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HUANG-Chap02 - HC12 Assembler

1. Mnemonic Function Equation or Operation BRA Branch always 1 1 BRN Branch never 1 0 Simple Branches Mnemonic Function Equation or Operation BCC Branch if carry clear C 0 BCS Branch if carry set C 1 BEQ Branch if equal Z 1 BMI Branch if minus N 1 BNE Branch if not equal Z 0 BPL Branch if plus N 0 BVC Branch if overflow clear V 0 BVS Branch if overflow set V 1 Unsigned Branches Mnemonic Function Equation or Operation BHI Branch if higher C Z 0 BHS Branch if higher or same C 0 BLO Branch if lower C 1 BLS Branch if lower or same C Z 1 Signed Branches Mnemonic Function Equation or Operation BGE Branch if greater than or equal N V 0 BGT Branch if greater than Z N V 0 BLE Branch if less than or equal Z N V 1 BLT Branch if less than N V 1 Table 2 2 m Summary of short branch instructions The numeric range of long branch offset values is 8000 32768 to 7FFF 32767 from the instruction immediately after the branch instruction This permits branching from any location in the standard 64 KB address map to any other location in the map A summary of the long branch instructions appears in Table 2 3 2 6 m Program Loops 55 Unary Branches Mnemonic Function Equation or Operation LBRA Long branch always 1 1 LBRN Long branch never 1 0 Simple Branches Mnemonic Function Equation or Operation LBCC Long branch if carry clear C 0 LBCS Long branch if carry
2. The input output box is used to represent data that are either read or displayed by the computer The decision box contains a question that can be answered either yes or no A decision box has two exits also marked yes or no The computer will take one action if the answer is yes and will take a different action if the answer is no The on page connector indicates that the flowchart continues elsewhere on the same page The place where it is continued will have the same label as the on page connector The off page connector indicates that the flowchart continues on another page To determine where the flow chart continues you need to look at the following pages of the flowchart to find the matching off page connector Normal flow on a flowchart is from top to bottom and from left to right Any line that does not follow this normal flow should have an arrowhead on it When the program gets complicated the flowchart that documents the logic flow of the program also becomes difficult to follow This is the limitation of the flowchart In this book we will use both the flowchart and the algorithm procedure to describe the solution to a problem After you are satisfied with the algorithm or the flowchart convert it to source code in one of the assembly or high level languages Each statement in your algorithm or each block of your flowchart will be converted into one or multiple assembly instructions or high level lan guage statements If you find
3. tracted from the second to most significant byte of the result In other words we need a sub tract with borrow instruction There is such an instruction but it is called subtract with carry There are two versions the SBCA instruction for accumulator A and the SBCB instruction for accumulator B The instructions to subtract the second to most significant bytes are Idaa 76 sbca 75 We also need to save the second to most significant byte of the result at 801 with the fol lowing instruction staa 801 The most significant bytes can be subtracted using similar instructions and the complete program with comments is as follows org 1000 _ starting address of the program Idd 5432 place the lower two bytes of the minuend in D subd 7284 subtract the lower bytes of the subtrahend from D std 802 save the lower two bytes of the difference Idaa 76 place the second to most significant byte of the minuend in A sbca 75 subtract the second to most significant byte of the subtrahend and the borrow from A staa 801 save the second to most significant byte of the difference Idaa 98 put the most significant byte of the minuend in A sbca 16 subtract the most significant byte of the subtrahend and the borrow from A staa 800 save the most significant byte of the difference end Example 2 8 v 2 5 m Writing Programs to do Arithmetic 43 Write a program to subtract the hex numbers stored at 804 807
4. Step 2 Generate the partial product M N in Y D and save it at locations P 4 P 7 Step 3 Generate the partial product M Ny in Y D and save it at locations P P 3 48 Chapter 2 m 68HC12 Assembly Programming Step 4 Generate the partial product MN in Y D and add it to memory locations P 2 P 5 The C flag may be set to 1 after this addition Step 5 Add the C flag to memory location P 1 using the ADCA or ADCB instruction This addition may also set the C flag to 1 So again add the C flag to memory location P Step 6 Generate the partial product MNu in Y D and add it to memory locations P 2 P 5 The carry flag may be set to 1 So add the C flag to memory location P 1 and then add it to memory location P A Example 2 12 v Write a program to multiply the 32 bit unsigned integers stored at M M 3 and N N 3 gt respectively and store the product at memory locations P P 7 Solution The following program is a direct translation of the previous multiplication algorithm org 800 M rmb 4 reserved to hold the multiplicand N rmb 4 reserved to hold the multiplier P rmb 8 reserved to hold the product org 1000 Idd M 2 place ML in D Idy N 2 place NL in Y emul compute MLNL sty P 4 save the upper 16 bits of the partial product MLNL std P 6 save the lower 16 bits of the partial product MLNL Idd M place MH in D Idy N place NH inY emul compute MHNH sty P save the upper 16 bits of the partial product MHNH
5. Z V and C in the CCR register after executing each of the following instructions independently given that A 00 and the initial condition codes are N 0 C 0 Z 1 and V 0 a TSTA b ADDA 40 c SUBA 78 d LSLA e ROLA f ADDA CF E2 21 Write an instruction sequence to toggle the odd number bits and clear the even number bits of the memory location at 66 E2 22 Write a program to shift the 8 byte number located at 800 807 to the left four places E2 23 Write a program to shift the 6 byte number located at 900 905 to the right three places E2 24 Write a program to create a time delay of 100 seconds by using program loops E2 25 Write a program to create a time delay of five seconds using program loops
6. an algorithmic step or a block in the flowchart requires many assembly instructions or high level language statements to implement then it might be bene ficial to either 1 convert this step or block into a subroutine and just call the subroutine or 2 further divide the algorithmic step or flowchart block into smaller steps or blocks so that it can be coded with just a few assembly instructions or high level language statements The next major step is to test your program Testing a program means testing for anomalies Here you will first test for normal inputs that you always expect If the result is what you expected then you go on to test the borderline inputs Test for the maximum and minimum values of the input When your program passes this test then test for illegal input values If your algorithm includes several branches then you must use enough values to exercise all the possible branches This is to make sure that your program will operate correctly under all possible circumstances In the rest of this book most of the examples are well defined Therefore our focus is on how to design the algorithm that solves the specified problem as well as how to convert the algorithm into source code 2 5 m Writing Programs to do Arithmetic 37 2 5 Writing Programs to do Arithmetic In this section we will use small programs that perform simple computations to demon strate how a program is written Example 2 3 v Write a program t
7. bit conditional branches Branch instructions can also be classified by the type of condition that must be satisfied in order for a branch to be taken Some instructions belong to more than one category Unary unconditional branch instructions always execute Simple branches are taken when a specific bit in the CCR register is in a specific state as a result of a previous operation Unsigned branches are taken when a comparison or test of unsigned quantities results in a specific combination of condition code register bits Signed branches are taken when a comparison or test of signed quantities results in a specific combination of condition code register bits When a short branch instruction is executed a signed 8 bit offset is added to the value in the program counter when a specified condition is met Program execution continues at the new address The numeric range of the short branch offset value is 80 128 to 7F 127 from the address of the instruction immediately following the branch instruction A summary of the short branch instructions is shown in Table 2 2 When a long branch instruction is executed a signed 16 bit offset is added to the value in the program counter when a specified condition is met Program execution continues at the new address Long branch instructions are used when large displacements between decision making steps are necessary Chapter 2 m 68HC12 Assembly Programming Unary Branches
8. branch if not zero yet forever 1 3 5 6 19 41 53 28 13 42 76 14 20 54 64 74 29 33 41 45 A 2 6 5 Decrementing amp Incrementing Instructions We often need to add or subtract one from a variable in our program Although we can use one of the ADD or SUB instructions to achieve this it would be more efficient to use a single instruction The 68HC12 has a few instructions for us to increment or decrement a variable by one A summary of decrement and increment instructions is listed in Table 2 6 60 Chapter 2 m 68HC12 Assembly Programming Decrement instructions Mnemonic Function Operation DEC Decrement memory by 1 M lt M 01 DECA Decrement A by 1 A lt A 01 DECB Decrement B by 1 B amp B 01 DES Decrement SP by 1 SP lt SP 01 DEX Decrement X by 1 X amp X 01 DEY Decrement Y by 1 Ye Y 01 Increment instructions Mnemonic Function Operation INC Increment memory by 1 M amp M 01 INCA Increment A by 1 A lt A 01 INCB Increment B by 1 B amp B 01 INS Increment SP by 1 SP lt SP 01 INX Increment X by 1 X lt X 01 INY Increment Y by 1 Ye Y 01 Table 2 6 m Summary of compare and test instructions Example 2 16 A c cc Use an appropriate increment or decrement instruction to replace the following instruction sequence Idaa i adda 1 staa i Solution The above three instructions can be replaced by the following instruction
9. instructions 57 Write a program to add an array of N 8 bit numbers and store the sum at memory location 800 801 Use the For i n1 to n2 do looping construct Solution We will use variable i as the array index This variable can also be used to keep track of the number of iterations remained to be performed We will use a two byte variable sum to hold the sum of array elements The logic flow of the program is illustrated in Figure 2 9 sum lt sum array i Figure 2 9 m Logic flow of example 2 14 58 Example 2 15 v gt Chapter 2 m 68HC12 Assembly Programming The program is a direct translation of the flowchart shown in Figure 2 9 N equ 20 array count org 800 starting address of on chip SRAM sum rmb 2 array sum i rmb 1 array index org 1000 starting address of the program Idaa 0 staa i initialize loop array index to O staa sum initialize sum to 0 staa sum 1 a loop Idab i cmpb N is N beq done if done then branch Idx array use index register X as a pointer to the array abx compute the address of array i Idab 0 x place array i in B dy sum place sum in Y aby compute sum lt sum array i sty sum update sum inc i increment the loop count by 1 bra loop done swi return to D Bug12 monitor the array is defined in the following statement array db 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 end It is a common mistake for an assembly language p
10. is described in Table 2 8 Mnemonic Function Operation ANDA lt opr gt AND A with memory A lt A M ANDB lt opr gt AND B with memory B amp B M ANDCC lt opr gt AND CCR with memory clear CCR bits CCR lt CCR M EORA lt opr gt Exclusive OR A with memroy A lt A M EORB lt opr gt Exclusive OR B with memory B lt B M ORAA lt opr gt OR A with memory A lt A M ORAB lt opr gt OR B with memory B amp B M ORCC lt opr gt OR CCR with memory CCR lt CCR M CLC Clear C bitin CCR c lt 0 CL Clear bitin CCR I0 CLV Clear V bit in CCR V lt 0 COM lt opr gt One s complement memory M lt FF M COMA One s complement A A lt FF A COMB One s complement B B lt FF B NEG lt opr gt Two s complement memory M amp 00 M NEGA Two s complement A A lt 00 A NEGB Two s complement B B lt 00 B Table 2 8 m Summary of Boolean logic instructions 2 9 m Bit Test amp Manipulate Instruction 69 The operand opr can be specified using all except the relative addressing modes Usually one would use the AND instruction to clear one or a few bits and use the OR instruction to set one or a few bits The exclusive OR instruction can be used to toggle change from 0 to 1 and from 1 to 0 one or a few bits For example the instruction sequence Idaa 56 anda 0F staa 56 clears the upper four pins of the I O port located at 56 The instruc
11. left memory T LSLA Logical shift left A I 0 C b7 b0 LSLB Logical shift left B lt lt lt _ i i Sa eo LSLD Logical shift left D c v K mo b7 A F LSR lt opr gt Logical shift right memory gt LSRA Logical shift right A o gt gt b7 bo c LSRB Logical shift right B gt gt Par o gt H Hi LSRD Logical shift right D o7 x bo b7 TE Arithmetic shift instructions Mnemonic Function Operation ASL lt opr gt Arithmetic shift left memory w ASLA Arithmetic shift left A A 0 C 7 b0 ASLB Arithmetic shift left B lt lt lt lt ASLD Arithmetic shift left D lt ie pad C b7 A bO b7 B b0 ASR lt opr gt Arithmetic shift right memory oS ASRA Arithmetic shift right A H b7 bo Cc ASRB Arithmetic shift right B Rotate instructions Mnemonic Function Operation ROL lt opr gt Rotate left memory thru carry ROLA Rotate left A through carry le Pi le C b7 b0 ROLB Rotate left B through carry ROR lt opr gt Rotate right memory thru carry RORA Rotate right A through carry gt gt gt b7 b0 cC RORB Rotate right B through carry Table 2 7 m Summary of shift and rotate instructions 64 Chapter 2 m 68HC12 Assembly Programming Example 2 18 v What are the values of acc
12. set Gat LBEQ Long branch if equal Z 1 LBMI Long branch if minus N 1 LBNE Long branch if not equal Z 0 LBPL Long branch if plus N 0 LBVC Long branch if overflow is clear V 0 LBVS Long branch if overflow set V 1 Unsigned Branches Mnemonic Function Equation or Operation LBHI Long branch if higher C Z 0 LBHS Long branch if higher or same C 0 LBLO Long branch if lower C 1 LBLS Long branch if lower or same C Z 1 Signed Branches Mnemonic Function Equation or Operation LBGE Long branch if greater than or equal N V 0 LBGT Long branch if greater than Z N V 0 LBLE Long branch if less than or equal Z N V 1 LBLT Long branch if less than N OV 1 Table 2 3 m Summary of long branch instructions Although there are many possibilities in writing a program loop the following one is a com mon format loop Bec or LBcc loop Where cc is one of the condition codes CC CS EQ MI NE PL VC VS HI HS LO LS GE GT LS and LT Usually there will be a comparison or arithmetic instruction to set up the condition code for use by the conditional branch instruction 56 Chapter 2 m 68HC12 Assembly Programming 2 6 3 Compare amp Test Instructions The 68HC12 has a set of compare instructions that are dedicated to the setting of condition flags The compare and test instructions perform subtraction between a pair of registers or between a register and a memory location The result is not stored but con
13. square of all elements of an array with thirty two 8 bit unsigned numbers The array is stored at 800 81F Store the result at 900 901 E2 13 Write a program to count the number of even elements of an array of N 16 bit elements The array is stored at memory locations starting from 900 74 Chapter 2 m 68HC12 Assembly Programming E2 14 Write an instruction sequence to shift the 32 bit number to the left four places The 32 bit number is located at 800 803 E2 15 Write a program to count the number of elements in an array that are smaller than 16 The array is stored at memory locations starting from 800 The array has N 8 bit unsigned elements E2 16 Write an instruction sequence to swap the upper four bits and the lower four bits of accu mulator A swap bit seven with bit three bit six with bit two and so on E2 17 Write a program to count the number of elements in an array whose bits three four and seven are zeroes The array has N 8 bit elements and is stored in memory locations starting from 800 E2 18 Write an instruction sequence to set bits three two one and zero to one and clear the upper four bits E2 19 Find the values of condition flags N Z V and C in the CCR register after the execution of each of the following instructions given that A 50 and the condition flags are N 0 Z 1 V 0 andC 1 a SUBA 40 b TESTA c ADDA 50 d LSRA e ROLA f LSLA E2 20 Find the values of condition flags N
14. std P 2 save the lower 16 bits of the partial product MHNH Idd M place MH in D Idy N 2 place NL in Y emul compute MHNL the following seven instructions add MHNL to memory locations P 2 P 5 addd P 4 add the lower half of MHNL to P 4 P 5 std P 4 G tfr y d transfer Y to D adcb P 3 stab P 3 adca P 2 staa P 2 the following six instructions propagate carry to the most significant byte Idaa P 1 adca 0 add C flag to location P 1 staa P 1 4 Idaa P adca 0 add C flag to location P staa P Example 2 13 v gt SS 2 5 m Writing Programs to do Arithmetic 49 the following three instructions compute MLNH Idd M 2 place MLin D Idy N place NH in Y emul compute MLNH the following seven instructions add MLNH to memory locations P 2 P 5 addd P 4 add the lower half of MLNH to P 4 P 5 std P 4 s tfr y d transfer Y to D adcb P 3 stab P 3 adca P 2 staa P 2 the following six instructions propagate carry to the most significant byte Idaa P 1 adca 0 add C flag to location P 1 staa P 1 Idaa P adca 0 add C flag to location P staa P end Write a program to convert the 16 bit binary number stored at 800 801 to BCD format and store the result at 900 904 Convert each BCD digit into its ASCII code and store it in one byte Solution A binary number can be converted to BCD format using repeated division by 10 The largest 16 bit binary number corresponds to the 5 digit decimal number 65535
15. system 19 plus 47 equals 66 not 60 this example involves a carry from the lower nibble to the higher nibble In summary a sum in the BCD format is incorrect if it is greater than 9 or if there is a carry to the next higher nibble Incorrect BCD sums can be adjusted by adding 6 to them To correct the examples do the following 1 Add 6 to every sum digit greater than nine 2 Add 6 to every sum digit that had a carry of one to the next higher digit Here are the problems with their sums adjusted 18 35 19 47 847 47 5F 7C 60 6 6 6 65 82 66 The fifth bit of the condition code register is the half carry or H flag A carry from the lower nibble to the higher nibble during the addition operation is a half carry A half carry of one during addition sets the H flag to one and a half carry of zero during addition clears it to zero If there is a carry from the high nibble during addition the C flag is set to one which indi cates that the high nibble is incorrect A 6 must be added to the high nibble to adjust it to the correct BCD sum Fortunately we don t need to write instructions to detect the illegal BCD sum following a BCD addition The 68HC12 provides a decimal adjust accumulator A instruction DAA which takes care of all these detailed detection and correction operations The DAA instruction mon itors the sums of BCD additions and the C and H flags and automatically adds 6 to any nibble that requires
16. that are stored at 800 803 and 804 807 and store the sum at 810 813 Solution The addition should start from the least significant byte and proceed to the most sig nificant byte The program is as follows org 1000 _ starting address of the program Idd 802 place the lowest two bytes of the first operand in D addd 806 add the lowest two bytes std 812 save the sum of the lowest two bytes Idaa 801 place the second to most significant byte of the first operand in A adca 805 add the second to most significant byte of the second operand and carry to A staa 811 save the sum of the second to most significant bytes Idaa 800 place the most significant byte of the first operand in A adca 804 add the most significant byte of the second operand and carry to A staa 810 save the sum of the most significant bytes end A 2 5 3 Subtraction amp the C Flag The C flag also enables the 68HC12 to borrow from the high byte to the low byte during a multiprecision subtraction Consider the following subtraction problem 39 74 We are attempting to subtract a larger number from a smaller one Subtracting 4 from 9 is not a problem 39 74 5 Now we need to subtract 7 from 3 To do this we need to borrow from somewhere The 68HC12 borrows from the C flag thus setting the C flag When we borrow from the next higher digit of a hex number the borrow has a value of decimal 16 After the borrow from the C flag th
17. the contents of index register X to index register Y and exe cute the EMUL instruction If index register Y holds useful information then we need to save it before the data transfer sty 810 save Y in a temporary location tfr XY transfer the contents of X to Y emul perform the multiplication sty 800 save the upper 16 bits of the product std 802 save the lower 16 bits of the product Idy 810 restore the value of Y Write an instruction sequence to divide the signed 16 bit number stored at memory loca tions 805 806 by the 16 bit unsigned number stored at memory locations 820 821 and store the quotient and remainder at 900 901 and 902 903 respectively Solution Before we can perform the division we need to place the dividend and divisor in D and X respectively Idd 805 place the dividend in D Idx 820 place the divisor in X idivs perform the signed division stx 900 save the quotient std 902 save the remainder 2 5 m Writing Programs to do Arithmetic 47 Because most arithmetic operations can be performed only on accumulators we need to transfer the contents of index register X to D so that further division on the quotient can be per formed The 68HC12 provides two exchange instructions in addition to the TFR instruction for this purpose The XGDX instruction exchanges the contents of accumulator D and index register X The XGDY instruction exchanges the contents of accumulator D and index
18. the period of the E clock signal are 8 MHz and 125 ns respectively The above instruction sequence will take 5 us to execute Example 2 25 v Example 2 26 v 2 10 m Program Execution Time Write an instruction sequence to create a 100 ms time delay 71 Solution In order to create a 100 ms time delay we need to repeat the above instruction sequence 20 000 times 100 ms 5 us 20 000 The following instruction sequence will create the desired delay loop Write an instruction sequence to create a delay of 10 seconds Idx psha pula psha pula psha pula psha pula psha pula psha pula psha pula nop nop dbne 20000 x loop 2 E cycles 2 E cycles 3 E cycles 2 E cycles 3 E cycles 2 E cycles 3 E cycles 2 E cycles 3 E cycles 2 E cycles 3 E cycles 2 E cycles 3 E cycles 2 E cycles 3 E cycles 1 E cycle 1 E cycle 3 E cycles Solution The instruction sequence in Example 2 25 can create no more than 327 ms delay In order to create a longer time delay we need to use a two layer loop For example the following instruction sequence will create a 10 second delay out_loop inner_loop Idab Idx psha pula psha pula psha pula psha pula psha pula psha pula psha pula 100 20000 1 E cycle 2 E cycles 72 Chapter 2 m 68HC12 Assembly Programming psha pula psha pula nop nop dbne x inner_loop dbne b out_loop 3 E cycles The time de
19. 68HC12 Assembly Programming 2 1 Objectives After completing this chapter you should be able to Use assembler directives to allocate memory blocks define constants and create a message to be output Write assembly programs to perform simple arithmetic operations Write program loops to perform repetitive operations Use program loops to create time delays Use Boolean and bit manipulation instructions to perform bit field manipulations 30 Chapter 2 m 68HC12 Assembly Programming 2 2 Assembly Language Program Structure An assembly language program consists of a sequence of statements that tells the computer to perform the desired operations From a global point of view a 68HC12 assembly program consists of three sections In some cases these sections can be mixed to provide better algorithm design The three sections are Assembler directives Assembler directives instruct the assembler how to process subsequent assembly language instructions Directives also provide a way to define program constants and reserve space for dynamic variables Some directives may also set a location counter Assembly language instructions These instructions are 68HC12 instructions Some instructions are defined with labels Comments There are two types of comments in an assembly program The first type is used to explain the function of a single instruction or directive The second type explains the function of a group of instruction
20. Program Structure 31 print jsr hexout label is terminated by a colon jmp begin instruction references the label begin The following instructions contain invalid labels hereis adda 5 a space is included in the label loop deca labels begins at column 2 A 2 2 2 The Operation Field This field contains the mnemonic names for machine instructions and assembler directives If a label is present the opcode or directive must be separated from the label field by at least one space If there is no label the operation field must be at least one space from the left margin Example 2 2 v Examples of operation fields adda 02 adda is the instruction mnemonic true equ 1 equate directive equ occupies the operation field A 2 2 3 The Operand Field If an operand field is present it follows the operation field and is separated from the oper ation field by at least one space The operand field may contain operands for instructions or arguments for assembler directives The following instructions include the operand field TCNT equ 0084 0084 is the operand field TCO equ 0090 0090 is the operand field 2 2 4 The Comment Field The comment field is optional and is added mainly for documentation purposes The com ment field is ignored by the assembler Here are the rules for comments Any line beginning with an is a comment Any line beginning with a semi colon is a comment In this book we will use to st
21. The first divi sion by 10 computes the least significant digit and should be stored in the memory location 904 the second division by 10 operation computes the ten s digit and so on The ASCII code of a BCD digit can be obtained by adding 30 to each BCD digit The program is as follows org data fdb org result rmb org Idd Idy Idx idiv addb stab xgdx Idx idiv addb stab xgdx Idx 800 12345 place a number for testing 900 5 reserve five bytes to store the result 1000 data make a copy of the number to be converted result 10 divide the number by 10 30 convert to ASCII code 4Y save the least significant digit swap the quotient to D 10 30 convert to ASCII code 3 Y save the second to least significant digit 10 50 Chapter 2 m 68HC12 Assembly Programming idiv addb stab xgdx Idx idiv addb stab xgdx addb stab end 2 6 Program Loops Many applications require repetitive operations We can write programs to tell computers to perform the same operation over and over A finite loop is a sequence of instructions that will be executed by the computer for a finite number of times while an endless loop is a sequence of 30 2 Y 10 30 LY 30 0 Y save the middle digit separate the most significant and second to most significant digits save the second to most significant digit swap the most significant digit to B convert to ASCII code save th
22. a given value The syntax of this directive is as follows fill value count where the number of bytes to be filled is indicated by count and the value to be filled is indi cated by value For example the statement space_line fill 20 40 will fill 40 bytes with the value of 20 starting from the memory location referred to by the label space_line ds define storage rmb reserve memory byte ds b define storage bytes Each of these three directives reserves a number of bytes given as the arguments to the directive The location counter will be incremented by the number that follows the direc tive mnemonic 34 Chapter 2 m 68HC12 Assembly Programming For example the statement buffer ds 100 reserves 100 bytes in memory starting from the location represented by the label buffer After this directive the location counter will be incremented by 100 The content s of the reserved memory location s are not defined ds w define storage word rmw reserve memory word Each of these directives increments the location counter by the value indicated in the num ber of words argument multiplied by two In other words if the ds w expression evaluates to k then the location counter is advanced by 2k These directives are often used with a label For example the statement dbuf ds w 20 reserves 40 bytes starting from the memory location represented by the label dbuf None of these 40 bytes are initialized equ equate T
23. and bit two of accumulator A and updates Z and N flags of CCR register accord ingly The V flag in CCR register is cleared The instruction bitb 22 tests the bit five and bit one of accumulator B and updates the Z and N flags of CCR register accordingly The V flag in CCR register is cleared Finally the instruction bset 0 y 33 sets the bits five four one and zero of the memory location pointed to by index register Y 2 10 Program Execution Time The 68HC12 uses the ECLK we will call it E clock from now on signal as a timing refer ence The frequency of the E clock is equal to one half of the frequency of the crystal oscillator out of reset The execution times of instructions are also measured in E cycles There are many applications that require the generation of time delays Program loops are often used to create some amount of delay unless the time delay needs to be very accurate The creation of a time delay involves two steps 1 Select a sequence of instructions that takes a certain amount of time to execute 2 Repeat the instruction sequence for the appropriate number of times For example the following instruction sequence takes 40 E clock cycles to execute 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 If the 68HC12 runs under the control of a 16 MHz crystal oscillator then the frequency and
24. apter 2 m 68HC12 Assembly Programming A amp A 802 A A 5805 Figure 2 2 m Logic flow of program 2 4 Write a program to subtract five from four memory locations at 800 801 802 and 803 Solution In the 68HC12 a memory location cannot be the destination of an ADD or SUB instruction Therefore three steps must be followed to add or subtract a number to or from a memory location Step 1 Load the memory contents into an accumulator Step 2 Add or subtract the number to from the accumulator Step 3 Store the result at the specified memory location The program is as follows org 1000 Idaa 800 copy the contents of memory location 800 to A suba 5 subtract 5 from A staa 800 store the result back to memory location 800 Idaa 801 suba 5 staa 801 Idaa 802 suba 5 staa 802 Example 2 6 v 2 5 m Writing Programs to do Arithmetic 39 Idaa 803 suba 5 staa 803 end Write a program to add two 16 bit numbers that are stored at 800 801 and 802 803 and store the sum at 900 901 Solution This program is very straightforward org 1000 Idd 800 place the 16 bit number at 800 801 in D addd 802 add the 16 bit number at 802 803 to D std 900 save the sum at 900 901 end A 2 5 1 Carry Borrow Flag The 68HC12 can add and subtract either 8 bit or 16 bit numbers and place the result in either 8 bit accumulators A or B or the double accumulator D The 8 b
25. art a comment You must have a semi colon prefixing any comment on a line with mnemonics Examples of comments appear in the following instructions this program computes the square root of N 8 bit integers org 1000 set the location counter to 1000 dec Ip_cnt decrement the loop count In this chapter we will use the Motorola Freeware cross assembler as12 as the standard to explain every aspects of the assembly programming 32 Chapter 2 m 68HC12 Assembly Programming 2 3 Assembler Directives Assembler directives look just like instructions in an assembly language program but they tell the assembler to do something other than creating the machine code for an instruction The available assembler directives vary with the assembler Interested readers should refer to the user s manual of the specific assembler for details We will discuss assembler directives supported by the as12 in detail In the following dis cussion statements enclosed in brackets are optional All directives and assembly instruc tions can be in either upper or lowercase End The end directive is used to end a program to be processed by the assembler In general an assembly program looks like this your program end The end directive indicates the logical end of the source program Any statement following the end directive is ignored A warning message will be raised if the end directive is missing from the source code however the program
26. dition codes are set by the operation In the 68HC12 most instructions update condition code flags automatically so it is often unnecessary to include a separate test or compare instruction Table 2 4 is a sum mary of compare and test instructions Compare instructions Mnemonic Function Operation CBA Compare A to B A B CMPA Compare A to memory A M CMPB Compare B to memory B M CPD Compare D to memory D M M 1 CPS Compare SP to memory SP M M 1 CPX Compare X to memory X M M 1 CPY Compare Y to memory Y M M 1 Test instructions Mnemonic Function Operation TST Test memory for zero or minus M 00 TSTA Test A for zero or minus A 00 TSTB Test B for zero or minus B 00 Table 2 4 m Summary of compare and test instructions 2 6 4 Loop Primitive Instructions A lot of the program loops are implemented by incrementing or decrementing a loop count The branch is taken when either the loop count is equal to zero or not equal to zero depending on the applications The 68HC12 provides a set of loop primitive instructions for implement ing this type of looping mechanism These instructions test a counter value in a register or accu mulator A B D X Y or SP for zero or nonzero values as a branch condition There are predecrement preincrement and test only versions of these instructions The range of the branch is from 80 128 to 7F 127 from the instruction i
27. e most significant digit instructions that the computer will execute forever There are four major loop constructs Do statement S forever This is an endless loop in which statement S is repeated forever In some applica tions we might add the statement If C then exit to leave the infinite loop An infinite loop is shown in Figure 2 4 Figure 2 4 m An infinite loop For i n1 to n2 do S or For i n2 to n1 do S Here the variable i is the Joop counter which keeps track of the number of remaining times statement S is to be executed The loop counter can be incre mented the first case or decremented the second case Statement S is repeated n2 n1 1 times The value of n2 is assumed to be no smaller than n1 If there is concern that the relationship n1 lt n2 may not hold then it must be checked at the beginning of the loop Four steps are required to implement a FOR loop 2 6 m Program Loops 51 Step 1 Initialize the loop counter and other variables Step 2 Compare the loop counter with the limit to see if it is within bounds If it is then perform the specified operations Otherwise exit the loop Step 3 Increment or decrement the loop counter Step 4 Go to step 2 A For loop is illustrated in Figure 2 5 a For i toi DOS b For I i downto i DO S Figure 2 5 m For looping construct While C Do S Whenever a While construct is executed the logical expression C is evaluated first If it yields a
28. e problem can be completed 39 74 C5 42 Chapter 2 m 68HC12 Assembly Programming When the 68HC12 executes a subtract instruction it always borrows from the C flag The borrow is either 1 or 0 The C flag operates as follows during a subtraction If the 68HC12 borrows a 1 from the C flag during a subtraction the C flag is set to 1 If the 68HC12 borrows a 0 from the C flag during a subtraction the C flag is set to 0 2 5 4 Multiprecision Subtraction For a 16 bit microcontroller multiprecision subtraction is the subtraction of numbers that are larger than 16 bits To subtract the hex number 16753284 from 98765432 the 68HC12 has to perform multiprecision subtraction 98765432 16757284 Like multiprecision addition multiprecision subtraction is performed one byte at a time beginning with the least significant byte The 68HC12 does allow us to subtract two bytes at a time because it has the SUBD instruction The following two instructions can be used to sub tract the least significant two bytes of the subtrahend from the minuend Idd 5432 subd 7284 Since a larger number is subtracted from a smaller one there is a need to borrow from the higher byte causing the C flag to be set to 1 The contents of double accumulator D should be saved before the higher bytes are subtracted Let s save these two bytes at 802 803 std 802 When the second to most significant bytes are subtracted the borrow 1 has to be sub
29. false value statement S will not be executed The action of a While construct is illustrated in Figure 2 6 Four steps are required to implement a While loop Step 1 Initialize the logical expression C Step 2 Evaluate the logical expression C Step 3 Perform the specified operations if the logical expression C evaluates to true Update the logical expression C and go to step 2 Note The logical expression C may be updated by external conditions or by an interrupt service routine Step 4 Exit the loop 52 Chapter 2 m 68HC12 Assembly Programming false Figure 2 6 m The While Do looping construct Repeat S Until C Statement S is first executed then the logical expression C is evaluated If C is false the next statement will be executed Otherwise statement S will be exe cuted again The action of this construct is illustrated in Figure 2 7 Statement S will be executed at least once Three steps are required to implement this construct Step 1 Initialize the logical expression C Step 2 Execute statement S Step 3 Go to Step 2 if the logical expression C evaluates to true Otherwise exit Figure 2 7 m The Repeat Until looping construct To implement one of the looping constructs we need to use unconditional branch or one of the conditional instructions When executing conditional branch instructions the 68HC12 checks the condition flags in the CCR register 2 6 1 Condition Code Register The contents of the co
30. from the hex number stored at 800 803 and save the difference at 900 903 Solution We will perform the subtraction from the least significant byte towards the most sig nificant byte as follows org 1000 _ starting address of the program Idd 802 place the lowest two bytes of the minuend in D subd 806 subtract the lowest two bytes of the subtrahend from D std 902 save the lowest two bytes of the difference Idaa 801 put the second to most significant byte of the minuend in A sbca 805 subtract the second to most significant byte of the subtrahend and the borrow from A staa 901 save the second to most significant byte of the difference Idaa 800 put the most significant byte of the minuend in A sbca 804 subtract the most significant byte of the subtrahend and the borrow from A staa 900 save the most significant byte of the difference end A 2 5 5 Binary Coded Decimal BCD Addition Although virtually all computers work internally with binary numbers the input and out put equipment generally uses decimal numbers Since most logic circuits only accept two val ued signals the decimal numbers must be coded in terms of binary signals In the simplest form of binary code each decimal digit is represented by its binary equivalent For example 2 538 is represented by 0010 0101 0011 1000 This representation is called binary coded decimal If the BCD format is used it must be pre served during arithmetic p
31. his directive assigns a value to a label Using equ to define constants will make our pro gram more readable For example the statement loop_cnt equ 40 informs the assembler that whenever the symbol Joop_cnt is encountered it should be replaced with the value of 40 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 you can write program seg ments like the following example without thinking up new labels loc Idaa 2 loop deca bne loop loc loop brelr 0 x 55 loop This code segment will work perfectly fine because the first loop label will be seen as loop001 whereas the second loop label will be seen as loop002 The assembler actually sees this loc Idaa 2 loop001 deca bne loop001 loc loop002 brelr 0 x 55 loop002 You can also set the loc directive with a valid expression or number by putting that expres sion or number in the operand field The resultant number will be used to increment the suffix to the label 2 4 m Software Development Issues 35 2 4 Software Development Issues A complete discussion of issues involved in software development is out of the scope of this text However we do need to take a serious look at some software development issues because embedded system designers must spend a significant amount of time on software development As we all know software development starts with pr
32. inc l m ISA 2 6 6 Bit Condition Branch Instructions In certain applications we need to make branch decisions based on the value of a few bits The 68HC12 provides two special conditional branch instructions for this purpose The syntax of the first special conditional branch instruction is lt label gt BRCLR opr msk rel where opr specifies the memory location to be checked and can be specified using direct extended and all indexed addressing modes Example 2 17 gt vy d 2 6 m Program Loops 61 msk is an 8 bit mask that specifies the bits of the memory location to be checked The bits to be checked correspond to those bit positions that are ones in the mask rel is the branch offset and is specified in 8 bit relative mode This instruction tells the 68HC12 to perform bitwise logical AND on the contents of the specified memory location and the mask supplied with the instruction then branch if the result is zero For example for the instruction sequence here brelr 66 80 here Idd 70 the 68HC12 will continue to execute the first instruction if the most significant bit of the mem ory location at 66 is 0 Otherwise the next instruction will be executed The syntax of the second special conditional branch instruction is lt label gt BRSET opr msk rel where opr specifies the memory location to be checked and can be specified using direct extended and all indexed addressing modes msk i
33. ing instructions a bne not_done b loop brelr 0 x 01 loop wait until the least significant bit is set c here dec Ip_cnt decrement the variable Ip_cnt E2 3 Write a sequence of assembler directives to reserve 10 bytes starting from 800 E2 4 Write a sequence of assembler directives to build a table of ASCII codes of lower case let ters a z The table should start from memory location 2000 E2 5 Write a sequence of assembler directives to store the message welcome to the robot demonstration starting from the memory location at 1050 E2 6 Write an instruction sequence to add the two 24 bit numbers stored at 810 812 and 813 815 and save the sum at 900 902 E2 7 Write an instruction sequence to subtract the 6 byte number stored at 800 805 from the 6 byte number stored at 810 805 and save the result at 900 905 E2 8 Write a sequence of instructions to add the BCD numbers stored at 800 and 801 and store the sum at 803 E2 9 Write an instruction sequence to add the 4 digit BCD numbers stored at 800 801 and 802 803 and store the sum at 900 901 E2 10 Write a program to compute the average of an array of N 8 bit numbers and store the result at 900 The array is stored at memory locations starting from 800 N is no larger than 255 E2 11 Write a program to multiply two 3 byte numbers that are stored at 800 802 and 803 805 and save the product at 900 905 E2 12 Write a program to compute the average of the
34. it The rules for using the DAA instruction are as follows 1 The DAA instruction can only be used for BCD addition It does not work for sub traction or hex arithmetic 2 The DAA instruction must be used immediately after one of the three instruc tions that leave their sum in accumulator A These three instructions are ADDA ADCA ABA 3 The numbers added must be legal BCD numbers to begin with Write an instruction sequence to add the BCD numbers stored at memory locations 800 and 801 and store the sum at 810 2 5 m Writing Programs to do Arithmetic 45 Solution Idaa 800 load the first BCD number in A adda 801 perform addition daa decimal adjust the sum in A staa 810 save the sum A 2 5 6 Multiplication amp Division The 68HC12 provides three multiply and five divide instructions A brief description of these instructions appears in Table 2 1 Mnemonic Function Operation EMUL unsigned 16 by 16 multiply D x Y gt YD EMULS signed 16 by 16 multiply D x Y gt YD MUL unsigned 8 by 8 multiply A x B gt AB EDIV unsigned 32 by 16 divide Y D X quotient Y remainder D EDIVS signed 32 by 16 divide Y D X quotient Y remainder D FDIV 16 by 16 fractional divide D X gt X remainder D IDIV unsigned 16 by 16 integer divide D X gt X remainder D IDIVS signed 16 by 16 integer divide D X gt X remainder D Table 2 1 m Summary of 68HC12 multiply a
35. it number stored in accumulator B can also be added to index register X However programs can also be written to add numbers larger than 16 bits Arithmetic performed in a 16 bit microproces sor microcontroller on numbers that are larger than 16 bits is called multiprecision arith metic Multiprecision arithmetic makes use of the carry flag C flag of the condition code register CCR Bit O of the CCR register is the C flag It can be thought of as a temporary 9 bit that is appended to any 8 bit register or 17 bit that is appended to any 16 bit register The C flag allows us to write programs to add and subtract hex numbers that are larger than 16 bit For example consider the following two instructions Idd 8645 addd 9978 These two instructions add the numbers 8645 and 9978 8645 9978 11FBD The result is 11FBD a 17 bit number which is too large to fit into the 16 bit double accumu lator D When the 68HC12 executes these two instructions the lower sixteen bits of the answer 1FBD are placed in double accumulator D This part of the answer is called the sum The leftmost bit is called a carry A carry of 1 following an addition instruction sets the C flag of the CCR register to 1 A carry of 0 following an addition clears the C flag to 0 This applies to both 8 bit and 16 bit additions for the 68HC12 For example execution of the following two instructions Idd 1245 addd 4581 40 Chapter 2 m 68HC12 Assembly P
36. lay created by using program loops is not accurate Some overhead is required to set up the loop count For example the one layer loop has a 2 E cycle overhead while the two layer loop has much more overhead overhead 1 E cycle caused by the Idab 100 instruction 100 x 2 E cycles caused by the out_loop Idx 20000 instruction 100 x 3 E cycles caused by the dbne b out_loop instruction 501 E cycles 62 625 us at 8 MHz E clock This overhead can be reduced by placing a larger value in index register X and a smaller value in accumulator B For example by placing 50 000 in X and 40 in B the overhead can be reduced to 25 125 us If higher accuracy is required then you should use one of the timer func tions to create the desired time delay 2 11 Summary An assembly language program consists of three major parts assembler directives assem bly language instructions and comments A statement of an assembly language program con sists of four fields Jabel operation code operand and comment Assembly directives supported by the freeware as12 are all discussed in this chapter The 68HC12 instructions are explained category by category Simple program examples are used to demonstrate the applications of different instructions The 68HC12 is a 16 bit micro controller Therefore it can perform 16 bit arithmetic Numbers greater than 16 bits must be manipulated using multiprecision arithmetic Microcontrollers are designed to perfor
37. m repetitive operations Repetitive operations are implemented by program loops There are two types of program loops infinite loop and finite loop There are four major variants of the looping constructs Do statement S forever Fori n ton do S or For i n downto n do S While C do S m Repeat S until C In general the implementation of program loops requires m the initialization of a loop counter or condition m performing the specified operation m comparing the loop count with the loop limit or evaluating the condition making a decision regarding whether the program loop should be continued The 68HC12 provides instructions to support the initialization of a loop counter decre menting or incrementing the loop counter and deciding whether looping should be continued 2 11 m Exercises 73 The shifting and rotating instructions are useful for bit field operations Integer multiplica tion by a power of two and division by a power of two can be sped up by using the shifting instructions The 68HC12 also provides many Boolean logical instructions that can be very useful for set ting clearing and toggling the I O port pins 2 12 Exercises E2 1 Find the valid and invalid labels in the following statements and explain why the invalid labels are invalid column 1 y a ABC decb b Ip adda 1 c too mul d not_true nega e star Idaa 10 f too_big dec count E2 2 Identify the four fields of the follow
38. mmediately following the loop primitive instruction Table 2 5 shows a summary of the loop primitive instructions Example 2 14 v gt SS 2 6 m Program Loops Mnemonic DBEQ cntr rel DBNE cntr rel IBEQ cntr rel IBNE cntr rel TBEQ cntr rel TBNE cntr rel Function Decrement counter and branch if 0 counter A B D X Y or SP Decrement counter and branch if 0 counter A B D X Y or SP Increment counter and branch if 0 counter A B D X Y or SP Increment counter and branch if 0 counter A B D X Y or SP Test counter and branch if 0 counter A B D X Y or SP Test counter and branch if 0 counter A B D X Y or SP Equation or Operation counter lt counter 1 If counter 0 then branch else continue to next instruction counter lt counter 1 f counter 0 then branch else continue to next instruction counter lt counter 1 f counter 0 then branch else continue to next instruction counter lt counter 1 f counter 0 then branch else continue to next instruction f counter 0 then branch else continue to next instruction f counter 0 then branch else continue to next instruction Note 1 entr is the loop counter and can be accumulator A B or D and register X Y or SP 2 rel is the relative branch offset and is usually a label Table 2 5 m Summary of loop primitive
39. n its immediate left with the exception of the most significant byte which will receive a zero Each byte will be shifted to the right by one bit The bit zero of the least signifi cant byte will be shifted out and lost The operation can therefore be implemented as follows Step 1 Shift the byte at Joc to the right one place using the LSR lt opr gt instruction Step 2 Rotate the byte at loc 1 to the right one place using the ROR lt opr gt instruction Step 3 Repeat step 2 for the remaining bytes 68 Chapter 2 m 68HC12 Assembly Programming By repeating this procedure the given k byte number can be shifted to the right as many bits as desired The operation to shift a multi byte number to the left should start from the least significant byte and rotate the remaining bytes toward the most significant byte A Example 2 24 v Write a program to shift the 32 bit number stored at 800 803 to the right four places Solution The most significant to the least significant bytes are stored at 800 803 The fol lowing instruction sequence implements the algorithm that we just described Idab 4 set up the loop count Idx 800 again Isr 0 x ror 1x ror 2 X ror 3 x dbne b again 2 8 Boolean Logic Instructions When dealing with input and output port pins we often need to change the values of a few bits For these types of applications Boolean logic instructions come in handy A summary of the 68HC12 Boolean logic instructions
40. nd divide instructions The EMUL instruction multiplies the 16 bit unsigned integers stored in accumulator D and index register Y and leaves the product in these two registers The upper 16 bits of the product are in Y whereas the lower 16 bits are in D The EMULS instruction multiplies the 16 bit signed integers stored in accumulator D and index register Y and leaves the product in these two registers The upper 16 bits of the product are in Y whereas the lower 16 bits are in D The MUL instruction multiplies the 8 bit unsigned integer in accumulator A by the 8 bit unsigned integer in accumulator B to obtain a 16 bit unsigned result in double accumulator D The upper byte of the product is in accumulator A whereas the lower byte of the product is in B The EDIV instruction performs an unsigned 32 bit by 16 bit division The dividend is the register pair Y and D with Y as the upper 16 bit of the dividend Index register X is the divisor After division the quotient and the remainder are placed in Y and D respectively The EDIVS instruction performs a signed 32 bit by 16 bit division using the same operands as the EDIV instruction does After division the quotient and the remainder are placed in Y and D respectively 46 Example 2 10 v gt n Example 2 11 v Chapter 2 m 68HC12 Assembly Programming The FDIV instruction divides an unsigned 16 bit dividend in double accumulator D by an unsigned 16 bit divisor in index register X p
41. nd shift it to the right sixteen times or until it becomes zero The algo rithm of this program is as follows Step 1 Initialize the loop count to 16 and the zero count to 0 Step 2 Place the 16 bit value in D Step 3 Shift D to the right one place Step 4 If the C flag is 0 increment zero count by 1 Step 5 Decrement loop count by 1 Step 6 If loop count is zero then stop Otherwise go to step 3 2 7 m Shift amp Rotate Instuctions 67 The program is as follows org 800 db 23 55 org 805 zero_cnt rmb 1 lp_cnt rmb 1 org 1000 clr zero_cnt initialize the zero count to O Idaa 16 staa Ip_cnt initialize loop count to 16 Idd 800 place the 16 bit number in D again Isrd bes chk_end branch if the Isb is a 1 inc zero_cnt chk_end dec Ip_cnt bne again have we tested all 16 bits yet forever bra forever end Sometimes we need to shift a number larger than 16 bits However the 68HC12 does not have an instruction that does this Suppose the number has k bytes and the most significant byte is located at loc The remaining k 1 bytes are located at loc 1 loc 2 loc k 1 as shown in Figure 2 16 loc loc 1 loc k 1 Figure 2 16 m k bytes to be shifted The logical shift one bit to the right operation is shown in Figure 2 17 loc loc 1 loc k 1 oS aE msb Isb Figure 2 17 m Shift one to the right operation As shown in Figure 2 17 The bit seven of each byte will receive the bit zero of the byte o
42. ndition code register are shown in Figure 2 8 The shaded characters are condition flags that reflect the status of an operation The meanings of these condition flags are as follows 7 6 5 4 3 2 1 0 Figure 2 8 m Condition code register 2 6 m Program Loops 53 C the carry flag Whenever a carry is generated as the result of an operation this flag will be set to 1 Otherwise it will be cleared to 0 V the overflow flag Whenever the result of a two s complement arithmetic operation is out of range this flag will be set to 1 Otherwise it will be set to 0 The V flag is set to 1 when the carry from the most significant bit and the second most significant bit differ as the result of an arithmetic operation Z the zero flag Whenever the result of an operation is zero this flag will be set to 1 Otherwise it will be set to 0 N the negative flag Whenever the most significant bit of the result of an operation is 1 this flag will be set to 1 Otherwise it will be set to 0 This flag indicates that the result of an operation is negative H the half carry flag Whenever there is a carry from the lower four bits to the upper four bits as the result of an operation this flag will be set to 1 Otherwise it will be set to 0 2 6 2 Branch Instructions Branch instructions cause program flow to change when specific conditions exist The 68HC12 has three kinds of branch instructions including short branches long branches and
43. o add the numbers stored at memory locations 800 801 and 802 and store the sum at memory location 900 Solution This problem can be solved by the following steps Step 1 Load the contents of the memory location at 800 into accumulator A Step 2 Add the contents of the memory location at 801 into accumulator A Step 3 Add the contents of the memory location at 802 into accumulator A Step 4 Store the contents of accumulator A at memory location 900 These steps can be translated into the as12 assembly program as follows org 1000 starting address of the program Idaa 800 place the contents of the memory location 800 into A adda 801 add the contents of the memory location 801 into A adda 802 add the contents of the memory location 802 into A staa 900 store the sum at the memory location 900 end A Example 2 4 v Write a program to subtract the contents of the memory location at 805 from the sum of the memory locations at 800 and 802 and store the result at the memory location 900 Solution The logic flow of this program is illustrated in Figure 2 2 The assembly program is as follows org 1000 starting address of the program Idaa 800 copy the contents of the memory location at 800 to A adda 802 add the contents of memory location at 802 to A suba 805 subtract the contents of memory location at 805 from A staa 900 store the contents of accumulator A to 805 end 38 Example 2 5 v Ch
44. oblem definition The problem pre sented by the application must be fully understood before any program can be written At the problem definition stage the most critical thing is to get you the programmer and your end user to agree upon what needs to be done To achieve this asking questions is very important For complex and expensive applications a formal written definition of the problem is formu lated and agreed upon by all parties Once the problem is known the programmer can begin to lay out an overall plan of how to solve the problem The plan is also called an algorithm Informally an algorithm is any well defined computational procedure that takes some value or set of values as input and produces some value or set of values as output An algorithm is thus a sequence of computational steps that transforms the input into the output We can also view an algorithm as a tool for solving a well specified computational problem The statement of the problem specifies in general terms the desired input output relationship The algorithm describes a specific computational proce dure for achieving that input output relationship An algorithm is expressed in pseudocode that is very much like C or PASCAL What sepa rates pseudocode from real code is that in pseudocode we employ whatever expressive method that is most clear and concise to specify a given algorithm Sometimes the clearest method is English so do not be surprised if you c
45. ome across an English phrase or sentence embedded within a section of real code An algorithm provides not only the overall plan for solving the problem but also documen tation to the software to be developed In the rest of this book all algorithms will be presented in the following format Step 1 Step 2 An earlier alternative for providing the overall plan for solving software problems was the use of flowcharts A flowchart shows the way a program operates It illustrates the logic flow of the program Therefore flowcharts can be a valuable aid in visualizing programs Flowcharts are not only used in computer programming they are also used in many other fields such as busi ness and construction planning The flowchart symbols used in this book are shown in Figure 2 1 The terminal symbol is used at the beginning and end of each program When it is used at the beginning of a pro gram the word Start is written inside it When it is used at the end of a program it contains the word Stop The process box indicates what must be done at this point in the program execution The operation specified by the process box could be shifting the contents of one general purpose reg ister to a peripheral register decrementing a loop count and so on 36 Chapter 2 m 68HC12 Assembly Programming Terminal Process Input or output ssi off page connector on page connector Figure 2 1 m Flowchart symbols used in this book
46. register Y The 68HC12 provides instructions for performing unsigned 8 bit by 8 bit and both signed and unsigned 16 bit by 16 bit multiplications Since the 68HC12 is a 16 bit microcontroller we expect that it will be used to perform complicated operations in many sophisticated applica tions Performing 32 bit by 32 bit multiplication will be one of them Since there is no 32 bit by 32 bit multiplication instructions we will have to break a 32 bit number into two 16 bit halves and use the 16 bit by 16 bit multiply instruction to synthesize the operation Assume M and N are the multiplicand and the multiplier respec tively These two numbers can be broken down as follows M M M N N N where My and Ny are the upper 16 bits and M and N are the lower 16 bits of M and N respec tively Four 16 bit by 16 bit multiplications are performed and then their partial products are added together as shown in Figure 2 3 16 bit 16 bit 16 bit 16 bit upper half lower half partial product M N upper half lower half partial product M N upper half lower half partial product M Ny upper half lower half partial product M N Address P 2 P 3 P 4 P 5 P 6 P 7 Final product M x N msb lsb Figure 2 3 m Unsigned 32 bit by 32 bit multiplication The procedure is as follows Step 1 Allocate eight bytes to hold the product Assume these eight bytes are located at P P 1 and P 7
47. rever array db 2 3 4 8 12 13 19 24 33 32 20 18 53 52 80 82 90 94 100 102 end 2 6 7 Instructions for Variable Initialization We often need to initialize a variable to zero when writing a program The 68HC12 has three instructions for this purpose They are lt label gt clr opr where opr is a memory location specified using the extended mode and all indexed addressing direct and indirect modes The memory location is initialized to zero by this instruction lt label gt clra Accumulator A is cleared to 0 by this instruction lt label gt clrb Accumulator B is cleared to 0 by this instruction 2 7 Shift amp Rotate Instructions Shift and rotate instructions are useful for bit field manipulation They can be used to speed up the integer multiply and divide operations if one of the operands is a power of two A shift rotate instruction shifts rotates the operand by one bit The 68HC12 has shift instructions that can operate on accumulators A B and D or on a memory location A memory operand must be specified using the extended or indexed direct or indirect addressing modes A sum mary of shift and rotate instructions is shown in Table 2 7 Ww 2 7 m Shift amp Rotate Instuctions 6 Logical shift instructions Mnemonic Function Operation LSL lt opr gt Logical shift
48. rocessing The principal advantage of the BCD encoding method is the simplicity of input output con version its major disadvantage is the complexity of arithmetic processing The choice between binary and BCD depends on the type of problems the system will be handling The 68HC12 microcontroller can add only binary numbers not decimal numbers The fol lowing instruction sequence appears to cause the 68HC12 to add the decimal numbers 25 and 31 and store the sum at the memory location 800 Idaa 25 adda 31 staa 800 This instruction sequence performs the following addition 25 31 56 When the 68HC12 executes this instruction sequence it adds the numbers according to the rules of binary addition and produces the sum 56 This is the correct BCD answer because the 44 Example 2 9 v Chapter 2 m 68HC12 Assembly Programming result represents the decimal sum of 25 31 56 In this example the 68HC12 gives the appear ance of performing decimal addition However a problem occurs when the 68HC12 adds two BCD digits and generates a sum greater than nine Then the sum is incorrect in the decimal number system as the following three examples illustrate 18 35 19 47 847 47 5F 7C 60 The answers to the first two problems are obviously wrong in the decimal number system because the hex digits F and C are not between zero and nine The answer to the third example appears to contain valid BCD digits but in the decimal
49. roducing an unsigned 16 bit quotient in X and an unsigned 16 bit remainder in D The dividend must be less than the divisor The radix point of the quotient is to the left of the bit 15 In the case of overflow the denominator is less than or equal to the nominator or division by zero the quotient is set to FFFF and the remain der is indeterminate The IDIV instruction divides an unsigned 16 bit dividend in double accumulator D by the unsigned 16 bit divisor in index register X producing an unsigned 16 bit quotient in X and an unsigned 16 bit remainder in D If both the divisor and the dividend are assumed to have radix points in the same positions to the right of bit 0 the radix point of the quotient is to the right of bit 0 In the case of division by zero the quotient is set to FFFF and the remain der is indeterminate The IDIVS instruction divides the signed 16 bit dividend in double accumulator D by the signed 16 bit divisor in index register X producing a signed 16 bit quotient in X and a signed 16 bit remainder in D If division by zero is attempted the values in D and X are not changed but the values of the N Z and V status bits are undefined Write an instruction sequence to multiply the contents of index register X and double accu mulator D and store the product at memory locations 800 803 Solution There is no instruction to multiply the contents of double accumulator D and index register X However we can transfer
50. rogrammer to forget to update the vari able in memory For example we will not get the correct value for sum if we did not add the instruction sty sum in the program shown in Example 2 14 Loop primitive instructions are especially suitable for implementing the repeat S until C looping construct as demonstrated in the following example 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 We will use the variable i as the array index and also as the loop count The variable max_val will be used to hold the array maximum The logic flow of the program is shown in Figure 2 10 2 6 m Program Loops 59 max_val lt array 0 ic N l max_val lt array i max_val lt array i Figure 2 10 m Logic flow of example 2 15 The program is as follows N equ org max_val rmb org Idaa staa Idx Idab loop Idaa cmpa bge Idaa staa chk_end dex dbne forever bra array db end 20 array count 800 Starting address of on chip SRAM 1 memory location to hold array max 1000 starting address of program array set array 0 as the temporary array max max_val g array N 1 start from the end of the array N 1 use B to hold variable i and initialize it to N 1 max_val 0 x compare max_val with array i chk_end no update if max_val is larger 0 x update max_val max_val move the array pointer b loop decrement the loop count
51. rogramming will clear the C flag to 0 because the carry resulting from this addition is 0 In summary If the addition produces a carry of 1 the carry flag is set to 1 If the addition produces a carry of 0 the carry flag is cleared to 0 2 5 2 Multiprecision Addition For a 16 bit microcontroller like the 68HC12 multiprecision addition is the addition of numbers that are larger than 16 bits To add the hex number 1A598183 to 76548290 the 68HC12 has to perform multiprecision addition 1 1 1 1A598183 76548290 90AE0413 Multiprecision addition is performed one byte at a time beginning with the least significant byte The 68HC12 does allow us to add 16 bit numbers at a time because it has the ADDD instruc tion The following two instructions can be used to add the least significant 16 bit numbers together Idd 8183 addd 8290 Since the sum of the most significant digit has a sum greater than 16 it generates a carry that must be added to the next more significant digit causing the C flag to be set to 1 The contents of double accumulator D must be saved before the higher bytes are added Let s save these two bytes at 802 803 std 802 When the second to most significant bytes are added the carry from the lower byte must be added in order to obtain the correct sum In other word we need an add with carry instruc tion There are two versions of this instruction the ADCA instruction for accumulator A and
52. ruction The result is shown in Figure 2 13b Original value New value 800 11100111 800 01110011 C 1 C 1 Figure 2 13b m Execution result of LSR 800 A Example 2 21 vy Compute the new values of accumulator B and the C flag after executing the instruction ROLB Assume the original value of B is BD and C flag is 1 Solution The operation of this instruction is illustrated in Figure 2 14a Figure 2 14a m Operation of the instruction ROLB The result is shown in Figure 2 14b Original value New value B 10111101 B 01111011 Ce C 1 Figure 2 14b m Execution result of ROLB 66 Chapter 2 m 68HC12 Assembly Programming Example 2 22 v What are the values of accumulator A and the C flag after executing the instruction RORA Assume the original value of A is BE and C 1 Solution The operation of this instruction is illustrated in Figure 2 15a C flag Figure 2 15a m Operation of the instruction RORA The result is shown in Figure 2 15b Original value New value A 10111110 A 11011111 C 1 Cap Figure 2 15b m Execution result of RORA Example 2 23 v 1 1 1 1 ke accumulator A gt Write a program to count the number of zeros contained in memory locations 800 801 and save the result at memory location 805 Solution The logical shift right instruction is available for double accumulator D We can load this 16 bit value into D a
53. s 33 org 800 array db 11 22 33 44 55 initializes the contents of memory locations at 800 801 802 803 and 804 to 11 22 33 44 and 55 respectively dw define word dc w define constant word fdb form double bytes These three directives define the value of a word or words that will be placed at a given address The value can be specified by an integer or an expression For example the statement vect_tab dw 1234 5678 initializes the two words starting from the current location counter to 1234 and 5678 respec tively After this statement the location will be incremented by four fcc form constant character This directive allows us to define a string of characters a message The first character in the string is used as the delimiter The last character must be the same as the first character because it will be used as the delimiter The delimiter must not appear in the string The space character cannot be used as the delimiter Each character is encoded by its correspond ing ASCII code For example the statement alpha fcc def will generate the following values in memory 64 65 66 and the assembler will use the label alpha to refer to the address of the first letter which is stored as the byte 64 A character string to be output to the LCD display is often defined using this directive fill fill memory This directive allows a user to fill a certain number of memory locations with
54. s an 8 bit mask that specifies the bits of the memory location to be checked The bits to be checked correspond to those bit positions that are ones in the mask rel is the branch offset and is specified in 8 bit relative mode This instruction tells the 68HC12 to perform the logical AND of the contents of the spec ified memory location inverted and the mask supplied with the instruction then branch if the result is zero this occurs only when all bits corresponding to ones in the mask byte are ones in the tested byte For example for the following instruction sequence loop inc count brset 66 e0 loop the branch will be taken if the most significant three bits of the memory location at 66 are all ones Write a program to count the number of elements that are divisible by 4 in an array of N 8 bit numbers Use the repeat S until C looping construct Solution The lowest two bits of a number divisible by four are 00 By checking the lowest two bits of a number we can determine if a number is divisible by four The program is as follows N equ 20 org 800 total rmb 1 org 1000 starting address of the program Idaa 0 staa total initialize total to 0 62 Chapter 2 m 68HC12 Assembly Programming Idx array use index register X as the array pointer Idab N use accumulator B as the loop count loop brelr 0 x 03 yes bra chkend yes inc total add 1 to the total chkend _ inx move the array pointer dbne b loop forever bra fo
55. s or directives or a whole rou tine Adding comments makes a program more readable Each line of a 68HC12 assembly program excluding certain special constructs is com prised of four distinct fields Some of the fields may be empty The order of these fields is 1 Label 2 Operation 3 Operand 4 Comment 2 2 1 The Label Field Labels are symbols defined by the user to identify memory locations in the programs and data areas of the assembly module For most instructions and assembler directives the label is optional The rules for forming a label are as follows A label must start at column one and begin with a letter A Z a z and the letter can be followed by letters digits or special symbols Some assemblers permit spe cial symbols to be used For example the assembler from IAR Inc allows a symbol to start with a question mark at character and underscore _ in addition to letters Digits and the dollar character can also be used after the first character in the IAR assembler Most assemblers restrict the number of characters in a label name The as12 assembler reference manual does not mention the limit The IAR assembler allows a user defined symbol to have up to 255 characters The asl2 assembler allows a label to be terminated by Example 2 1 v Valid and Invalid labels The following instructions contain valid labels begin Idaa 10 label begins in column 1 2 2 m Assembly Language
56. the ADCB instruction for accumulator B The instructions for adding the second to most sig nificant bytes are Idaa 59 adca 54 We also need to save the second to most significant byte of the result at 801 with the fol lowing instruction staa 801 The most significant bytes can be added using similar instructions and the complete pro gram with comments appears as follows Idd 8183 place the lowest two bytes of the first number in D addd 8290 add the lowest two bytes of the second number to D std 802 store the lowest two bytes of the sum at 802 803 Idaa 59 place the second to most significant byte of the first number in A adca 54 add the second to most significant byte of the second number and carry to A staa 801 store the second to most significant byte of the sum at 801 Idaa 1A place the most significant byte of the first number in A adca 76 add the most significant byte of the second number and carry to A staa 800 store the most significant byte of the sum end Example 2 7 v 2 5 m Writing Programs to do Arithmetic 41 Note that the LOAD and STORE instructions do not affect the value of the C flag other wise the program would not work The 68HC12 does not have a 16 bit instruction with the carry flag as an operand Whenever the carry needs to be added we must use the 8 bit instruc tion ADCA or ADCB This is shown in the previous program Write a program to add two 4 byte numbers
57. tion sequence Idaa 56 oraa 01 staa 56 sets the bit 0 of the I O port at 56 The instructions sequence Idaa 56 eora 0F staa 56 toggles the lower four bits of the I O port at 56 The instructions COMA and COMB that per form one s complementing can be used if all of the port pins need to be toggled 2 9 Bit Test amp Manipulate Instruction These instructions use a mask value to test or change the value of individual bits in an accumulator or in a memory location BITA and BITB provide a convenient means of testing bits without altering the value of either operand Table 2 9 shows a summary of bit test and manip ulation instructions Mnemonic Function Operation BCLR lt opr gt msk8 Clear bits in memory M amp M mm BITA lt opr gt t Bit test A A M BITB lt opr gt Bit test B B M BSET lt opr gt msk8 Set bits in memory M lt M mm Note 1 lt opr gt can be specified using all except relative addressing modes for BITA and BITB 2 lt opr gt can be specified using direct extended and indexed exclude indiriect addressing modes 3 msk8 is an 8 bit value Table 2 9 m Summary of Boolean logic instructions For example the instruction belr 0 x 81 clears the most significant and least significant bits of the memory location pointed to by index register X 70 Chapter 2 m 68HC12 Assembly Programming The instruction bita 44 tests the bit six
58. umulator A and the C flag after executing the ASLA instruction Assume that originally A contains 95 and the C flag is 1 Solution The operation of this instruction is shown in Figure 2 11a accumulator A 1 Jo fo 1 o ifofi 0 C flag Figure 2 114a m Operation of the ASLA instruction The result is shown in Figure 2 11b 10010101 A 00101010 C A C Figure 2 11b m Execution result of the ASLA instruction Example 2 19 vy What are the new values of the memory location at 800 and the C flag after executing the instruction ASR 800 Assume that the memory location 800 originally contains the value of ED and the C flag is 0 Solution The operation of this instruction is shown in Figure 2 12a 1fifilfo Haaa AA A 1 TAE 1 Figure 2 12a m a of the ASR 800 instruction The result is shown in Figure 2 12b 800 11101101 800 11110110 C 0 C 1 Figure 2 12b m Result of the ASR 800 instruction 2 7 m Shift amp Rotate Instuctions 65 Example 2 20 vy What are the new values of the memory location at 800 and the C flag after executing the instruction LSR 800 Assume the memory location 800 originally contains E7 and the C flag is 1 Solution The operation of this instruction is illustrated in Figure 2 13a 1 1 1 0 0O 1 171 memory location C flag 800 Figure 2 13a m Operation of the LSR 800 inst
59. will still be assembled correctly org origin The assembler uses a Jocation counter to keep track of the memory location where the next machine code byte should be placed If the programmer wants to force the program or data array to start from a certain memory location then this directive can be used For example the statement org 1000 forces the location counter to be set to 1000 The org directive is mainly used to force a data table or a segment of instructions to start with a certain address As a general rule this directive should be used as infrequently as possi ble Using too many orgs will make your program less reusable db define byte dc b define constant byte fcb form constant byte These three directives define the value of a byte or bytes that will be placed at a given mem ory location The db or fcb or dc b directive assigns the value of the expression to the mem ory location pointed to by the location counter Then the location counter is incremented Multiple bytes can be defined at a time by using commas to separate the arguments For example the statement array db 11 22 33 44 55 initializes five bytes in memory to 11 22 33 44 55 and the assembler will use array as the symbolic address of the first byte whose initial value is 11 The program can also force these five bytes to a particular address by adding the org direc tive For example the sequence 2 3 m Assembler Directive

HUANG-Chap02 - HC12 Assembler

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