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Chapter 2 HCS12 Assembly ProgrammingPowerPoint Presentation

Chapter 2 HCS12 Assembly Programming

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### Chapter 2HCS12 Assembly Programming

Three Sections of a HCS12/MC9S12 Assembly Program

- 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

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

Fields of a HCS12 InstructionIdentify the Four Fields of an Instruction

Example

loop ADDA #$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

movb 0,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

Assembler Directives

- 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.

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

array dc.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_tab dc.w $1234, abc-20

fcc (form constant character)

- 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
msg fcc “Please enter 1, 2 or 3:”

- 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

buffer ds 100

reserves 100 bytes

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

dbuf ds.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:
- loc loc
- ldaa #2 ldaa #2
- loop` deca same as loop001 deca
- bne loop` bne loop001
- loc loc
- loop` brclr 0,x,$55,loop` loop002 brclr 0,x,$55,loop002

- Macro
- A name assigned to a group of instructions
- Use macro and endm to define a macro.
- Example of macro
- sumOf3 macro arg1,arg2,arg3
- ldaa arg1
- adda arg2
- adda arg3
- 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

Software Development Process

- 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

Programs to Do Simple Arithmetic (1 of 5)

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

BCD Numbers and Addition

- 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 (1 of 2)

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

Illustration of 32-bit by 32-bit Multiplication

- 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

M ds.b 4

N ds.b 4

P ds.b 8

org $1500

ldd M+2

ldy N+2

emul ; compute MLNL

sty P+4

std P+6

ldd M

ldy N

emul ; compute MHNH

sty P

std P+2

ldd M

ldy N+2

emul ; compute MHNL

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

addd P+4

std P+4

tfr Y,D

adcb P+3

stab P+3

adca P+2

staa P+2

; propagate carry to the most significant byte

ldaa P+1

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

staa P+1 ; “

ldaa P ; add carry to the location at P

adca #0 ; “

staa P ; “

; compute MLNH

ldd M+2

ldy N

emul

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

addd P+4

std P+4

tfr Y,D

adcb P+3

stab P+3

adca P+2

staa P+2

; propagate carry to the most significant byte

clra

adca P+1

staa P+1

ldaa P

adca #0

staa P

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

data dc.w 12345 ; data to be tested

org $1010

result ds.b 5 ; reserve bytes to store the result

org $1500

ldd data

ldy #result

ldx #10

idiv

addb #$30 ; convert the digit into ASCII code

stab 4,Y ; save the least significant digit

xgdx

ldx #10

adcb #$30

stab 3,Y ; save the second to least significant digit

xgdx

ldx #10

idiv

addb #$30

stab 2,Y ; save the middle digit

xgdx

ldx #10

idiv

addb #$30

stab 1,Y ; save the second most significant digit

xgdx

addb #$30

stab 0,Y ; save the most significant digit

end

Program Loops

- 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.

Condition Code 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

Compare and Test Instructions

- 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.

Loop Primitive Instructions

- 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:

N equ 20

org $1000

sum rmb 2

i rmb 1

org $1500

ldaa #0

staa i

staa sum ; sum 0

staa sum+1 ; “

loop ldab i

cmpb #N ; is i = N?

beq done

ldx #array

abx

ldab 0,X ; sum sum + array[i]

ldy sum ; “

aby ; “

sty sum ; “

inc i ; increment the loop count by 1

bra loop

done swi

array dc.b 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20

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:

org $1000

max_val ds.b 1

org $1500

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

staa max_val ; “

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

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

loop ldaa max_val

cmpa 0,x

bge chk_end

ldaa 0,x

staa max_val

chk_end dex

dbne b,loop ; finish all the comparison yet?

forever bra forever

array db 1,3,5,6,19,41,53,28,13,42,76,14

db 20,54,64,74,29,33,41,45

end

Bit Condition Branch Instructions

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

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

where

opr specifies the memory location to be checked and must be specified using either

the direct, extended, or index addressing mode.

msk is 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.

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

For example, in the sequence

loop inc 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.

N equ 20

org $1000

total ds.b 1

org $1500

clr total ; initialize total to 0

ldx #array

ldab #N ; use B as the loop count

loop brclr 0,x,$03,yes ; check bits 1 and 0

bra chkend

yes inc total

chkend inx

dbne b,loop

forever bra forever

array db 2,3,4,8,12,13,19,24,33,32,20,18,53,52,80,82,90,94,100,102

end

Instructions for Variable Initialization

- [<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

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

C

Shift and Rotate InstructionsThe 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

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

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

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

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_cnt rmb 1

lp_cnt rmb 1

org $1500

clr zero_cnt ; initialize the 0s count to 0

ldaa #16

staa lp_cnt

ldd $1000 ; place the number in D

loop lsrd ; shift the lsb of D to the C flag

bcs chkend ; is the C flag a 0?

inc zero_cnt ; increment 1s count if the lsb is a 1

chkend dec lp_cnt ; check to see if D is already 0

bne loop

forever bra forever

end

Shift a Multi-byte Number (1 of 3)

- 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

Shift a Multi-byte Number (2 of 3)

- 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

Shift a Multi-byte Number (3 of 3)

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

again lsr 0,X

ror 1,X

ror 2,X

ror 3,X

dbne b,again

end

Boolean Logic Instructions

- 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.

Program Execution Time (1 of 2)

- 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)

loop psha ; 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

dbne x,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

loop psha ; 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

dbne x,loop ; 3 E cycles

out_loop ldx #20000

in_loop psha ; 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

dbne x,in_loop ; 3 E cycles

dbne b,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.

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