Z80 Routines:Math:Division
Contents
Introduction
All these routines use the restoring division algorithm, adapted to the z80 architecture to maximize speed. They can easily be unrolled to gain some speed.
8/8 division
The following routine divides d by e and places the quotient in d and the remainder in a
div_d_e: xor a ld b, 8 _loop: sla d rla cp e jr c, $+4 sub e inc d djnz _loop ret
16/8 division
The following routine divides hl by c and places the quotient in hl and the remainder in a
div_hl_c: xor a ld b, 16 _loop: add hl, hl rla jr c, $+5 cp c jr c, $+4 sub c inc l djnz _loop ret
16/16 division
The following routine divides ac by de and places the quotient in ac and the remainder in hl
div_ac_de: ld hl, 0 ld b, 16 _loop: sll c rla adc hl, hl sbc hl, de jr nc, $+4 add hl, de dec c djnz _loop ret
24/8 division
The following routine divides ehl by d and places the quotient in ehl and the remainder in a
div_ehl_d: xor a ld b, 24 _loop: add hl, hl rl e rla jr c, $+5 cp d jr c, $+4 sub d inc l djnz _loop ret
32/8 division
The following routine divides dehl by c and places the quotient in dehl and the remainder in a
div_dehl_c: xor a ld b, 32 _loop: add hl, hl rl e rl d rla jr c, $+5 cp c jr c, $+4 sub c inc l djnz _loop ret
32/16 division
The following routine divides acix by de and places the quotient in acix and the remainder in hl
Div32By16: ; IN: ACIX=dividend, DE=divisor ; OUT: ACIX=quotient, DE=divisor, HL=remainder, B=0 ld hl,0 ld b,32 Div32By16_Loop: add ix,ix rl c rla adc hl,hl jr c,Div32By16_Overflow sbc hl,de jr nc,Div32By16_SetBit add hl,de djnz Div32By16_Loop ret Div32By16_Overflow: or a sbc hl,de Div32By16_SetBit: .db $DD,$2C ; inc ixl, change to inc ix to avoid undocumented djnz Div32By16_Loop ret
32/16 division, Speed Optimized
At the cost of 30 bytes (exactly double the size), we can save 600cc on average, a 19.5% speed up. However, the inputs are slightly different here!
div32_16: ;DEIX/BC -> HLIX remainder DE ;170+4*div32_16_sub8 ;min: 2182cc ;max: 2790cc ;avg: 2462cc ;60 bytes ; Negate BC to allow add instead of sbc xor a ; 4 ; Need to set HL to 0 anyways, so save 2cc and a byte ld h,a ; 4 ld l,a ; 4 sub c ; 4 ld c,a ; 4 sbc a,a ; 4 sub b ; 4 ld b,a ; 4 ld a,d ; 4 call div32_16_sub8 ; 17 rla ; 4 ld d,a ; 4 ld a,e ; 4 call div32_16_sub8 ; 17 rla ; 4 ld e,a ; 4 ld a,ixh ; 8 call div32_16_sub8 ; 17 rla ; 4 ld ixh,a ; 8 ld a,ixl ; 8 call div32_16_sub8 ; 17 rla ; 4 ld ixl,a ; 8 ex de,hl ; 4 ret ; 10 div32_16_sub8: ;119+8*div32_16_sub ;min: 503cc ;max: 655cc ;avg: 573cc call +_ _: ;17+2(17+2(div32_16_sub))) call +_ _: ;17+2(div32_16_sub) call div32_16_sub div32_16_sub: ;48+{8,0+{0,19}} ;min: 48cc ;max: 67cc ;avg: 56.75cc rla ; 4 adc hl,hl ; 15 jr c,+_ ;12/7 add hl,bc ; 11 ret c ;11/5 sbc hl,bc ; 15 ret ; 10 _: add hl,bc ; 11 scf ; 4 ret ; 10
Rounded 16/8 division
The following routine divides hl by c and places the rounded quotient in hl and twice the prerounded remainder in a.
RoundHL_Div_C: xor a ld b, 16 _loop: add hl, hl rla jr c, $+5 cp c jr c, $+4 sub c inc l djnz _loop ;This part is the rounding add a,a cp c ret c inc hl ret