Re: divide by 10 with Z80

On Jan 22, 5:14 pm, Herb Johnson <herbrjohn...@xxxxxxxxx> wrote:
On Jan 22, 2:29 pm, "dkel...@xxxxxxxxxxx" <dkel...@xxxxxxxxxxx> wrote:

; divide number between 0 and 800 by 10
; hl = dividend
; returns:
;  a = quotient
;  h = remainder

Joseph carr "z-80 microcomputer handbook" p. 190. Put
the 16-bit value in HL, put the 8 bit value in C and zero in B.
Then shift HL left and subtract BC successively. In effect you
are shifting BC right relative to HL.and doing "long division"
Feed the results into the lower bits of HL as you shift HL to the
left.

divide8:
;B has 8 bit divider, HL the "16-bit" dividend
;quotient in H, remainder in L
;successive shift and subtract operations
        LD   C,0
        LD   D,8        ;shift counter
LOOP    ADD  HL,HL      ;shift HL left
        XOR  A,A        ;clear carry
        SBC  HL,BC      ;subtact divisor from UPPER byte
        INC  HL         ;presume result "good"
        JP   P,JUMP1    ;jump if positive result
        ADD  HL,BC      ;if negative, restore first
JUMP1:  RES  0,L        ;and clear bit 0 in L
        DEC  D          ;count down 8 shifts
        JR   NZ,LOOP    ;and loop until none

Note from Herb. I don't think this works if HL has a value
above 32K decimal. Looks to me like the first "ADD HL,HL" effectively
left-shifts bit 15 to oblivion.  But for the required value of 0 to
800,
that's not a problem. Like any code in a book, check it throughly
for problems and marginal conditions; make sure you avoid "feeding"
it values it won't handle. (What happens if B=0?)

Herb Johnson
"don't try to be a programmer, Herb" - French Luser
retrotechnology.com

Hi Herb
I'd still need to load B. That would push it to 20 bytes ( If I count
right ). It is clearly better than mine for numbers larger than
1023. Still, see my latest post. I have the count down to 16
bytes and fewer clocks. Yours is 20 bytes, including the ret
and ld bc,0a00h.
Thanks
Dwight
.

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