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; Shift Product Down with STATUS, C Product + 4, Product + 3, Product + 2, Product + 1, Product, f f f f f
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the Reset Carry from the Multiplier shift down or the result of the sixteen bit addition.
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decfsz BitCount goto Loop
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Both of the multiplication routines shown here will work with positive and negative numbers. For the PIC microcontrollers that have built-in 8 8 multipliers, the code for 16-bit multiplication uses the techniques taught in high school mathematics for multiplying together to variable factors. Instead of thinking of a 16-bit number as just a contiguous set of 16 bits, I can represent it as two 8-bit numbers with the high byte multiplied by 256 (0x100): A = (Ah * 0x100) + Al Breaking 16 numbers up into this format, I can FOIL ( First, Outside, Inside, Last ) the two factors together using the equation A x B = ((Ah * 0x100) + Al) x ((Bh * 0x100) + Bl) = ((Ah * 0x100) x (Bh * 0x100)) + ((Ah * 0x100) x Bl) + (Al x (BH * 0x100)) + (Al x Bl) = (Ah x Bh * 0x10000) + (Ah x Bl * 0x100) + (Al x Bh * 0x100) + (Al x Bl) Multiplying numbers by factors of 256 is accomplished simply by putting them in the next higher signi cant byte. Using this formula and the built-in multiplier, 16-bit multiplication with a 32-bit result can be accomplished in the PIC18 using the code
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clrf clrf Product + 2 Product + 3 ; Clear the High-Order Bits
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APPENDIX G
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movf Al, w mulwf Bl movf PRODL, w movwf Product movf PRODH, w movwf Product + 1 movf Al, w mulwf Bh movf PRODL, w addwf Product + 1, f movf PRODH, w addwfc Product + 2, f movf Ah, w mulwf Bl movf PRODL, w addwf Product + 1, f movf PRODH, w addwfc Product + 2, f btfsc STATUS, C incf Product + 3, f movf Ah, w mulwf Bh movf PORDL, w addwf Product + 2, f movf PRODH, w addwfc Product + 3, f
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; Do the L Multiplication rst ; Save result
; Do the I Multiplication ; Save the Most Signi cant Byte First
; Add to the Last Result ; Do the O Multiplication ; Add the Lower Byte Next ; Add the High Byte First ; Add the Carry ; Do the F Multiplication
DIVISION
The division routine provided here rst nds how far the divisor can be shifted up before comparing to the quotient. The Count variable in this routine is a 16-bit variable that is used to both count the bits and add to the quotient. Temp is an 8-bit temporary storage variable. At the end of the division routine, Dividend will contain the remainder of the operation.
clrf Quotient clrf Quotient + 1 movlw 1 movwf Count clrf Count + 1 StartLoop: btfsc Divisor + 1, 7 goto Loop bcf rlf STATUS, C Count, f ; Initialize Count
; ; ; ;
Find How Large Divisor can be If at the top , then do the Division
; Shift Count and Divisor Up
REUSE, RETURN, AND RECYCLE
rlf rlf rlf
Count + 1, f Divisor, f Divisor + 1, f
goto StartLoop Loop: ; Now, Take Away Divisor ; from Dividend movf Divisor + 1, w ; If Divisor < Dividend then subwf Dividend + 1, w ; Don t Take Away movwf Temp movf Divisor, w subwf Dividend, w btfss STATUS, C decf Temp, f btfsc Temp, 7 ; If Temp Negative then goto Skip ; Divisor < Dividend movwf Dividend movf Temp, w movwf Dividend + 1 ; Save the New Dividend
movf Count, w ; Add Count to the Quotient addwf Quotient + 1, f movf Count, w addwf Quotient + 1, f ; No Opportunity for Carry Skip: bcf STATUS, C rrf Divisor + 1, f rrf Divisor, f rrf rrf Count + 1, f Count, f ; Shift Divisor/Count Down
; If Carry Set after Count ; Shift, Finished ; If Carry NOT Set, then ; Process next Bit
btfss STATUS, C goto Loop
This division routine is designed to handle only positive numbers there is not a general algorithm that handles both positive and negative numbers and passes back both the quotient and remainder with the correct polarity ef ciently. Along with the problems with negative values and returning the required value, handling zero can cause problems. In many processors, division by zero causes a system fault. In my experience, I have found that implementing a division routine is very dependent on the expected values. There are some very ef cient algorithms for speci c divisors, the most obvious being how to divide by multiples of 2 (simply shift the dividend value to the right an appropriate number of times).
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