This application note presents programming techniques for performing commonly found arithmetic operations, such as multi-byte binary addition and subtraction, multi-digit BCD addition and subtraction, multiplication and division.
The default configuration of SX is to ignore the carry flag in additions and subtractions even the results of those operations do affect that flag. For multi-byte arithmetic operations, it is often desirable to the result of lower bytes to propagate to higher bytes by the means of the carry flag.
To enable the effect of the carry flag, carryx must be included in the list of device directives which are specified before the instructions, to make the carry flag an input to add, sub instruction.
The carry flag should be set to zero first before any addition
The SX SUB instruction will set the carry flag to zero if there is an underflow. Therefore, it is necessary for us to set it to one before any subtraction is performed.
The following program segment illustrates 32 bit binary addition. The 4 byte operand1 and the 4 byte operand2 are added together. The result is put back into operand2.
Note that operand1 is located at locations 8,9,a,b, hence 10xx binary and operand2 is at locations c,d,e,f or 11xx binary. Therefore toggling bit 2 of the FSR register effectively enable us to switch back and forth among the two operands. With that in mind, the indirect addressing of SX help us a lot in saving code by just using IND as the register pointed to by FSR.
This routine assumes that the two operands are adjacent to one another and operand1 starts at the 08 location. To relocate the operands to other locations, make sure that they are still adjacent to one another, thus occupying a contiguous 8 bytes, and that operand1 is aligned to x0 or x8. The only change needed in the code will be the ending condition. Note that in the example, we tested bit 4 which will be toggled after the inc fsr instruction if fsr was $f, and therefore pointing to the last byte. To make the routine work with operands located in $10-$17, for example, would need the ending condition be changed from sb fsr.4 to sb fsr.3 since the inc fsr instruction will change the address of last byte from $17 to $18 (%00011000) and set bit 3. Using this technique, we can save the need to store the count separately in order to keep track of the number of bytes added.
;32 bit addition ;entry = 32 bit operand1 and 32 bit operand2 in binary form ;exit = operand2 become operand1+operand2, carry flag=1 for overflow from MSB add32 clc ; clear carry, prepare for addition mov fsr,#operand1 ; points to operand 1 first add_more clrb fsr.2 ; toggle back to operand 1 mov w,ind ; get contents into the work register setb fsr.2 ; points to operand 2 add ind,w ; operand2=operand2+operand1 inc fsr ; next byte sb fsr.4 ; done? (fsr=$10?) jmp add_more ; not yet ret ; done, return to calling routine
The 32 bit subtraction routine is very similar to addition, except that we set the carry flag first to indicate no underflow. Note that the result is in operand2 and it is operand2-operand1, not the other way around. When the carry flag is 0 on return, it means that the result is negative and is therefore in 2’s complement form.
;32 bit subtraction ;entry = 32 bit operand1 and 32 bit operand2 in binary form ;exit = operand2 become operand2-operand1, carry flag=0 for underflow from MSB sub32 stc ; set carry, prepare for subtraction mov fsr,#operand1 ; points to operand 1 first sub_more clrb fsr.2 ; toggle back to operand 1 mov w,ind ; get contents into the work register setb fsr.2 ; points to operand 2 sub ind,w ; operand2=operand2-operand1 inc fsr ; next byte sb fsr.4 ; done? (fsr=$10?) jmp sub_more ; not yet ret ; done, return to calling routine
The next topic will be BCD addition. In a lot of application where calculation result needs to be displayed, BCD or binary coded decimal can be much more easily converted into visual form, as in the case of adding machine or calculator.
The algorithm here for BCD addition is very similar to binary addition except for 1 important difference: decimal adjustment or correction. The need for such operation will be evident as we examine the follow simple addition:
85+15 = 9A
Obviously the correction result should be 100 in BCD. We can see that by adding 6 to the least significant digit (LSD), in this case, $9A+6=$A0, will correct the LSD. Finally, by adding a $60 to the whole number (equal to adding 6 to the most significant digit, MSD), the entire number is corrected to $00 with a carry of 1, which can be propagated into the next byte.
By looking at another example: 19+19=32. After the addition, the digit carry will be set to one, indicating an overflow in the LSD. The result then can be corrected by adding 6 to the LSD, giving us the correct answer of 38.
In general, we will do a correction on LSD of the result if the digit carry is set or the LSD is greater than 9. The same is true for the MSD. It will be corrected, i.e., added with 6, when the carry bit is set or the MSD is greater than 9.
The tricky part now is how to check if the digit is greater than 9. A straight implementation will require masking 1 nibble off at a time and do a subtraction. This will require additional storage if we do not want the operands (and the result) to be changed. The way it is implemented here is a bitwise comparison.
Let us look at a 4 bit number, if bit number 3 is 0, the number must be then %0xxx, and therefore ranges from 0-7, hence less than 9. If that’s not the case, then we go on to check bit 2. If it is a one, then we have %11xx, and the number is definitely bigger than 9, since the minimum is already %1100 or 12. If bit 2 is a zero, we proceed to check bit 1. If this bit is a zero, then we have %100x, which means the number is either 8 or 9, and no correction is needed. But if bit 1 is an one, then we have %101x, which is higer than 9 and correction will be needed.
This method of detecting whether the digit is greater than 9 or not, is used twice in the code. Once for LSD and once for MSD. The changes is only the bit number that is being checked on.
One more point worth noting is the carry bit. After the initial binary addition, we have to store the carry bit that is used to propagate the result to higher bytes. The reason for doing this is simple: the decimal correction process of adding 6 to the number will clear the carry bit.
Notice also that the ending condition has been changed to sb fsr.2 instead of sb fsr.4. This is simply because the code happens to point at operand 1 at that time and it just saves us code to check if fsr is pointing to the last byte of operand 1 at location $0b (%1011) or not. The fsr will be $0c (%1100) after the increment operation and therefore setting bit 2.
;8 BCD digit addition ;entry = 8 BCD digit operand1 and 8 BCD digit operand2 in BCD form ;exit = operand2 become operand2+operand1, carry flag=1 for overflow from MSB ; operand1 will be DESTROYED badd32 clc ; clear carry, prepare for addition mov fsr,#operand1 ; points to operand 1 first badd_more mov w,ind ; get contents into the working register clr ind setb fsr.2 ; points to operand 2 add ind,w ; operand2=operand2+operand1 clrb fsr.2 rl ind ; store carry bit which will be altered by decimal ; adjustment (adding 6) setb fsr.2 ; points back to operand 2 snb status.1 ; digit carry set? if so, need decimal correction jmp dcor jnb ind.3,ck_overflow ; if 0xxx, check MSD jb ind.2,dcor ; if 11xx, it's >9, thus need correction jnb ind.1,ck_overflow ; 100x, number is 8 or 9, no decimal correction ; here if 101x, decimal adjust dcor clc ; clear effect of previous carry add ind,#6 ; decimal correction by adding 6 ; finish dealing with least significant digit, proceed to MSD ck_overflow clrb fsr.2 ; points to operand1 jb ind.0,dcor_msd ; stored carry=1, decimal correct ; test if MSD > 9 setb fsr.2 ; points back to operand2 jnb ind.7,next_badd ; if 0xxx, it's <9, add next byte jb ind.6,dcor_msd ; if 11xx, it's >9, thus need correction jnb ind.5,next_badd ; if 100x, it's <9 ;here if 101x, decimal adjust dcor_msd clc ; clear effect of carry setb fsr.2 ; make sure that it's pointing at the result add ind,#$60 ; decimal correct next_badd clrb fsr.2 ; points to stored carry snb ind.0 ; skip if not set stc ; restore stored carry inc fsr ; next byte sb fsr.2 ; done? (fsr=$0c?) jmp badd_more ; not yet ret ; done, return to calling routine
BCD subtraction is very similar to addition except for a few notes, which are summarized below:
;8 BCD digit subtraction ;entry = 8 BCD digit operand1 and 8 BCD digit operand2 in BCD form ;exit = operand2 become operand2-operand1, carry flag=0 for underflow from MSB ; carry flag=1 for positive result ; operand1 will be DESTROYED bsub32 call bs32 ; do subtraction snc ; no carry=underflow? jmp bs_done ; carry=1 positive, done call neg_result ; yes, get the magnitude, 0-result call bs32 ; keep in mind that this result is a negative ; number (carry=0) bs_done ret bs32 stc ; set carry, prepare for subtraction mov fsr,#operand1 ; points to operand 1 first bsub_more mov w,ind ; get contents into the working register clr ind setb ind.7 ; set to 1 so that carry=1 after rl instruction setb fsr.2 ; points to operand 2 sub ind,w ; operand2=operand2+operand1 clrb fsr.2 rl ind ; store carry bit which will be altered by decimal ; adjustment (adding 6) setb fsr.2 ; points back to operand 2 sb status.1 ; digit carry set? if so, need decimal correction jmp dec_cor jnb ind.3,ck_underflow ; if 0xxx, check MSD jb ind.2,dec_cor ; if 11xx, it's >9, thus need correction jnb ind.1,ck_underflow ; 100x, number is 8 or 9, no decimal correction ; here if 101x, decimal adjust dec_corstc ; clear effect of previous carry sub ind,#6 ; decimal correction by subtracting 6 ; finish dealing with least significant digit, proceed to MSD ck_underflow clrb fsr.2 ; points to operand1 jnb ind.0,dadj_msd ; stored carry=0, decimal adjust ; test if MSD > 9 setb fsr.2 ; points back to operand2 jnb ind.7,next_bsub ; if 0xxx, it's <9, add next byte jb ind.6,dadj_msd ; if 11xx, it's >9, thus need correction jnb ind.5,next_bsub ; if 100x, it's <9 ;here if 101x, decimal adjust dadj_msd stc ; clear effect of carry setb fsr.2 ; make sure that it's pointing at the result sub ind,#$60 ; decimal correct next_bsub clrb fsr.2 ; points to stored carry sb ind.0 ; skip if not set clc ; restore stored carry inc fsr ; next byte sb fsr.2 ; done? (fsr=$0c?) jmp bsub_more ; not yet ret ; done, return to calling routine ; move the result to operand1 and change operand2 to 0 ; the intention is prepare for 0-result or getting the magnitude of a ; negative BCD number which is in complement form neg_result mov fsr,#operand2 ; points to mov_more ; operand2 setb fsr.2 mov w,ind ; temp. storage clr ind ; clear operand2 clrb fsr.2 ; points to operand1 mov ind,w ; store result inc fsr ; next byte sb fsr.2 ; done? jmp mov_more ; no ret ; yes, finish
In a lot of situations, we will find BCD representations very difficult to deal with, especially when anything more than addition and subtraction is needed, due to the need for decimal correction. This problem is alleviated by representing the numbers internally as binary to facilitate computation and convert it to BCD for display or printing purposes. In this section, we will discuss how that is implemented.
There are many different algorithms for binary to BCD conversions. We will only consider one of the easiest to implement, that is, shifting the binary number to the left and let the most significant bit be shifted into a BCD result. The result is then continuously decimally corrected to give a right answer.
In the following code segment, we have implemented a 32 bit binary number to 10 digit BCD conversion routine. With the RL instruction of the SX, the shift operation of both numbers together is a breeze.
Decimal correction is done here differently than before. Instead of checking the carry and digit carry, we check the BCD value before a shift and adjust it properly. This will save us both code and time. This was not possible before in our addition and subtraction routines since we were not doing shift operations.
To see how this is done, let’s look at some examples:
| Current value |
binary | Shifted value in binary |
Shifted value in hex |
What the shifted value should be in BCD |
| 0 | 0000 | 0000 | 0 | 0 |
| 1 | 0001 | 0010 | 2 | 2 |
| 2 | 0010 | 0100 | 4 | 4 |
| 3 | 0011 | 0110 | 6 | 6 |
| 4 | 0100 | 1000 | 8 | 8 |
| 5 | 0101 | 1010 | A | 10 |
| 6 | 0110 | 1100 | C | 12 |
| 7 | 0111 | 1110 | E | 14 |
| 8 | 1000 | 1 0000 | 10 | 16 |
| 9 | 1001 | 1 0010 | 12 | 18 |
From the table, we can see that whenever the current value is 4 or less, then it is okay. For all digits of 5 and above, decimal correction is needed. This can be done by adding 6 to the shifted value or by adding 3 to the current value. If we add 3 to all current values and check if they are greater than 7, all number satisfying this condition will need decimal correction and we will just keep that added number, otherwise we fall back to the original number.
This decimal correction process applies also to the most significant digit, except we use $30 instead of 3.
; 32 bit binary to BCD conversion ; entry: 32 bit binary number in $10-13 ; exit: 10 digit BCD number in $14-18 ; algorithm= shift the bits of binary number into the BCD number and ; decimal correct on the way bindec mov count,#32 mov fsr,#bcd_number ; points to the BCD result clr_bcd clr ind ; clear BCD number snb fsr.3 ; reached $18? jmp shift_both ; yes, begin algorithm inc fsr ; no, continue on next byte jmp clr_bcd ; loop to clear shift_both mov fsr,#bin_number ; points to the binary number input clc ; clear carry, prepare for shifting shift_loop rl ind ; shift the number left snb fsr.3 ; reached $18? (finish shifting both ; numbers) jmp check_adj ; yes, check if end of everything inc fsr ; no, next byte jmp shift_loop ; not yet check_adj decsz count ; end of 32 bit operation? jmp bcd_adj ; no, do bcd adj ret bcd_adj mov fsr,#bcd_number ; points to first byte of the BCD result bcd_adj_loop call digit_adj ; decimal adjust snb fsr.3 ; reached last byte? jmp shift_both ; yes, go to shift both number left again inc fsr ; no, next byte jmp bcd_adj_loop ; looping for decimal adjust digit_adj ; consider LSD first mov w,#3 ; 3 will become 6 on next shift add w,ind ; which is the decimal correct factor to be added mov temp,w snb temp.3 ; > 7? if bit 3 not set, then must be <=7, no adj. mov ind,w ; yes, decimal adjust needed, so store it ; now for the MSD mov w,#$30 ; 3 for MSD is $30 add w,ind ; add for testing mov temp,w snb temp.7 ; > 7? mov ind,w ; yes, store it ret
Input from keyboards can be easily rendered into BCD form. To let the CPU process the number effectively, however, binary representation is more desirable.
In this section we will discuss how the BCD to binary conversion process is implemented. It is basically a reversal of the binary to BCD conversion process: we shift the BCD number to the right and let the least significant bit be shifted into a binary result. The original BCD number is then continuously decimally corrected to maintain the BCD format.
In the following code segment, we have implemented a 10 digit BCD number to 32 bit binary number conversion routine. With the RR instruction of the SX, the shift operation of both numbers together can be very efficiently implemented.
Decimal correction is done again differently here since we are shifting right instead of shifting left.
To derive the algorithm, let’s look at the following table:
| Current value |
binary | Shifted value in binary |
Shifted value in hex |
What the shifted value should be in BCD |
|
| 0 | 0000 | 0000 | 0 | 0 | |
| 2 | 0010 | 0001 | 1 | 1 | |
| 4 | 0100 | 0010 | 2 | 2 | |
| 6 | 0110 | 0011 | 3 | 3 | |
| 8 | 1000 | 0100 | 4 | 4 | |
| 10 | 10000 | 1000 | 8 | 5 | |
| 12 | 10010 | 1001 | 9 | 6 | |
| 14 | 10100 | 1010 | A | 7 | |
| 16 | 10110 | 1011 | B | 8 | |
| 18 | 11000 | 1100 | C | 9 | |
As we can see, whenever the shifted value has a 1 on bit 3, the result should be subtracted with 3 to make it correct. And this is the algorithm that we have adopted in the following code: shift right both numbers and decimally adjust the BCD number along the way. Note that for the most significant digit in each BCD number, we subtract $30 instead of 3 to account for its position.
; 10 digit BCD to 32 bit binary conversion ; entry: 10 digit BCD number in $14-18 ; exit: 32 bit binary number in $10-13 ; algorithm= shift the bits of BCD number into the binary number and decimal ; correct on the way decbin mov count,#32 ; 32 bit number mov fsr,#bin_number ; points to the binary result clr_bin clr ind ; clear binary number inc fsr ; no, continue on next byte snb fsr.2 ; reached $13? (then fsr will be $14 here) jmp shift_b ; yes, begin algorithm jmp clr_bin ; loop to clear shift_b mov fsr,#bcd_number+4 ; points to the last BCD number clc ; clear carry, prepare for shifting right shft_loop rr ind ; shift the number right dec fsr ; reached $10? (finish shifting both numbers) sb fsr.4 ; then fsr will be $0f jmp chk_adj ; yes, check if end of everything jmp shft_loop ; not yet chk_adj decsz count ; end of 32 bit operation? jmp bd_adj ; no, do bcd adj ret bd_adj mov fsr,#bcd_number ; points to first byte of the BCD result bd_adj_loop call dgt_adj ; decimal adjust snb fsr.3 ; reached last byte? jmp shift_b ; yes, go to shift both number right again inc fsr ; no, next byte jmp bd_adj_loop ; looping for decimal adjust ; prepare for next shift right ; 0000 --> 00000 -->0 ; 0010 --> 0001 2 -->1 ; 0100 --> 0010 4 -->2 ; 0110 --> 0011 6 -->3 ; 1000 --> 0100 8 -->4 ; 1 0000 --> 1000 10-->8 !! it should be 5, so -3 ; 1 0010 --> 1001 12-->9 !! it should be 6, so -3 ; in general when the highest bit in a nibble is 1, it should be subtracted with 3 dgt_adj ; consider LSD first sb ind.3 ; check highest bit in LSD, =1? jmp ck_msd ; no, check MSD stc ; prepare for subtraction, no borrow sub ind,#3 ; yes, adjust ck_msd ; now for the MSD sb ind.7 ; highest bit in MSD, =1? ret ; no ; yes, do correction stc ; no borrow sub ind,#$30 ; this is a 2 word instruction, and cannot be skipped ret
The need for multiplication permeates through the use of microcontrollers. Here we will consider both 8 bit by 8 bit and 16 bit by 16 bit multiplications. As we can see, the basic algorithms are all the same regardless of the number of bits involved.
Let’s first discuss how the multiplier, multiplicand, and the result are generally organized.
| Multiplicand | |
| Upper product | Multiplier (lower product) |
The lower part of the result are initially occupied by the multiplier and the upper part is cleared to zero.
To summarize, the following steps are needed to do a multiplication by software:
This algorithm is amazingly elegant as we can see in the next program segment.
As implemented for 8 bit by 8 bit multiplication, this routine requires only 2 bytes of RAM provided the multiplicand is pre-loaded into the W, working register.
; 8 bit x 8 bit multiplication (RAM efficient, 2 bytes only) ; entry: multiplicand in W, multiplier at 09 ; exit : product at $0a,09 ; cycles mul88 mov upper_prdt,w ; 1 store W mov count,#9 ; 2 set number of times to shift mov w,upper_prdt ; 1 restore W (multiplicand) clr upper_prdt ; 1 clear upper product clc ; 1 clear carry ; the following are executed [count] times m88loop rr upper_prdt ; 1 rotate right the whole product rr multiplier ; 1 check lsb snc ; 1 skip addition if no carry add upper_prdt,w ; 1 add multiplicand to upper product no_add decsz count ; 1/2 loop 9 times to get proper product jmp m88loop ; 3 jmp to rotate the next half of product ret ; 3 done... ; one time instructions = 1+2+1+1+1+3= 9 cycles ; repetitive ones= (1+1+1+1+1+3)9-3+2=71 ; total worst case cycles=80 cycles
A faster implementation can be obtained if we unroll the loop and repeat the code using a macro:
; fast 8 bit x 8 bit multiplication (RAM efficient, 2 bytes only) ; entry: multiplicand in W, multiplier at 09 ; exit : product at $0a,09 ; macro to rotate product right and add rra MACRO rr upper_prdt ; 1 rotate right the whole product rr multiplier ; 1 check lsb snc ; 1 skip addition if no carry add upper_prdt,w ; 1 add multiplicand to upper product ENDM ; cycles fmul88 clr upper_prdt ; 1 clear upper product clc ; 1 clear carry ; the following are executed [count] times rra ; call the macro 9 times rra rra rra rra rra rra rra rra ret ; 3 done... ; one time instructions = 1+1+3= 5 cycles ; repetitive ones= (1+1+1+1)9=36 ; total worst case cycles=41 cycles
We have saved almost half of the time by using macros and eliminating the loop control. Notice that in both algorithms, 9 shifts are needed to obtain a correct result. The last shift is used to align the result properly.
The same algorithm has been implemented for 16 bit by 16 bit multiplication, which is included as follows:
; 16 bit x 16 bit multiplication ; entry: multiplicand in $09,08, multiplier at $0b,$0a ; exit : 32 bit product at $0d,$0c,$b,$a ; cycles mul1616 mov count,#17 ; 2 set number of times to shift clr upper_prdt ; 1 clear upper product clr upper_prdt+1 ; 1 higher byte of the 16 bit upeper product clc ; 1 clear carry ; the following are executed [count] times m1616loop rr upper_prdt+1 ; 1 rotate right the whole product rr upper_prdt ; 1 lower byte of the 16 bit upper product rr mr16+1 ; 1 high byte of the multiplier rr mr16 ; 1 check lsb sc ; 1 skip addition if no carry jmp no_add ; 3 no addition since lsb=0 clc ; 1 clear carry add upper_prdt,md16 ; 1 add multiplicand to upper product add upper_prdt+1,md16+1 ; 1 add the next 16 bit of multiplicand no_add decsz count ; 1/2 loop [count] times to get proper product jmp m1616loop ; 3 jmp to rotate the next half of product ret ; 3 done... ; one time instructions = 8 cycles ; repetitive ones= 15*16+11+2=253 ; total worst case cycles=261 cycles
Note that the only difference is the number of bits that we shift, and more bytes to add and rotate. Other than that, it is basically the same as a 8 x 8 multiplication. A fast version is also available but it is too lengthy to list here. Please see the program file for details. A saving of 26% is achieved here by unrolling the loop and reduced the cycles to 193.
Finally, we are going to tackle the most difficult arithmetic problem: that of divison. If the reader can recall how he or she was taught how to do division by long hand, then we are very close to understanding the algorithm.
In division by long hand, we examine the dividend digit by digit, and see if it is bigger than the divisor. If it is, then we subtract the divisor or the multiples of it from the dividend and write down that multiple as a digit in our quotient. This process is repeated until all digits of the dividend are exhausted.
This exact process is being implemented in the following code segment with one difference with our long hand division: we are dealing with binary numbers here. So we modify the algorithm as follows:
; 16 bit by 16 bit division (b/a) ; entry: 16 bit b, 16 bit a ; exit : result in b, remainder in remainder ; cycles div1616 mov count,#16 ; 2 no. of time to shift mov d,b ; 2 move b to make space mov d+1,b+1 ; 2 for result clr b ; 1 clear the result fields clr b+1 ; 1 one more byte clr rlo ; 1 clear remainder low byte clr rhi ; 1 clear remainder high byte ; subtotal=10 divloop clc ; 1 clear carry before shift rl d ; 1 check the dividend rl d+1 ; 1 bit by bit rl rlo ; 1 put it in the remainder for rl rhi ; 1 trial subtraction ; subtotal=5 stc ; 1 prepare for subtraction, no borrow mov w,a+1 ; 1 do trial subtraction mov w,rhi-w ; 1 from MSB first sz ; 1/2 if two MSB equal, need to check LSB jmp chk_carry ; 3 not equal, check which one is bigger ; ; if we are here, then z=1, so c must be 1 too, since there is no ; underflow, so we save a stc instruction mov w,a ; 1 equal MSB, check LSB mov w,rlo-w ; 1 which one is bigger? ; subtotal=7 chk_carry sc ; 1/2 partial dividend >a? jmp shft_quot ; 3 no, partial dividend < a, set a 0 into quotient ; if we are here, then c must be 1, again, we save another stc instruction ; yes, part. dividend > a, subtract a from it sub rlo,a ; 2 store part. dividend-a into a sub rhi,a+1 ; 2 2 bytes stc ; 1 shift a 1 into quotient ; subtotal=7 worst case shft_quot rl b ; 1 store into result rl b+1 ; 1 16 bit result, thus 2 rotates decsz count ; 1/2 jmp divloop ; 3 ; subtotal=6, 4 on last count ret ; 3 ; one time instructions=13 ; repetitive ones=(19+6)*15+19+4=398 ; total=411
The fast version of this division algorithm is implemented by unrolling the loop and repeat all the instructions inside it. It consumes 336 cycles and therefore saves 18% of time.
The SX instructions, namely, ADD (add), ADDB (add bit), SUB (subtract), SUBB (subtract bit), CLC (clear carry), STC (set carry), RL (rotate left 1 bit), RR (rotate right 1 bit), are very useful in implementing arithmetic routines. With careful planning and smart algorithm design, all normal arithmetic functions can be accomplished.
Modifications and further options
There exists a lot of literature on computer arithmetic and the implementations included in this application note is not the only way of doing it. It only serves as an example for the readers and help them to bring their product to the market faster by using existing routines.
To test the example programs, remember to set the equate options mentioned in the first sentence of the program listing properly (for example, to use BCD routines, set bcd_test equ 1 and reset all other options to 0). This will enable you to include only the code you need in a program.
; to test different modules, set the corresponding test to 1 and reset all others to 0 addsub_test equ 0 bcd_test equ 0 mul88_test equ 0 mul1616_test equ 0 div1616_test equ 1 ; mathpak for SX ; IF mul88_test=1 device pins28,pages4,banks8,turbo,oschs,optionx,stackx ELSE device pins28,pages4,banks8,turbo,oschs,optionx,carryx,stackx ENDIF ; use oschs only for debugging ; frequency does not matter much here except when incorporating these routines ; into a real program ;32 bit addition org 8 IF addsub_test=1 OR bcd_test=1 ;8,9,a,b=1000,1001,1010,1011=10xx operand1 ds 4 ;c,d,e,f=1100,1101,1110,1111=11xx operand2 ds 4 bin_number ds 4 bcd_number ds 5 count equ operand1 ; share storage in other routines temp equ operand1+1 ; share 1 more ENDIF IF mul88_test=1 ; for 8 bit x 8 bit multiplication count ds 1 multiplier ds 1 upper_prdt ds 1 ENDIF IF mul1616_test=1 ; for 16 bit x 16 bit multiplication md16 ds 2 mr16 ds 2 upper_prdt ds 2 count ds 1 ENDIF IF div1616_test=1 ; for 16 bit / 16 bit division a ds 2 b ds 2 rlo ds 1 rhi ds 1 d ds 2 count ds 1 ENDIF load32 MACRO 8 ; macro used to load the operands mov operand1,#\1 mov operand1+1,#\2 mov operand1+2,#\3 mov operand1+3,#\4 mov operand2,#\5 mov operand2+1,#\6 mov operand2+2,#\7 mov operand2+3,#\8 ENDM reset start ; goto 'start' on reset org 0 start ;test procedures IF addsub_test=1 ; $7fffffff+$01ffffff load32 $ff,$ff,$ff,$01,$ff,$ff,$ff,$7f call add32 ;result=$81fffffe ; $8ffffffe-$01ffffff call sub32 ;result=$7fffffff, carry=1-> result is positive ; $01ffffff-$7fffffff load32 $ff,$ff,$ff,$7f,$ff,$ff,$ff,$01 call sub32 ;result=$82000000, carry=0-> result is negative and in ;2's complement form, =-$7e000000 ENDIF IF bcd_test=1 load32 $78,$56,$89,$67,$78,$56,$89,$67 call badd32 ;result=35791356 carry=1 (overflow) load32 $12,$34,$56,$78,$78,$56,$34,$21 call badd32 ;result=99909090 ; 87654321-90123456= -2469135 load32 $56,$34,$12,$90,$21,$43,$65,$87 call bsub32 ; result= 2469135, no carry means result is negative ; 90123456-87654321 load32 $21,$43,$65,$87,$56,$34,$12,$90 call bsub32 ;result=02469135 carry=1 (no borrow), result is positive mov bin_number,#$ff ; largest 32 bit number mov bin_number+1,#$ff ; $ffffffff mov bin_number+2,#$ff ; let's see how big the number will be mov bin_number+3,#$ff ; in decimal call bindec ; result = 4,294,967,295 call decbin ; result = $ffffffff ENDIF IF mul88_test=1 mov multiplier,#$ff ; largest number mov W,#$ff ; largest 8 bit no. as multiplicand call mul88 ; result=$fe01 ; test fast 8 x 8 multiplication mov multiplier,#$ff ; largest number mov W,#$ff ; largest 8 bit no. as multiplicand call fmul88 ; result=$fe01 ENDIF IF mul1616_test=1 mov md16,#$ff ; largest 16 bit number mov md16+1,#$ff mov mr16,#$ff ; multiplied with largest 16 bit number mov mr16+1,#$ff call mul1616 ; result = fffe0001 mov md16,#$ff ; largest 16 bit number mov md16+1,#$01 mov mr16,#$ff ; multiplied with largest 16 bit number mov mr16+1,#$7f call mul1616 ; result = 00ff7e01 mov md16,#$ff ; largest 16 bit number mov md16+1,#$ff mov mr16,#$ff ; multiplied with largest 16 bit number mov mr16+1,#$ff call fmul1616 ; result = fffe0001 mov md16,#$ff ; largest 16 bit number mov md16+1,#$01 mov mr16,#$ff ; multiplied with largest 16 bit number mov mr16+1,#$7f call fmul1616 ; result = 00ff7e01 ENDIF IF div1616_test=1 mov a+1,#$01 mov a,#$ff ; 01ff mov b+1,#$7f mov b,#$ff ; 7fff call div1616 ; 7fff/01ff=$40 remainder $3f mov a+1,#$00 mov a,#$ff ; 00ff mov b+1,#$ff mov b,#$ff ; ffff call div1616 ; ffff/00ff=$101 remainder 0 mov a+1,#$01 mov a,#$ff ; 01ff mov b+1,#$7f mov b,#$ff ; 7fff call fdiv1616 ; 7fff/01ff=$40 remainder $3f mov a+1,#$00 mov a,#$ff ; 00ff mov b+1,#$ff mov b,#$ff ; ffff call fdiv1616 ; ffff/00ff=$101 remainder 0 ENDIF loop jmp loop IF addsub_test=1 ;32 bit addition ;entry = 32 bit operand1 and 32 bit operand2 in binary form ;exit = operand2 become operand2+operand1, carry flag=1 for overflow from MSB add32 clc ; clear carry, prepare for addition mov fsr,#operand1 ; points to operand 1 first add_more clrb fsr.2 ; toggle back to operand 1 mov w,ind ; get contents into the work register setb fsr.2 ; points to operand 2 add ind,w ; operand2=operand2+operand1 inc fsr ; next byte sb fsr.4 ; done? (fsr=$10?) jmp add_more ; not yet ret ; done, return to calling routine ;32 bit subtraction ;entry = 32 bit operand1 and 32 bit operand2 in binary form ;exit = operand2 become operand2-operand1, carry flag=0 for underflow from MSB sub32 stc ; set carry, prepare for subtraction mov fsr,#operand1 ; points to operand 1 first sub_more clrb fsr.2 ; toggle back to operand 1 mov w,ind ; get contents into the work register setb fsr.2 ; points to operand 2 sub ind,w ; operand2=operand2-operand1 inc fsr ; next byte sb fsr.4 ; done? (fsr=$10?) jmp sub_more ; not yet ret ; done, return to calling routine ENDIF IF bcd_test=1 ;8 BCD digit addition ;entry = 8 BCD digit operand1 and 8 BCD digit operand2 in BCD form ;exit = operand2 become operand2+operand1, carry flag=1 for overflow from MSB ; operand1 will be DESTROYED badd32 clc ; clear carry, prepare for addition mov fsr,#operand1 ; points to operand 1 first badd_more mov w,ind ; get contents into the working register clr ind setb fsr.2 ; points to operand 2 add ind,w ; operand2=operand2+operand1 clrb fsr.2 rl ind ; store carry bit which will be altered by decimal ; adjustment (adding 6) setb fsr.2 ; points back to operand 2 snb status.1 ; digit carry set? if so, need decimal correction jmp dcor jnb ind.3,ck_overflow ; if 0xxx, check MSD jb ind.2,dcor ; if 11xx, it's >9, thus need correction jnb ind.1,ck_overflow ; 100x, number is 8 or 9, no decimal correction ; here if 101x, decimal adjust dcor clc ; clear effect of previous carry add ind,#6 ; decimal correction by adding 6 ; finish dealing with least significant digit, proceed to MSD ck_overflow clrb fsr.2 ; points to operand1 jb ind.0,dcor_msd ; stored carry=1, decimal correct ; test if MSD > 9 setb fsr.2 ; points back to operand2 jnb ind.7,next_badd ; if 0xxx, it's <9, add next byte jb ind.6,dcor_msd ; if 11xx, it's >9, thus need correction jnb ind.5,next_badd ; if 100x, it's <9 ;here if 101x, decimal adjust dcor_msd clc ; clear effect of carry setb fsr.2 ; make sure that it's pointing at the result add ind,#$60 ; decimal correct next_badd clrb fsr.2 ; points to stored carry snb ind.0 ; skip if not set stc ; restore stored carry inc fsr ; next byte sb fsr.2 ; done? (fsr=$0c?) jmp badd_more ; not yet ret ; done, return to calling routine ;8 BCD digit subtraction ;entry = 8 BCD digit operand1 and 8 BCD digit operand2 in BCD form ;exit = operand2 become operand2-operand1, carry flag=0 for underflow from MSB ; carry flag=1 for positive result ; operand1 will be DESTROYED bsub32 call bs32 ; do subtraction snc ; no carry=underflow? jmp bs_done ; carry=1 positive, done call neg_result ; yes, get the magnitude, 0-result call bs32 ; keep in mind that this result is a negative number (carry=0) bs_done ret bs32 stc ; set carry, prepare for subtraction mov fsr,#operand1 ; points to operand 1 first bsub_more mov w,ind ; get contents into the working register clr ind setb ind.7 ; set to 1 so that carry=1 after rl instruction setb fsr.2 ; points to operand 2 sub ind,w ; operand2=operand2+operand1 clrb fsr.2 rl ind ; store carry bit which will be altered by decimal ; adjustment (adding 6) setb fsr.2 ; points back to operand 2 sb status.1 ; digit carry set? if so, need decimal correction jmp dec_cor jnb ind.3,ck_underflow ; if 0xxx, check MSD jb ind.2,dec_cor ; if 11xx, it's >9, thus need correction jnb ind.1,ck_underflow ; 100x, number is 8 or 9, no decimal correction ; here if 101x, decimal adjust dec_cor stc ; clear effect of previous carry sub ind,#6 ; decimal correction by subtracting 6 ; finish dealing with least significant digit, proceed to MSD ck_underflow clrb fsr.2 ; points to operand1 jnb ind.0,dadj_msd ; stored carry=0, decimal adjust ; test if MSD > 9 setb fsr.2 ; points back to operand2 jnb ind.7,next_bsub ; if 0xxx, it's <9, add next byte jb ind.6,dadj_msd ; if 11xx, it's >9, thus need correction jnb ind.5,next_bsub ; if 100x, it's <9 ;here if 101x, decimal adjust dadj_msd stc ; clear effect of carry setb fsr.2 ; make sure that it's pointing at the result sub ind,#$60 ; decimal correct next_bsub clrb fsr.2 ; points to stored carry sb ind.0 ; skip if not set clc ; restore stored carry inc fsr ; next byte sb fsr.2 ; done? (fsr=$0c?) jmp bsub_more ; not yet ret ; done, return to calling routine ; move the result to operand1 and change operand2 to 0 ; the intention is prepare for 0-result or getting the magnitude of a ; negative BCD number which is in complement form neg_result mov fsr,#operand2 ; points to mov_more setb fsr.2 ; operand2 mov w,ind ; temp. storage clr ind ; clear operand2 clrb fsr.2 ; points to operand1 mov ind,w ; store result inc fsr ; next byte sb fsr.2 ; done? jmp mov_more ; no ret ; yes, finish ; 32 bit binary to BCD conversion ; entry: 32 bit binary number in $10-13 ; exit: 10 digit BCD number in $14-18 ; algorithm= shift the bits of binary number into the BCD number and decimal ; correct on the way bindec mov count,#32 mov fsr,#bcd_number ; points to the BCD result clr_bcd clr ind ; clear BCD number snb fsr.3 ; reached $18? jmp shift_both ; yes, begin algorithm inc fsr ; no, continue on next byte jmp clr_bcd ; loop to clear shift_both mov fsr,#bin_number ; points to the binary number input clc ; clear carry, prepare for shifting shift_loop rl ind ; shift the number left snb fsr.3 ; reached $18? (finish shifting both numbers) jmp check_adj ; yes, check if end of everything inc fsr ; no, next byte jmp shift_loop ; not yet check_adj decsz count ; end of 32 bit operation? jmp bcd_adj ; no, do bcd adj ret bcd_adj mov fsr,#bcd_number ; points to first byte of the BCD result bcd_adj_loop call digit_adj ; decimal adjust snb fsr.3 ; reached last byte? jmp shift_both ; yes, go to shift both number left again inc fsr ; no, next byte jmp bcd_adj_loop ; looping for decimal adjust ; prepare for next shift ; 0000 --> 0000 0 -->0 ; 0001 --> 0010 1 -->2 ; 0010 --> 0100 2 -->4 ; 0011 --> 0110 3 -->6 ; 0100 --> 1000 4 -->8 ; 0101 --> 1010 5 -->A, correct result is 10, so need to add 3 ; so that 5+3=8, and 1000 will be shifted to be 1 0000 ; the same is true for 6-9 digit_adj ; consider LSD first mov w,#3 ; 3 will become 6 on next shift add w,ind ; which is the decimal correct factor to be added mov temp,w snb temp.3 ; > 7? if bit 3 not set, then must be <=7, no adj. mov ind,w ; yes, decimal adjust needed, so store it ; now for the MSD mov w,#$30 ; 3 for MSD is $30 add w,ind ; add for testing mov temp,w snb temp.7 ; > 7? mov ind,w ; yes, store it ret ; 10 digit BCD to 32 bit binary conversion ; entry: 10 digit BCD number in $14-18 ; exit: 32 bit binary number in $10-13 ; algorithm= shift the bits of BCD number into the binary number and decimal ; correct on the way decbin mov count,#32 ; 32 bit number mov fsr,#bin_number ; points to the binary result clr_bin clr ind ; clear binary number inc fsr ; no, continue on next byte snb fsr.2 ; reached $13? (then fsr will be $14 here) jmp shift_b ; yes, begin algorithm jmp clr_bin ; loop to clear shift_b mov fsr,#bcd_number+4 ; points to the last BCD number clc ; clear carry, prepare for shifting right shft_loop rr ind ; shift the number right dec fsr ; reached $10? (finish shifting both numbers) sb fsr.4 ; then fsr will be $0f jmp chk_adj ; yes, check if end of everything jmp shft_loop ; not yet chk_adj decsz count ; end of 32 bit operation? jmp bd_adj ; no, do bcd adj ret bd_adj mov fsr,#bcd_number ; points to first byte of the BCD result bd_adj_loop call dgt_adj ; decimal adjust snb fsr.3 ; reached last byte? jmp shift_b ; yes, go to shift both number right again inc fsr ; no, next byte jmp bd_adj_loop ; looping for decimal adjust ; prepare for next shift right ; 0000 --> 0000 0 -->0 ; 0010 --> 0001 2 -->1 ; 0100 --> 0010 4 -->2 ; 0110 --> 0011 6 -->3 ; 1000 --> 0100 8 -->4 ; 1 0000 --> 1000 10-->8 !! it should be 5, so -3 ; 1 0010 --> 1001 12-->9 !! it should be 6, so -3 ; in general when the highest bit in a nibble is 1, it should be subtracted with 3 dgt_adj ; consider LSD first sb ind.3 ; check highest bit in LSD, =1? jmp ck_msd ; no, check MSD stc ; prepare for subtraction, no borrow sub ind,#3 ; yes, adjust ; now for the MSD ck_msd sb ind.7 ; highest bit in MSD, =1? ret ; no ; yes, do correction stc ; no borrow sub ind,#$30 ; this is a 2 word instruction, and cannot be skipped ret ENDIF IF mul88_test=1 ; 8 bit x 8 bit multiplication (RAM efficient, 2 bytes only) ; entry: multiplicand in W, multiplier at 09 ; exit : product at $0a,09 ; cycles mul88 mov upper_prdt,w ; 1 store W mov count,#9 ; 2 set number of times to shift mov w,upper_prdt ; 1 restore W (multiplicand) clr upper_prdt ; 1 clear upper product clc ; 1 clear carry ; the following are executed [count] times m88loop rr upper_prdt ; 1 rotate right the whole product rr multiplier ; 1 check lsb snc ; 1 skip addition if no carry add upper_prdt,w ; 1 add multiplicand to upper product no_add decsz count ; 1/2 loop 9 times to get proper product jmp m88loop ; 3 jmp to rotate the next half of product ret ; 3 done... ; one time instructions = 1+2+1+1+1+3= 9 cycles ; repetitive ones = (1+1+1+1+1+3)9-3+2=71 ; total worst case cycles=80 cycles ; fast 8 bit x 8 bit multiplication (RAM efficient, 2 bytes only) ; entry: multiplicand in W, multiplier at 09 ; exit : product at $0a,09 ; macro to rotate product right and add rra MACRO rr upper_prdt ; 1 rotate right the whole product rr multiplier ; 1 check lsb snc ; 1 skip addition if no carry add upper_prdt,w ; 1 add multiplicand to upper product ENDM ; cycles fmul88 clr upper_prdt ; 1 clear upper product clc ; 1 clear carry ; the following are executed [count] times rra ; call the macro 9 times rra rra rra rra rra rra rra rra ret ; 3 done... ; one time instructions = 1+1+3= 5 cycles ; repetitive ones = (1+1+1+1)9=36 ; total worst case cycles=41 cycles ENDIF IF mul1616_test=1 ; 16 bit x 16 bit multiplication ; entry: multiplicand in $09,08, multiplier at $0b,$0a ; exit : 32 bit product at $0d,$0c,$b,$a ; cycles mul1616 mov count,#17 ; 2 set number of times to shift clr upper_prdt ; 1 clear upper product clr upper_prdt+1 ; 1 higher byte of the 16 bit upeper product clc ; 1 clear carry ; the following are executed [count] times m1616loop rr upper_prdt+1 ; 1 rotate right the whole product rr upper_prdt ; 1 lower byte of the 16 bit upper product rr mr16+1 ; 1 high byte of the multiplier rr mr16 ; 1 check lsb sc ; 1 skip addition if no carry jmp no_add ; 3 no addition since lsb=0 clc ; 1 clear carry add upper_prdt,md16 ; 1 add multiplicand to upper product add upper_prdt+1,md16+1 ; 1 add the next 16 bit of multiplicand no_add decsz count ; 1/2 loop [count] times to get proper product jmp m1616loop ; 3 jmp to rotate the next half of product ret ; 3 done... ; one time instructions = 8 cycles ; repetitive ones = 15*16+11+2=253 ; total worst case cycles=261 cycles ; fast 16 bit x 16 bit multiplication ; entry: multiplicand in $09,08, multiplier at $0b,$0a ; exit : 32 bit product at $0d,$0c,$b,$a ; cycles ; one time instructions = 6 ; repetitive ones = 11*17=187 ; total = 193 cycles fmul1616 clr upper_prdt ; 1 clear upper product clr upper_prdt+1 ; 1 higher byte of the 16 bit upeper product clc ; 1 clear carry rr upper_prdt+1 ; 1 rotate right the whole product rr upper_prdt ; 1 lower byte of the 16 bit upper product rr mr16+1 ; 1 high byte of the multiplier rr mr16 ; 1 check lsb sc ; 1 skip addition if no carry jmp no_add1 ; 3 no addition since lsb=0 clc ; 1 clear carry add upper_prdt,md16 ; 1 add multiplicand to upper product add upper_prdt+1,md16+1 ; 1 add the next 16 bit of multiplicand no_add1 rr upper_prdt+1 ; 1 rotate right the whole product rr upper_prdt ; 1 lower byte of the 16 bit upper product rr mr16+1 ; 1 high byte of the multiplier rr mr16 ; 1 check lsb sc ; 1 skip addition if no carry jmp no_add2 ; 3 no addition since lsb=0 clc ; 1 clear carry add upper_prdt,md16 ; 1 add multiplicand to upper product add upper_prdt+1,md16+1 ; 1 add the next 16 bit of multiplicand no_add2 rr upper_prdt+1 ; 1 rotate right the whole product rr upper_prdt ; 1 lower byte of the 16 bit upper product rr mr16+1 ; 1 high byte of the multiplier rr mr16 ; 1 check lsb sc ; 1 skip addition if no carry jmp no_add3 ; 3 no addition since lsb=0 clc ; 1 clear carry add upper_prdt,md16 ; 1 add multiplicand to upper product add upper_prdt+1,md16+1 ; 1 add the next 16 bit of multiplicand no_add3 rr upper_prdt+1 ; 1 rotate right the whole product rr upper_prdt ; 1 lower byte of the 16 bit upper product rr mr16+1 ; 1 high byte of the multiplier rr mr16 ; 1 check lsb sc ; 1 skip addition if no carry jmp no_add4 ; 3 no addition since lsb=0 clc ; 1 clear carry add upper_prdt,md16 ; 1 add multiplicand to upper product add upper_prdt+1,md16+1 ; 1 add the next 16 bit of multiplicand no_add4 rr upper_prdt+1 ; 1 rotate right the whole product rr upper_prdt ; 1 lower byte of the 16 bit upper product rr mr16+1 ; 1 high byte of the multiplier rr mr16 ; 1 check lsb sc ; 1 skip addition if no carry jmp no_add5 ; 3 no addition since lsb=0 clc ; 1 clear carry add upper_prdt,md16 ; 1 add multiplicand to upper product add upper_prdt+1,md16+1 ; 1 add the next 16 bit of multiplicand no_add5 rr upper_prdt+1 ; 1 rotate right the whole product rr upper_prdt ; 1 lower byte of the 16 bit upper product rr mr16+1 ; 1 high byte of the multiplier rr mr16 ; 1 check lsb sc ; 1 skip addition if no carry jmp no_add6 ; 3 no addition since lsb=0 clc ; 1 clear carry add upper_prdt,md16 ; 1 add multiplicand to upper product add upper_prdt+1,md16+1 ; 1 add the next 16 bit of multiplicand no_add6 rr upper_prdt+1 ; 1 rotate right the whole product rr upper_prdt ; 1 lower byte of the 16 bit upper product rr mr16+1 ; 1 high byte of the multiplier rr mr16 ; 1 check lsb sc ; 1 skip addition if no carry jmp no_add7 ; 3 no addition since lsb=0 clc ; 1 clear carry add upper_prdt,md16 ; 1 add multiplicand to upper product add upper_prdt+1,md16+1 ; 1 add the next 16 bit of multiplicand no_add7 rr upper_prdt+1 ; 1 rotate right the whole product rr upper_prdt ; 1 lower byte of the 16 bit upper product rr mr16+1 ; 1 high byte of the multiplier rr mr16 ; 1 check lsb sc ; 1 skip addition if no carry jmp no_add8 ; 3 no addition since lsb=0 clc ; 1 clear carry add upper_prdt,md16 ; 1 add multiplicand to upper product add upper_prdt+1,md16+1 ; 1 add the next 16 bit of multiplicand no_add8 rr upper_prdt+1 ; 1 rotate right the whole product rr upper_prdt ; 1 lower byte of the 16 bit upper product rr mr16+1 ; 1 high byte of the multiplier rr mr16 ; 1 check lsb sc ; 1 skip addition if no carry jmp no_add9 ; 3 no addition since lsb=0 clc ; 1 clear carry add upper_prdt,md16 ; 1 add multiplicand to upper product add upper_prdt+1,md16+1 ; 1 add the next 16 bit of multiplicand no_add9 rr upper_prdt+1 ; 1 rotate right the whole product rr upper_prdt ; 1 lower byte of the 16 bit upper product rr mr16+1 ; 1 high byte of the multiplier rr mr16 ; 1 check lsb sc ; 1 skip addition if no carry jmp no_add10 ; 3 no addition since lsb=0 clc ; 1 clear carry add upper_prdt,md16 ; 1 add multiplicand to upper product add upper_prdt+1,md16+1 ; 1 add the next 16 bit of multiplicand no_add10 rr upper_prdt+1 ; 1 rotate right the whole product rr upper_prdt ; 1 lower byte of the 16 bit upper product rr mr16+1 ; 1 high byte of the multiplier rr mr16 ; 1 check lsb sc ; 1 skip addition if no carry jmp no_add11 ; 3 no addition since lsb=0 clc ; 1 clear carry add upper_prdt,md16 ; 1 add multiplicand to upper product add upper_prdt+1,md16+1 ; 1 add the next 16 bit of multiplicand no_add11 rr upper_prdt+1 ; 1 rotate right the whole product rr upper_prdt ; 1 lower byte of the 16 bit upper product rr mr16+1 ; 1 high byte of the multiplier rr mr16 ; 1 check lsb sc ; 1 skip addition if no carry jmp no_add12 ; 3 no addition since lsb=0 clc ; 1 clear carry add upper_prdt,md16 ; 1 add multiplicand to upper product add upper_prdt+1,md16+1 ; 1 add the next 16 bit of multiplicand no_add12 rr upper_prdt+1 ; 1 rotate right the whole product rr upper_prdt ; 1 lower byte of the 16 bit upper product rr mr16+1 ; 1 high byte of the multiplier rr mr16 ; 1 check lsb sc ; 1 skip addition if no carry jmp no_add13 ; 3 no addition since lsb=0 clc ; 1 clear carry add upper_prdt,md16 ; 1 add multiplicand to upper product add upper_prdt+1,md16+1 ; 1 add the next 16 bit of multiplicand no_add13 rr upper_prdt+1 ; 1 rotate right the whole product rr upper_prdt ; 1 lower byte of the 16 bit upper product rr mr16+1 ; 1 high byte of the multiplier rr mr16 ; 1 check lsb sc ; 1 skip addition if no carry jmp no_add14 ; 3 no addition since lsb=0 clc ; 1 clear carry add upper_prdt,md16 ; 1 add multiplicand to upper product add upper_prdt+1,md16+1 ; 1 add the next 16 bit of multiplicand no_add14 rr upper_prdt+1 ; 1 rotate right the whole product rr upper_prdt ; 1 lower byte of the 16 bit upper product rr mr16+1 ; 1 high byte of the multiplier rr mr16 ; 1 check lsb sc ; 1 skip addition if no carry jmp no_add15 ; 3 no addition since lsb=0 clc ; 1 clear carry add upper_prdt,md16 ; 1 add multiplicand to upper product add upper_prdt+1,md16+1 ; 1 add the next 16 bit of multiplicand no_add15 rr upper_prdt+1 ; 1 rotate right the whole product rr upper_prdt ; 1 lower byte of the 16 bit upper product rr mr16+1 ; 1 high byte of the multiplier rr mr16 ; 1 check lsb sc ; 1 skip addition if no carry jmp no_add16 ; 3 no addition since lsb=0 clc ; 1 clear carry add upper_prdt,md16 ; 1 add multiplicand to upper product add upper_prdt+1,md16+1 ; 1 add the next 16 bit of multiplicand no_add16 rr upper_prdt+1 ; 1 rotate right the whole product rr upper_prdt ; 1 lower byte of the 16 bit upper product rr mr16+1 ; 1 high byte of the multiplier rr mr16 ; 1 check lsb sc ; 1 skip addition if no carry jmp no_add17 ; 3 no addition since lsb=0 clc ; 1 clear carry add upper_prdt,md16 ; 1 add multiplicand to upper product add upper_prdt+1,md16+1 ; 1 add the next 16 bit of multiplicand no_add17 ret ; 3 done... ENDIF IF div1616_test=1 ; 16 bit by 16 bit division (b/a) ; entry: 16 bit b, 16 bit a ; exit : result in b, remainder in remainder ; cycles div1616 mov count,#16 ; 2 no. of time to shift mov d,b ; 2 move b to make space mov d+1,b+1 ; 2 for result clr b ; 1 clear the result fields clr b+1 ; 1 one more byte clr rlo ; 1 clear remainder low byte clr rhi ; 1 clear remainder high byte ; subtotal=10 divloop clc ; 1 clear carry before shift rl d ; 1 check the dividend rl d+1 ; 1 bit by bit rl rlo ; 1 put it in the remainder for rl rhi ; 1 trial subtraction ; subtotal=5 stc ; 1 prepare for subtraction, no borrow mov w,a+1 ; 1 do trial subtraction mov w,rhi-w ; 1 from MSB first sz ; 1/2 if two MSB equal, need to check LSB jmp chk_carry ; 3 not equal, check which one is bigger ; ; if we are here, then z=1, so c must be 1 too, since there is no ; underflow, so we save a stc instruction mov w,a ; 1 equal MSB, check LSB mov w,rlo-w ; 1 which one is bigger? ; subtotal=7 chk_carry sc ; 1/2 partial dividend >a? jmp shft_quot ; 3 no, partial dividend < a, set a 0 into quotient ; if we are here, then c must be 1, again, we save another stc instruction ; yes, part. dividend > a, subtract a from it sub rlo,a ; 2 store part. dividend-a into a sub rhi,a+1 ; 2 2 bytes stc ; 1 shift a 1 into quotient ; subtotal=7 worst case shft_quot rl b ; 1 store into result rl b+1 ; 1 16 bit result, thus 2 rotates decsz count ; 1/2 jmp divloop ; 3 ; subtotal=6, 4 on last count ret ; 3 ; one time instructions=13 ; repetitive ones=(19+6)*15+19+4=398 ; total=411 ; fast 16 bit by 16 bit division (b/a) ; entry: 16 bit b, 16 bit a ; exit : result in b, remainder in remainder ; cycles=347 ; one time=11 ; repetitive=21*16=336 fdiv1616 mov d,b ; 2 move b to make space mov d+1,b+1 ; 2 for result clr b ; 1 clear the result fields clr b+1 ; 1 one more byte clr rlo ; 1 clear remainder low byte clr rhi ; 1 clear remainder high byte ; subtotal=8 clc ; 1 clear carry before shift rl d ; 1 check the dividend rl d+1 ; 1 bit by bit rl rlo ; 1 put it in the remainder for rl rhi ; 1 trial subtraction ; subtotal=5 stc ; 1 prepare for subtraction, no borrow mov w,a+1 ; 1 do trial subtraction mov w,rhi-w ; 1 from MSB first sz ; 1/2 if two MSB equal, need to check LSB jmp chk_carry1 ; 3 not equal, check which one is bigger ; ; if we are here, then z=1, so c must be 1 too, since there is no ; underflow, so we save a stc instruction mov w,a ; 1 equal MSB, check LSB mov w,rlo-w ; 1 which one is bigger? ; subtotal=7 chk_carry1 sc ; 1/2 partial dividend >a? jmp shft_quot1 ; 3 no, partial dividend < a, set a 0 into quotient ; if we are here, then c must be 1, again, we save another stc instruction ; yes, part. dividend > a, subtract a from it sub rlo,a ; 2 store part. dividend-a into a sub rhi,a+1 ; 2 2 bytes stc ; 1 shift a 1 into quotient ; subtotal=7 worst case shft_quot1 rl b ; 1 store into result rl b+1 ; 1 16 bit result, thus 2 rotates clc ; 1 clear carry before shift rl d ; 1 check the dividend rl d+1 ; 1 bit by bit rl rlo ; 1 put it in the remainder for rl rhi ; 1 trial subtraction ; subtotal=5 stc ; 1 prepare for subtraction, no borrow mov w,a+1 ; 1 do trial subtraction mov w,rhi-w ; 1 from MSB first sz ; 1/2 if two MSB equal, need to check LSB jmp chk_carry2 ; 3 not equal, check which one is bigger ; ; if we are here, then z=1, so c must be 1 too, since there is no ; underflow, so we save a stc instruction mov w,a ; 1 equal MSB, check LSB mov w,rlo-w ; 1 which one is bigger? ; subtotal=7 chk_carry2 sc ; 1/2 partial dividend >a? jmp shft_quot2 ; 3 no, partial dividend < a, set a 0 into quotient ; if we are here, then c must be 1, again, we save another stc instruction ; yes, part. dividend > a, subtract a from it sub rlo,a ; 2 store part. dividend-a into a sub rhi,a+1 ; 2 2 bytes stc ; 1 shift a 1 into quotient ; subtotal=7 worst case shft_quot2 rl b ; 1 store into result rl b+1 ; 1 16 bit result, thus 2 rotates clc ; 1 clear carry before shift rl d ; 1 check the dividend rl d+1 ; 1 bit by bit rl rlo ; 1 put it in the remainder for rl rhi ; 1 trial subtraction ; subtotal=5 stc ; 1 prepare for subtraction, no borrow mov w,a+1 ; 1 do trial subtraction mov w,rhi-w ; 1 from MSB first sz ; 1/2 if two MSB equal, need to check LSB jmp chk_carry3 ; 3 not equal, check which one is bigger ; ; if we are here, then z=1, so c must be 1 too, since there is no ; underflow, so we save a stc instruction mov w,a ; 1 equal MSB, check LSB mov w,rlo-w ; 1 which one is bigger? ; subtotal=7 chk_carry3 sc ; 1/2 partial dividend >a? jmp shft_quot3 ; 3 no, partial dividend < a, set a 0 into quotient ; if we are here, then c must be 1, again, we save another stc instruction ; yes, part. dividend > a, subtract a from it sub rlo,a ; 2 store part. dividend-a into a sub rhi,a+1 ; 2 2 bytes stc ; 1 shift a 1 into quotient ; subtotal=7 worst case shft_quot3 rl b ; 1 store into result rl b+1 ; 1 16 bit result, thus 2 rotates clc ; 1 clear carry before shift rl d ; 1 check the dividend rl d+1 ; 1 bit by bit rl rlo ; 1 put it in the remainder for rl rhi ; 1 trial subtraction ; subtotal=5 stc ; 1 prepare for subtraction, no borrow mov w,a+1 ; 1 do trial subtraction mov w,rhi-w ; 1 from MSB first sz ; 1/2 if two MSB equal, need to check LSB jmp chk_carry4 ; 3 not equal, check which one is bigger ; ; if we are here, then z=1, so c must be 1 too, since there is no ; underflow, so we save a stc instruction mov w,a ; 1 equal MSB, check LSB mov w,rlo-w ; 1 which one is bigger? ; subtotal=7 chk_carry4 sc ; 1/2 partial dividend >a? jmp shft_quot4 ; 3 no, partial dividend < a, set a 0 into quotient ; if we are here, then c must be 1, again, we save another stc instruction ; yes, part. dividend > a, subtract a from it sub rlo,a ; 2 store part. dividend-a into a sub rhi,a+1 ; 2 2 bytes stc ; 1 shift a 1 into quotient ; subtotal=7 worst case shft_quot4 rl b ; 1 store into result rl b+1 ; 1 16 bit result, thus 2 rotates clc ; 1 clear carry before shift rl d ; 1 check the dividend rl d+1 ; 1 bit by bit rl rlo ; 1 put it in the remainder for rl rhi ; 1 trial subtraction ; subtotal=5 stc ; 1 prepare for subtraction, no borrow mov w,a+1 ; 1 do trial subtraction mov w,rhi-w ; 1 from MSB first sz ; 1/2 if two MSB equal, need to check LSB jmp chk_carry5 ; 3 not equal, check which one is bigger ; ; if we are here, then z=1, so c must be 1 too, since there is no ; underflow, so we save a stc instruction mov w,a ; 1 equal MSB, check LSB mov w,rlo-w ; 1 which one is bigger? ; subtotal=7 chk_carry5 sc ; 1/2 partial dividend >a? jmp shft_quot5 ; 3 no, partial dividend < a, set a 0 into quotient ; if we are here, then c must be 1, again, we save another stc instruction ; yes, part. dividend > a, subtract a from it sub rlo,a ; 2 store part. dividend-a into a sub rhi,a+1 ; 2 2 bytes stc ; 1 shift a 1 into quotient ; subtotal=7 worst case shft_quot5 rl b ; 1 store into result rl b+1 ; 1 16 bit result, thus 2 rotates clc ; 1 clear carry before shift rl d ; 1 check the dividend rl d+1 ; 1 bit by bit rl rlo ; 1 put it in the remainder for rl rhi ; 1 trial subtraction ; subtotal=5 stc ; 1 prepare for subtraction, no borrow mov w,a+1 ; 1 do trial subtraction mov w,rhi-w ; 1 from MSB first sz ; 1/2 if two MSB equal, need to check LSB jmp chk_carry6 ; 3 not equal, check which one is bigger ; ; if we are here, then z=1, so c must be 1 too, since there is no ; underflow, so we save a stc instruction mov w,a ; 1 equal MSB, check LSB mov w,rlo-w ; 1 which one is bigger? ; subtotal=7 chk_carry6 sc ; 1/2 partial dividend >a? jmp shft_quot6 ; 3 no, partial dividend < a, set a 0 into quotient ; if we are here, then c must be 1, again, we save another stc instruction ; yes, part. dividend > a, subtract a from it sub rlo,a ; 2 store part. dividend-a into a sub rhi,a+1 ; 2 2 bytes stc ; 1 shift a 1 into quotient ; subtotal=7 worst case shft_quot6 rl b ; 1 store into result rl b+1 ; 1 16 bit result, thus 2 rotates clc ; 1 clear carry before shift rl d ; 1 check the dividend rl d+1 ; 1 bit by bit rl rlo ; 1 put it in the remainder for rl rhi ; 1 trial subtraction ; subtotal=5 stc ; 1 prepare for subtraction, no borrow mov w,a+1 ; 1 do trial subtraction mov w,rhi-w ; 1 from MSB first sz ; 1/2 if two MSB equal, need to check LSB jmp chk_carry7 ; 3 not equal, check which one is bigger ; ; if we are here, then z=1, so c must be 1 too, since there is no ; underflow, so we save a stc instruction mov w,a ; 1 equal MSB, check LSB mov w,rlo-w ; 1 which one is bigger? ; subtotal=7 chk_carry7 sc ; 1/2 partial dividend >a? jmp shft_quot7 ; 3 no, partial dividend < a, set a 0 into quotient ; if we are here, then c must be 1, again, we save another stc instruction ; yes, part. dividend > a, subtract a from it sub rlo,a ; 2 store part. dividend-a into a sub rhi,a+1 ; 2 2 bytes stc ; 1 shift a 1 into quotient ; subtotal=7 worst case shft_quot7 rl b ; 1 store into result rl b+1 ; 1 16 bit result, thus 2 rotates clc ; 1 clear carry before shift rl d ; 1 check the dividend rl d+1 ; 1 bit by bit rl rlo ; 1 put it in the remainder for rl rhi ; 1 trial subtraction ; subtotal=5 stc ; 1 prepare for subtraction, no borrow mov w,a+1 ; 1 do trial subtraction mov w,rhi-w ; 1 from MSB first sz ; 1/2 if two MSB equal, need to check LSB jmp chk_carry8 ; 3 not equal, check which one is bigger ; ; if we are here, then z=1, so c must be 1 too, since there is no ; underflow, so we save a stc instruction mov w,a ; 1 equal MSB, check LSB mov w,rlo-w ; 1 which one is bigger? ; subtotal=7 chk_carry8 sc ; 1/2 partial dividend >a? jmp shft_quot8 ; 3 no, partial dividend < a, set a 0 into quotient ; if we are here, then c must be 1, again, we save another stc instruction ; yes, part. dividend > a, subtract a from it sub rlo,a ; 2 store part. dividend-a into a sub rhi,a+1 ; 2 2 bytes stc ; 1 shift a 1 into quotient ; subtotal=7 worst case shft_quot8 rl b ; 1 store into result rl b+1 ; 1 16 bit result, thus 2 rotates clc ; 1 clear carry before shift rl d ; 1 check the dividend rl d+1 ; 1 bit by bit rl rlo ; 1 put it in the remainder for rl rhi ; 1 trial subtraction ; subtotal=5 stc ; 1 prepare for subtraction, no borrow mov w,a+1 ; 1 do trial subtraction mov w,rhi-w ; 1 from MSB first sz ; 1/2 if two MSB equal, need to check LSB jmp chk_carry9 ; 3 not equal, check which one is bigger ; ; if we are here, then z=1, so c must be 1 too, since there is no ; underflow, so we save a stc instruction mov w,a ; 1 equal MSB, check LSB mov w,rlo-w ; 1 which one is bigger? ; subtotal=7 chk_carry9 sc ; 1/2 partial dividend >a? jmp shft_quot9 ; 3 no, partial dividend < a, set a 0 into quotient ; if we are here, then c must be 1, again, we save another stc instruction ; yes, part. dividend > a, subtract a from it sub rlo,a ; 2 store part. dividend-a into a sub rhi,a+1 ; 2 2 bytes stc ; 1 shift a 1 into quotient ; subtotal=7 worst case shft_quot9 rl b ; 1 store into result rl b+1 ; 1 16 bit result, thus 2 rotates clc ; 1 clear carry before shift rl d ; 1 check the dividend rl d+1 ; 1 bit by bit rl rlo ; 1 put it in the remainder for rl rhi ; 1 trial subtraction ; subtotal=5 stc ; 1 prepare for subtraction, no borrow mov w,a+1 ; 1 do trial subtraction mov w,rhi-w ; 1 from MSB first sz ; 1/2 if two MSB equal, need to check LSB jmp chk_carry10 ; 3 not equal, check which one is bigger ; ; if we are here, then z=1, so c must be 1 too, since there is no ; underflow, so we save a stc instruction mov w,a ; 1 equal MSB, check LSB mov w,rlo-w ; 1 which one is bigger? ; subtotal=7 chk_carry10 sc ; 1/2 partial dividend >a? jmp shft_quot10 ; 3 no, partial dividend < a, set a 0 into quotient ; if we are here, then c must be 1, again, we save another stc instruction ; yes, part. dividend > a, subtract a from it sub rlo,a ; 2 store part. dividend-a into a sub rhi,a+1 ; 2 2 bytes stc ; 1 shift a 1 into quotient ; subtotal=7 worst case shft_quot10 rl b ; 1 store into result rl b+1 ; 1 16 bit result, thus 2 rotates clc ; 1 clear carry before shift rl d ; 1 check the dividend rl d+1 ; 1 bit by bit rl rlo ; 1 put it in the remainder for rl rhi ; 1 trial subtraction ; subtotal=5 stc ; 1 prepare for subtraction, no borrow mov w,a+1 ; 1 do trial subtraction mov w,rhi-w ; 1 from MSB first sz ; 1/2 if two MSB equal, need to check LSB jmp chk_carry11 ; 3 not equal, check which one is bigger ; ; if we are here, then z=1, so c must be 1 too, since there is no ; underflow, so we save a stc instruction mov w,a ; 1 equal MSB, check LSB mov w,rlo-w ; 1 which one is bigger? ; subtotal=7 chk_carry11 sc ; 1/2 partial dividend >a? jmp shft_quot11 ; 3 no, partial dividend < a, set a 0 into quotient ; if we are here, then c must be 1, again, we save another stc instruction ; yes, part. dividend > a, subtract a from it sub rlo,a ; 2 store part. dividend-a into a sub rhi,a+1 ; 2 2 bytes stc ; 1 shift a 1 into quotient ; subtotal=7 worst case shft_quot11 rl b ; 1 store into result rl b+1 ; 1 16 bit result, thus 2 rotates clc ; 1 clear carry before shift rl d ; 1 check the dividend rl d+1 ; 1 bit by bit rl rlo ; 1 put it in the remainder for rl rhi ; 1 trial subtraction ; subtotal=5 stc ; 1 prepare for subtraction, no borrow mov w,a+1 ; 1 do trial subtraction mov w,rhi-w ; 1 from MSB first sz ; 1/2 if two MSB equal, need to check LSB jmp chk_carry12 ; 3 not equal, check which one is bigger ; ; if we are here, then z=1, so c must be 1 too, since there is no ; underflow, so we save a stc instruction mov w,a ; 1 equal MSB, check LSB mov w,rlo-w ; 1 which one is bigger? ; subtotal=7 chk_carry12 sc ; 1/2 partial dividend >a? jmp shft_quot12 ; 3 no, partial dividend < a, set a 0 into quotient ; if we are here, then c must be 1, again, we save another stc instruction ; yes, part. dividend > a, subtract a from it sub rlo,a ; 2 store part. dividend-a into a sub rhi,a+1 ; 2 2 bytes stc ; 1 shift a 1 into quotient ; subtotal=7 worst case shft_quot12 rl b ; 1 store into result rl b+1 ; 1 16 bit result, thus 2 rotates clc ; 1 clear carry before shift rl d ; 1 check the dividend rl d+1 ; 1 bit by bit rl rlo ; 1 put it in the remainder for rl rhi ; 1 trial subtraction ; subtotal=5 stc ; 1 prepare for subtraction, no borrow mov w,a+1 ; 1 do trial subtraction mov w,rhi-w ; 1 from MSB first sz ; 1/2 if two MSB equal, need to check LSB jmp chk_carry13 ; 3 not equal, check which one is bigger ; ; if we are here, then z=1, so c must be 1 too, since there is no ; underflow, so we save a stc instruction mov w,a ; 1 equal MSB, check LSB mov w,rlo-w ; 1 which one is bigger? ; subtotal=7 chk_carry13 sc ; 1/2 partial dividend >a? jmp shft_quot13 ; 3 no, partial dividend < a, set a 0 into quotient ; if we are here, then c must be 1, again, we save another stc instruction ; yes, part. dividend > a, subtract a from it sub rlo,a ; 2 store part. dividend-a into a sub rhi,a+1 ; 2 2 bytes stc ; 1 shift a 1 into quotient ; subtotal=7 worst case shft_quot13 rl b ; 1 store into result rl b+1 ; 1 16 bit result, thus 2 rotates clc ; 1 clear carry before shift rl d ; 1 check the dividend rl d+1 ; 1 bit by bit rl rlo ; 1 put it in the remainder for rl rhi ; 1 trial subtraction ; subtotal=5 stc ; 1 prepare for subtraction, no borrow mov w,a+1 ; 1 do trial subtraction mov w,rhi-w ; 1 from MSB first sz ; 1/2 if two MSB equal, need to check LSB jmp chk_carry14 ; 3 not equal, check which one is bigger ; ; if we are here, then z=1, so c must be 1 too, since there is no ; underflow, so we save a stc instruction mov w,a ; 1 equal MSB, check LSB mov w,rlo-w ; 1 which one is bigger? ; subtotal=7 chk_carry14 sc ; 1/2 partial dividend >a? jmp shft_quot14 ; 3 no, partial dividend < a, set a 0 into quotient ; if we are here, then c must be 1, again, we save another stc instruction ; yes, part. dividend > a, subtract a from it sub rlo,a ; 2 store part. dividend-a into a sub rhi,a+1 ; 2 2 bytes stc ; 1 shift a 1 into quotient ; subtotal=7 worst case shft_quot14 rl b ; 1 store into result rl b+1 ; 1 16 bit result, thus 2 rotates clc ; 1 clear carry before shift rl d ; 1 check the dividend rl d+1 ; 1 bit by bit rl rlo ; 1 put it in the remainder for rl rhi ; 1 trial subtraction ; subtotal=5 stc ; 1 prepare for subtraction, no borrow mov w,a+1 ; 1 do trial subtraction mov w,rhi-w ; 1 from MSB first sz ; 1/2 if two MSB equal, need to check LSB jmp chk_carry15 ; 3 not equal, check which one is bigger ; ; if we are here, then z=1, so c must be 1 too, since there is no ; underflow, so we save a stc instruction mov w,a ; 1 equal MSB, check LSB mov w,rlo-w ; 1 which one is bigger? ; subtotal=7 chk_carry15 sc ; 1/2 partial dividend >a? jmp shft_quot15 ; 3 no, partial dividend < a, set a 0 into quotient ; if we are here, then c must be 1, again, we save another stc instruction ; yes, part. dividend > a, subtract a from it sub rlo,a ; 2 store part. dividend-a into a sub rhi,a+1 ; 2 2 bytes stc ; 1 shift a 1 into quotient ; subtotal=7 worst case shft_quot15 rl b ; 1 store into result rl b+1 ; 1 16 bit result, thus 2 rotates clc ; 1 clear carry before shift rl d ; 1 check the dividend rl d+1 ; 1 bit by bit rl rlo ; 1 put it in the remainder for rl rhi ; 1 trial subtraction ; subtotal=5 stc ; 1 prepare for subtraction, no borrow mov w,a+1 ; 1 do trial subtraction mov w,rhi-w ; 1 from MSB first sz ; 1/2 if two MSB equal, need to check LSB jmp chk_carry16 ; 3 not equal, check which one is bigger ; ; if we are here, then z=1, so c must be 1 too, since there is no ; underflow, so we save a stc instruction mov w,a ; 1 equal MSB, check LSB mov w,rlo-w ; 1 which one is bigger? ; subtotal=7 chk_carry16 sc ; 1/2 partial dividend >a? jmp shft_quot16 ; 3 no, partial dividend < a, set a 0 into quotient ; if we are here, then c must be 1, again, we save another stc instruction ; yes, part. dividend > a, subtract a from it sub rlo,a ; 2 store part. dividend-a into a sub rhi,a+1 ; 2 2 bytes stc ; 1 shift a 1 into quotient ; subtotal=7 worst case shft_quot16 rl b ; 1 store into result rl b+1 ; 1 16 bit result, thus 2 rotates ret ; 3 ENDIF
| file: /Techref/new/letter/news0305.htm, 74KB, , updated: 2004/1/14 14:51, local time: 2024/12/24 22:31,
3.12.73.221:LOG IN
|
| ©2024 These pages are served without commercial sponsorship. (No popup ads, etc...).Bandwidth abuse increases hosting cost forcing sponsorship or shutdown. This server aggressively defends against automated copying for any reason including offline viewing, duplication, etc... Please respect this requirement and DO NOT RIP THIS SITE. Questions? <A HREF="http://techref.massmind.org/Techref/new/letter/news0305.htm"> May 2003 SXList.com newsletter - SX arithmetic routines</A> |
| Did you find what you needed? |
Welcome to massmind.org! |
Welcome to techref.massmind.org! |
.