On Fri, 6 Apr 2001, [Iso-8859-1] K=FCbek Tony wrote:

> Hi,
>=20
> Scott Dattalo wrote:

Actually Bob. And it's a damned clever idea.

> >Tony,
> >I hate to do this to you, but here is simpler(?) way to round to nearest=
 5:
>=20
> Thankfully accepted :)
>=20
> >Given 25 bit number HI:MID:LO
> >Its value is
> >LO + 256*MID + 65536*HI
> >Which is equal to
> >LO+255*MID+MID+65535*HI+HI
> >But 255*MID and 65535*HI are both multiples of 5, so:
> >LO+MID+HI =3D=3D HI:MID:LO (mod 5)
> >This can be at most a 10 bit number, let us call it H2:L2
> >now we can:
> >    movf    L2,W,A
> >    andlw  0x0F
> >    movwf ONES,A
> >
> >    swapf L2,W,A
> >    andlw 0x0F
> >    btfsc   L2,4
> >    addlw 5        ; worth 6, not 1
> >    btfsc    H2,0
> >    addlw    6
> >    btfsc    H2,1
> >    addlw    2
>=20
> I'll be darned, I certanily hope there are no such shortcuts for
> the 10's multiple, that would make my latest 3 hours a total waste :).

actually the 10's aren't much harder:

lo + 256*mid + 65536*hi =3D lo +250*mid + 65530*hi + 6*(mid + hi)

   clrf   h2
   movf   mid, w
   addwf  hi,w
   rlfc   h2,f
   movwf  l2       ;mid+hi
   addwf  l2,f     ;2*mid+2*hi
   skpnc
    incf  h2,f
   addwf  l2,f     ;3*mid+3*hi
   skpnc
    incf  h2,f

   clrc  =20
   rlf    temp,w   ;6*mid+6*hi
   skpnc
    incf  h2,f

   addwf  lo,w
   skpnc
    incf  h2,f

at this point, h2:l2 is congruent to the original 24-bit number modulo 10 a=
nd is
no more than 14-bits. But this is off the top of my head and not necessaril=
y
optimized...


>=20
> Brilliant !

Yeah! Let's by Bob a beer.

Scott

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