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Chapter 2 HCS12 Assembly Programming

Chapter 2 HCS12 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

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Chapter 2 HCS12 Assembly Programming

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  1. Chapter 2HCS12 Assembly Programming

  2. 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

  3. 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 Fields of a HCS12 Instruction

  4. Identify 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

  5. 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.

  6. 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 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

  7. 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:”

  8. 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 buffer ds 100 reserves 100 bytes

  9. 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 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

  10. 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

  11. 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

  12. 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. • Programming. Convert the algorithm or flowchart into programs. • Program testing • Program maintenance

  13. Symbols of Flowchart

  14. 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

  15. 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

  16. 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

  17. 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

  18. 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

  19. 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.

  20. Multiplication and Division (1 of 2)

  21. 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

  22. 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

  23. 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

  24. ; 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

  25. ; 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

  26. 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

  27. idiv 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

  28. 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.

  29. 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

  30. 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.

  31. 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).

  32. 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 ; “

  33. 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:

  34. N equ 20 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

  35. 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.

  36. 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

  37. 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

  38. 0 b7 ----------------- b0 C Shift and Rotate Instructions 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

  39. 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

  40. 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

  41. 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

  42. 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

  43. 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:

  44. 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:

  45. 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:

  46. 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

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