Tutorial Mult 24 x 24 out 48 (max)

;------------------------------------------------------------------ ; Tutorial Mult 24 x 24 out 48 (max) ; Fred Maher 3rd Nov 2002 ;------------------------------------------------------------------ LIST P=16F628 ; 16F628 Runs at 20 MHz INCLUDE "p16F628.inc" __CONFIG _CP_OFF & _WDT_OFF & _HS_OSC & _PWRTE_ON & _LVP_OFF & _BODEN_OFF ERRORLEVEL -224 ; suppress annoying message because of tris ERRORLEVEL -302 ; suppress message because of page change CBLOCK 20H top1 top2 top3 top4 top5 top6 bot1 bot2 bot3 res1 res2 res3 res4 res5 res6 count temp c1 c2 ENDC ; Top and bottom registers are not symetrical. The bottom registers will never need to ; handle more than 24 bits. The top registers can, when the numbers are near the maximum, ; have to handle 48 bits due to left shifting. Now we can reuse registers but it makes ; following these "tutor" multiply routines harder to follow. ;There is no ORG, ISR, as this is a sub routine of a main program. However we ;want it to behave like a program to debug and emulate it running with real number inputs. ;When it is all working we remove ;the header and place the variables with the rest of ;the variables of the existing main ;program code. ;So we can start by putting values into the top and bottom, values that will not cause ;overflow in the result. ;Example numbers :top 00FFFF bot 000024 txb = 23FFDC(inside 24 bit limit)decimal = 2,359,260 Testnums: movlw 0x0B movwf top1 movlw 0x00 movwf top2 movlw 0x00 movwf top3 movlw 0x00 movwf top4 movlw 0x00 movwf top5 movlw 0x00 movwf top6 movlw 0xBA movwf bot1 movlw 0xDC movwf bot2 movlw 0xFE movwf bot3 clrf res1 clrf res2 clrf res3 clrf res4 clrf res5 clrf res6 ; In example top x bot = B044 F100 ; we will use top as multiplicand and bot as the multiplier Note:- the smaller number is as we ; discussed in the multiplier, bottom. ; A word about the code we are going to write. The shifts and the checks produce a lot of trees ;making it easy to lose sight of the main actions. These are. ; THE MULTIPLY SEQUENCE ;---------------------- ;Step 1. Use bot bit 0 to decide if we add tops to reses or to skip directly to 4. ; ;Step 2. Add the rotated left 3top to 3res if bot bit not 0, else skip add. ; Please note, the First Time as there is no shifting, ; add directly all the top bits to result. If the first ( rightmost digit of the ;bot.. multiplier ) had been zero and not four then we skip the add and move on ;to the next action. ( bot bit tells us when to skip the add) ;Step 2 actions:. The addition of res and top ; top1 add to res1, carry flag? yes, add 1 to res2 , clear carry flag ; top2 add to res2, carry flag? yes, add 1 to res3 , clear carry flag ; top3 add to res3 , carry flag? yes, ; ;Step 3. Rotate left the three top registers and check overflows in this sequence... ; top1 rotate left, carry flag? yes, add 1 to top2 , clear carry flag ; top2 rotate left, carry flag? yes, add 1 to top3 , clear carry flag ; top3 rotate left, carry flag? yes, ; not this time ---> "add 1 to top4, clear carry flag"<-- ; ;Step 4. Bot,bit0( value 1 or 0 )decides if we add or don't add so we need to ; right shift the three bots or read a bit in the corresponding register ; using the loop counter count. This counter would loop from 1 to 24 and ;needs checks ;to move from bot1 reg to bot2 reg then to bot3 reg. ;This first attempt will use the same actions as 3. and 4. So we have after 1st read ;bot1 rotate right, Carry flag? yes , clear carry flag ;bot2 rotate right, Carry flag? yes add b'1000000' = hex80 to bot1 , clear carry flag ;bot3 rotate right, Carry flag? yes add b'1000000' = hex80 to bot2 , clear carry flag ; These are the four steps explained in words.... ; The execution time of the multiplication can be quickened considerably if ; we know when the multiplier has finished. ; e.g. look at bot3, bot2, bot1 00 00 24 = 00000000 00000000 0010 0100 ; now after five shifts right what remains are zeros. ; So calculation finished at count 6, in 243 cycles; yet we continue ; until we reach count 24, which takes 836 cycles ; What we need to add is a top bottom swap, and detect bot_first_1 for count needed ; routine, to optimize the 24x24 multiply a little more. ;----------------------------------------------------------------------- ; Mult24 is the main mult routine start point ;------------------------------------------------------------------------ Mult24: ; bcf STATUS,DC call Regcomp ; Puts smaller number into multiplier and finds count minimum. nop incf count,f clrf res1 ; may have been used for storage in swap clrf res2 clrf res3 call X24 ; main 24 x 24 mult Mult24end: ; here back to principal program goto Mult24end ; dummy loop for testing ; return ; back to main program ;-------------------------------------------------------------------------- ; End of multiply 24 x 24 ; out 48, routine ;-------------------------------------------------------------------------- Regcomp: ; compare top and bot regs 3 2 1 to decide if we need to swap. bcf STATUS,2 bcf STATUS,1 bcf STATUS,0 ; These blocks test for:- "0 "... "="... ">" ... using STATUS flags Test33: movf bot3,w subwf top3,w btfsc STATUS,C ; goto $+ 3 ; top is equal or bigger call Swap goto Leftbit ; find leftmost 1 for count. ; end case top>bot ; Think carefully about this bit. We get here because bot is ; equal or less than top. So we test top for zero, then if zero ; bot is also zero because we have no negative numbers ; (e.g. bot can't be -7) ; if STATUS not zero bot is less than top, then find ; leftmost "1" for count. clrw ; test top=0 (if top=0 then bot=0 (can't be minus)) xorwf top3,w btfsc STATUS,Z goto Test22 ; both zero, continue test with top2,bot2 goto Leftbit Test22: movf bot2,w subwf top2,w btfsc STATUS,C ; goto $+ 3 ; top is equal or bigger call Swap goto Leftbit ; find leftmost 1 for count. ; end case top>bot clrw ; test top=0 (if top=0 then bot=0 (can't be minus)) xorwf top2,w btfsc STATUS,Z goto Test11 ; both zero, continue test with top2,bot2 goto Leftbit Test11: movf bot1,w subwf top1,w btfsc STATUS,C ; goto $+ 3 ; top is equal or bigger call Swap goto Leftbit ; find leftmost 1 for count. ; end case top>bot clrw ; test top=0 (if top=0 then bot=0 (can't be minus)) xorwf top1,w btfsc STATUS,Z goto Test11 ; both zero, continue test with top2,bot2 movf bot1,w ;put bot in temp movwf temp movlw d'08' ; bot is smaller so set count to 24 movwf count ; do Leftbit goto Leftbit Swap: movf top3,w movwf res3 movf bot3,w movwf top3 movf res3,w movwf bot3 movf top2,w movwf res2 movf bot2,w movwf top2 movf res2,w movwf bot2 movf top1,w movwf res1 movf bot1,w movwf top1 movf res1,w movwf bot1 return Leftbit: ;Now find in bot3,2,1 that which is not zero, find leftmost ONE ; We leave the routine with count holding the minimum necessary number ;for the multiplier adds and with multiplier as bot, the smaller number clrw xorwf bot3,w btfsc STATUS, Z goto $+6 movf bot3,w movwf temp movlw d'24' movwf count goto Oneloop clrw xorwf bot2,w btfsc STATUS, Z goto $+6 movf bot2,w movwf temp movlw d'16' movwf count goto Oneloop clrw xorwf bot1,w btfsc STATUS, Z goto $+4 movf bot1,w movwf temp movlw d'08' movwf count goto Oneloop Oneloop: ; We move through the 8 bits of temp from 8 to 1 btfsc temp,7 return decf count,f rlf temp,f goto Oneloop return ; to main mult ;---------------------------------------------------------------- ; End of the compare / count routine ;---------------------------------------------------------------- X24: ; mult loop proper starts here with entered values Step1: ; To add or not to add? that is the question. btfss bot1,0 goto Step3 ; 0... don't add top to res goto Step2 ; 1... add top to res Step2: ; Adding top to res movfw top1 addwf res1,f movfw top2 skpnc incfsz top2,w addwf res2,f movfw top3 skpnc incfsz top3,w addwf res3,f movfw top4 skpnc incfsz top4,w addwf res4,f movfw top5 skpnc incfsz top5,w addwf res5,f movfw top6 skpnc incfsz top6,w addwf res6,f Step3: ; Left shifting top ; This looks the same as an add but it is different, when we shift ; we never sum, but rather drop the carried 1 into a space vacated ; So we start with top6, check C,; top5 check C,.....; top1 check C bcf STATUS,C rlf top6,f btfsc STATUS,C nop ;incf top4,f bcf STATUS,C rlf top5,f btfsc STATUS,C incf top6,f bcf STATUS,C rlf top4,f btfsc STATUS,C incf top5,f bcf STATUS,C rlf top3,f btfsc STATUS,C incf top4,f bcf STATUS,C rlf top2,f btfsc STATUS,C incf top3,f bcf STATUS,C rlf top1,f btfsc STATUS,C incf top2,f bcf STATUS,C Stepend: nop Step4: ;Right shifting bottom and terminate loop for count = MAX, here 24. decf count,f ; we start with count =24 OR speeded up count clrw xorwf count,w btfsc STATUS,Z RETURN ; count finished ; count not finished, continue Step4 rrf bot1,f bcf STATUS,C rrf bot2,f btfss STATUS,C goto $ + 4 ; there was no carry, jump forward 4 movlw 0x80 addwf bot1,f bcf STATUS,C rrf bot3,f btfss STATUS,C goto $ + 4 ; there was no carry, jump forward 4 movlw 0x80 addwf bot2,f bcf STATUS,C goto Step1 ;--------------------------------------------------------------------- ; End of reg multiplication ;--------------------------------------------------------------------- end ;not used when this is a subroutine ;------------------------------------------------------------------ ; End Mult 24 x 24 ;------------------------------------------------------------------

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