006;“

std$1012;“

ldaa$1001; add and save the second most significant bytes

adca$1005;“

staa$1011;“

ldaa$1000; add and save the most significant bytes

adca$1004;“

staa$1010;“

end

Programs to Do Simple Arithmetic (5 of 5)

- Example 2.8 Write a program to subtract the hex number stored at $1004-$1007 from the the hex number stored at $1000-$1003 and save the result at $1100-$1103.
- Solution: The subtraction starts from the LSBs and proceeds toward the MSBs.
- org$1500
- ldd$1002; subtract and save the least significant two bytes
- subd$1006;“
- std$1102;“
- ldaa$1001; subtract and save the difference of the second to most
- sbca$1005; significant bytes
- staa$1001;“
- ldaa$1000; subtract and save the difference of the most significant
- sbca$1004; bytes
- staa$1100;“
- end

- Each digit is encoded by 4 bits.
- Two digits are packed into one byte
- The addition of two BCD numbers is performed by binary addition and an adjust operation using the DAA instruction.
- The instruction DAA can be applied after the instructions ADDA, ADCA, and ABA.
- Simplifies I/O conversion
- For example, the instruction sequence
- LDAA$1000
- ADDA$1001
- DAA
- STAA$1002
adds the BCD numbers stored at $1000 and $1001 and saves the sum at $1002.

Multiplication and Division (2 of 2)

Example 2.10 Write an instruction sequence to multiply the 16-bit numbers stored at $1000-$1001 and $1002-$1003 and store the product at $1100-$1103.

Solution:

ldd$1000

ldy$1002

emul

sty$1100

std$1102

Example 2.11 Write an instruction sequence to divide the 16-bit number stored at $1020-$1021 into the 16-bit number stored at $1005-$1006 and store the quotient and remainder at $1100 and $1102, respectively.

Solution:

ldd$1005

ldx$1020

idiv

stx$1100; store the quotient

std$1102 ; store the remainder

- Two 32-bit numbers M and N are divided into two 16-bit halves
- M = MHML
- N = NHNL

Example 2.12 Write a program to multiply two unsigned 32-bit numbers stored at M~M+3 and N~N+3, respectively and store the product at P~P+7.

Solution:

org$1000

Mds.b4

Nds.b4

Pds.b8

org$1500

lddM+2

ldyN+2

emul; compute MLNL

styP+4

stdP+6

lddM

ldyN

emul; compute MHNH

styP

stdP+2

lddM

ldyN+2

emul; compute MHNL

; add MHNL to memory locations P+2~P+5

adddP+4

stdP+4

tfrY,D

adcbP+3

stabP+3

adcaP+2

staaP+2

; propagate carry to the most significant byte

ldaaP+1

adca#0; add carry to the location at P+1

staaP+1;“

ldaaP; add carry to the location at P

adca#0;“

staaP;“

; compute MLNH

lddM+2

ldyN

emul

; add MLNH to memory locations P+2 ~ P+5

adddP+4

stdP+4

tfrY,D

adcbP+3

stabP+3

adcaP+2

staaP+2

; propagate carry to the most significant byte

clra

adcaP+1

staaP+1

ldaaP

adca#0

staaP

end

Example 2.13 Write a program to convert the 16-bit number stored at $1000-$1001 to BCD format and store the result at $1010-$1014. Convert each BCD digit into its ASCII code and store it in one byte.

Solution:

-A binary number can be converted to BCD format by using repeated division by 10.

-The largest 16-bit binary number is 65535 which has five decimal digits.

-The first division by 10 generates the least significant digit, the second division by 10 obtains the second least significant digit, and so on.

org$1000

datadc.w12345; data to be tested

org$1010

resultds.b5; reserve bytes to store the result

org$1500

ldddata

ldy#result

ldx#10

idiv

addb#$30; convert the digit into ASCII code

stab4,Y; save the least significant digit

xgdx

ldx#10

idiv

adcb#$30

stab3,Y; save the second to least significant digit

xgdx

ldx#10

idiv

addb#$30

stab2,Y; save the middle digit

xgdx

ldx#10

idiv

addb#$30

stab1,Y; save the second most significant digit

xgdx

addb#$30

stab0,Y; save the most significant digit

end

- Types of program loops: finite and infinite loops
- Looping mechanisms:
- dostatement S forever
- For i = n1ton2dostatement S or For i = n2downton1dostatement S
- WhileCdostatement S
- Repeatstatement S untilC

- Program loops are implemented by using the conditional branch instructions and the execution of these instructions depends on the contents of the CCR register.

- Four types of branch instructions
- Unary (unconditional) branch: always execute
- Simple branches: branch is taken when a specific bit of CCR is in a specific status
- Unsigned branches: branches are taken when a comparison or test of unsigned numbers results in a specific combination of CCR bits
- Signed branches: branches are taken when a comparison or test of signed quantities are in a specific combination of CCR bits

- Two categories of branches
- Short branches: in the range of -128 ~ +127 bytes
- Long branches: in the range of 64KB

- Condition flags need to be set up before conditional branch instruction should be executed.
- The HCS12 provides a group of instructions for testing the condition flags.

- HCS12 provides a group of instructions that either decrement or increment a loop count to determine if the looping should be continued.
- The range of the branch is from $80 (-128) to $7F (+127).

Example 2.14 Write a program to add an array of N 8-bit numbers and store the sum at memory locations $1000~$1001. Use the For i = n1 to n2 do looping construct.

Solution:

Nequ20

org$1000

sumrmb2

irmb1

org$1500

ldaa#0

staai

staasum; sum 0

staa sum+1;“

loopldabi

cmpb#N; is i = N?

beqdone

ldx#array

abx

ldab0,X; sum sum + array[i]

ldysum;“

aby;“

stysum;“

inci; increment the loop count by 1

braloop

doneswi

arraydc.b1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20

end

Example 2.15 Write a program to find the maximum element from an array of N 8-bit elements using the repeat S until C looping construct.

Solution:

Nequ20

org$1000

max_valds.b1

org$1500

ldaaarray ; set array[0] as the temporary max max

staamax_val ; “

ldx#array+N-1 ; start from the end of the array

ldab#N-1 ; set loop count to N - 1

loopldaamax_val

cmpa0,x

bgechk_end

ldaa0,x

staamax_val

chk_enddex

dbneb,loop ; finish all the comparison yet?

foreverbraforever

arraydb1,3,5,6,19,41,53,28,13,42,76,14

db20,54,64,74,29,33,41,45

end

[<label>] brclr(opr),(msk),(rel) [<comment>]

[<label>] brset(opr),(msk),(rel) [<comment>]

where

oprspecifies the memory location to be checked and must be specified using either

the direct, extended, or index addressing mode.

mskis an 8-bit mask that specifies the bits of the memory location to be checked.

The bits of the memory byte to be checked correspond to those bit positions

that are 1s in the mask.

relis the branch offset and is specified in the 8-bit relative mode.

For example, in the sequence

loopinc count

…

brclr $66,$e0,loop

…

the branch will be taken if the most significant three bits at $66 are all ones.

Example 2.17 Write a program to compute the number of elements that are divisible by 4 in an array of N 8-bit elements. Use the repeat S until C looping construct.

Solution: A number divisible by 4 would have the least significant two bits equal 0s.

Nequ20

org$1000

totalds.b1

org$1500

clrtotal; initialize total to 0

ldx#array

ldab#N; use B as the loop count

loopbrclr0,x,$03,yes; check bits 1 and 0

brachkend

yesinctotal

chkendinx

dbneb,loop

foreverbraforever

arraydb2,3,4,8,12,13,19,24,33,32,20,18,53,52,80,82,90,94,100,102

end

- [<label>] CLRopr [<comment>]where opr is specified using the extended or index addressing modes. The specified memory location is cleared.
- [<label>] CLRA[<comment>]Accumulator A is cleared to 0
- [<label>] CLRB[<comment>]Accumulator B is cleared to 0

0

b7 ----------------- b0

C

The HCS12 has shift and rotate instructions that apply to a memory location, accumulators A, B, and D. A memory operand must be specified using the extended or index addressing modes.

There are three 8-bit arithmetic shift left instructions:

[<label>]asl opr [<comment>]-- memory location opr is shifted left one place

[<label>]asla[<comment>]-- accumulator A is shifted left one place

[<label>]aslb[<comment>]-- accumulator B is shifted left one place

The operation is

b7 ----------------- b0

b7 ----------------- b0

0

C

accumulator A

accumulator B

b7 ----------------- b0

C

The HCS12 has one 16-bit arithmetic shift left instruction:

[<label>]asld[<comment>]

The operation is

The HCS12 has arithmetic shift right instructions that apply to a memory location and accumulators A and B.

[<label>]asr opr[<comment>]-- memory location opr is shifted right one place

[<label>]asra[<comment>]-- accumulator A is shifted right one place

[<label>]asrb[<comment>]-- accumulator B is shifted right one place

The operation is

0

C

b7 ----------------- b0

b7 ----------------- b0

b7 ----------------- b0

0

C

accumulator B

accumulator A

The HCS12 has logical shift left instructions that apply to a memory location and accumulators A and B.

[<label>]lsl opr[<comment>]-- memory location opr is shifted left one place

[<label>]lsla[<comment>]-- accumulator A is shifted left one place

[<label>]lslb[<comment>]-- accumulator B is shifted left one place

The operation is

The HCS12 has one 16-bit logical shift left instruction:

[<label>]lsld[<comment>]

The operation is

b7 ----------------- b0

C

0

C

0

b7 ----------------- b0

b7 ----------------- b0

accumulator B

accumulator A

The HCS12 has three logical shift right instructions that apply to 8-bit operands.

[<label>]lsr opr[<comment>]-- memory location opr is shifted right one place

[<label>]lsra[<comment>]-- accumulator A is shifted right one place

[<label>]lsrb[<comment>]-- accumulator B is shifted right one place

The operation is

The HCS12 has one 16-bit logical shift right instruction:

[<label>]lsrd[<comment>]

The operation is

b7 ----------------- b0

C

b7 ----------------- b0

C

The HCS12 has three rotate left instructions that operate on 9-bit operands.

[<label>]rol opr[<comment>]-- memory location opr is rotated left one place

[<label>]rola[<comment>]-- accumulator A is rotated left one place

[<label>]rolb[<comment>]-- accumulator B is rotated left one place

The operation is

The HCS12 has three rotate right instructions that operate on 9-bit operands.

[<label>]ror opr[<comment>]-- memory location opr is rotated right one place

[<label>]rora[<comment>]-- accumulator A is rotated right one place

[<label>]rorb[<comment>]-- accumulator B is rotated right one place

The operation is

Example 2.18 Suppose that [A] = $95 and C = 1. Compute the new values of A and C after the execution of the instruction asla.

Solution:

Example 2.19 Suppose that m[$800] = $ED and C = 0. Compute the new values of m[$800] and the C flag after the execution of the instruction asr $1000.

Solution:

Example 2.20 Suppose that m[$1000] = $E7 and C = 1. Compute the new contents of m[$1000] and the C flag after the execution of the instruction lsr $1000.

Solution:

Example 2.21 Suppose that [B] = $BD and C = 1. Compute the new values of

B and the C flag after the execution of the instruction rolb.

Solution:

Example 2.22 Suppose that [A] = $BE and C = 1. Compute the new values of mem[$00] after the execution of the instruction rora.

Solution:

Example 2.23 Write a program to count the number of 0s in the 16-bit number stored at $1000-$1001 and save the result in $1005.

Solution:

* The 16-bit number is shifted to the right 16 time.

* If the bit shifted out is a 0 then increment the 0s count by 1.

org$1000

db$23,$55; test data

org$1005

zero_cntrmb1

lp_cntrmb1

org$1500

clrzero_cnt; initialize the 0s count to 0

ldaa#16

staalp_cnt

ldd$1000; place the number in D

looplsrd; shift the lsb of D to the C flag

bcschkend; is the C flag a 0?

inczero_cnt; increment 1s count if the lsb is a 1

chkenddeclp_cnt; check to see if D is already 0

bneloop

foreverbraforever

end

- For shifting right
- The bit 7 of each byte will receive the bit 0 of its immediate left byte with the exception of the most significant byte which will receive a 0.
- Each byte will be shifted to the right by 1 bit. The bit 0 of the least significant byte will be lost.

- Suppose there is a k-byte number that is stored at loc to loc+k-1.
- Method for shifting right
- Step 1: Shift the byte at loc to the right one place.
- Step 2: Rotate the byte at loc+1 to the right one place.
- Step 3: Repeat Step 2 for the remaining bytes.

- Method for shifting right

- For shifting left
- The bit 0 of each byte will receive the bit 7 of its immediate right byte with the exception of the least significant byte which will receive a 0.
- Each byte will be shifted to the left by 1 bit. The bit 7 of the most significant byte will be lost.

- Suppose there is a k-byte number that is stored at loc to loc+k-1.
- Method for shifting left
- Step 1: Shift the byte at loc+k-1 to the left one place.
- Step 2: Rotate the byte at loc+K-2 to the left one place.
- Step 3: Repeat Step 2 for the remaining bytes.

- Method for shifting left

Example 2.24 Write a program to shift the 32-bit number stored at $820-$823 to the right four places.

Solution:

ldab#4; set up the loop count

ldx#$820; use X as the pointer to the left most byte

againlsr0,X

ror1,X

ror2,X

ror3,X

dbneb,again

end

- Changing a few bits are often done in I/O applications.
- Boolean logic operation can be used to change a few I/O port pins easily.

- The HCS12 uses the E clock as a timing reference.
- The frequency of the E clock is half of that of the crystal oscillator.
- There are many applications that require the generation of time delays.
- The creation of a time delay involves two steps:
- Select a sequence of instructions that takes a certain amount of time to execute.
- Repeat the selected instruction sequence for an appropriate number of times.
- For example, the instruction sequence on the next page takes 40 E cycles to execute. By repeating this instruction sequence a certain number of times, any time delay can be created.

- Assume that the HCS12 runs under a crystal oscillator with a frequency of 16 MHz, then the E frequency is 8 MHz and, hence, its clock period is 125 ns.
- Therefore, the instruction sequence on the next page will take 5 ms to execute.

Program Execution Time (2 of 2)

looppsha; 2 E cycles

pula; 3 E cycles

psha

pula

psha

pula

psha

pula

psha

pula

psha

pula

psha

pula

nop; 1 E cycle

nop; 1 E cycle

dbnex,loop; 3 E cycles

Example 2.25 Write a program loop to create a delay of 100 ms.

Solution: A delay of 100 ms can be created by repeating the previous loop 20,000 times.

The following instruction sequence creates a delay of 100 ms.

ldx#20000

looppsha; 2 E cycles

pula; 3 E cycles

psha

pula

psha

pula

psha

pula

psha

pula

psha

pula

psha

pula

nop; 1 E cycle

nop; 1 E cycle

dbnex,loop; 3 E cycles

ldab#100

out_loopldx#20000

in_looppsha; 2 E cycles

pula; 3 E cycles

psha

pula

psha

pula

psha

pula

psha

pula

psha

pula

psha

pula

nop; 1 E cycle

nop; 1 E cycle

dbnex,in_loop; 3 E cycles

dbneb,out_loop; 3 E cycles

Example 2.26 Write an instruction sequence to create a delay of 10 seconds.

Solution: By repeating the previous instruction sequence 100 times, we can create a delay of 10 seconds.