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16 bit integer for the 18Cxxx
Author: Scott Dattalo
This code was originally located @ http://www.piclist.com
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list p=18c242,t=ON,c=132,n=80
radix dec
include "p18c242.inc"
cblock 0
N_hi,N_lo
mask
x,y
s_hi,s_lo
endc
ORG 0 ;Reset Vector
Main
CLRF status
CLRF x
CLRF s_hi
CLRF s_lo
INCF x,f
; Here's some test code that goes through all 2^16 cases
;
; psuedo code:
:
; int s=0;
; int x=0;
;
; do {
;
; if(x*x == s)
; x++;
;
; n = s;
; w = sqrt(n);
; if(w+1 != x)
; failed();
;
; s++;
;
; } while (x<256);
; for(i=0; i< 256; i++)
xxx
MOVF x,W ;W = x*x
MULWF x
movf prodh,w,0 ;if x*x is equal to the current
cpfseq s_hi ;test case, s, then we need to advance
bra L1 ;x. Note that x will always be 1 greater
movf prodl,w,0 ;than the sqrt(s)
cpfseq s_lo
bra L1
infsnz x,f
here bra here
L1:
; N = s
movf s_hi,w
movwf N_hi
movf s_lo,w
movwf N_lo
; find the sqrt of N (N_hi:N_lo)
RCALL sqrt
infsnz s_lo,f ;s++ advance for the next test case
incf s_hi,f
addlw 1 ;if w (sqrt(N)) is one less than x
cpfseq x ; then we ve passed
bra failed
bra xxx
failed:
bra xxx ;Set a break point here to catch failures
;----------------------------------------------------------
;sqrt
;
; The purpose of this routine is take the square root of a 16 bit
;unsigned integer.
;Inputs: N_hi - High byte of the 16 bit number to be square rooted
; N_lo - Low byte " " "
;Outputs: W register returned with the 8 bit result
;
;Memory used:
; mask
;
;Minimum 68 cycles
;Maximum 108 cycles
sqrt
MOVLW 0xc0 ;Initialize value for mask
MOVWF mask
MOVLW 0x40 ;Initial value for the root
sq1 CPFSLT N_hi
BRA sq6
;Subtract the root developed so far
sq3 RLCF N_lo,F ;Shift N left one position
RLCF N_hi,F
BC sq4
RRCF mask,F ;mov the 2-bit mask down 1
XORWF mask,W ;clear the last and set the next bit
BNC sq1
;We are almost done. In the last iteration, we have 7 bits of the root. When
;"01" is appended to it, we will have a 9-bit number that must be subtracted
;from N.
SUBWF N_hi,F ;
SKPC ;If the upper 7 bits cause a borrow, then
RETURN ;the appended "01" will as well: We're done.
SKPNZ ;If the result of the subtraction is zero
BTFSC N_lo,7 ;AND the msb of N_lo is set then the LSB of the
;root is zero.
XORLW 1 ;Otherwise, it is one.
RETURN ;
sq4 BTFSC mask,0 ;Need to unconditionally set the current bit of
the root.
RETURN ;However, if we're through iterating, then
leave.
RRNCF mask,F
XORWF mask,W ;Append "01" to the root developed
so far.
sq6 SUBWF N_hi,F
IORWF mask,W ;Set the current bit
bra sq3 ;Go unconditionally set the current bit.
END |
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