Divide by 3: Difference between revisions
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== 16-bit dividend, no remainder == | == 16-bit dividend, no remainder == | ||
<source lang=" | <source lang="6502"> | ||
; divide 16 bit number by 3 by multiplying by 1/3 | ; divide 16 bit number by 3 by multiplying by 1/3 | ||
; enter with | ; enter with | ||
| Line 52: | Line 52: | ||
== 8-bit dividend, no remainder == | == 8-bit dividend, no remainder == | ||
<source lang=" | <source lang="6502"> | ||
; enter with number to be divided in A | ; enter with number to be divided in A | ||
; answer returned in A | ; answer returned in A | ||
Revision as of 07:42, 25 August 2013
A lot of times, you need to divide something by 3. One way is to multiply by $55.
16-bit dividend, no remainder
<source lang="6502">
; divide 16 bit number by 3 by multiplying by 1/3 ; enter with ; A containing the hi byte of the number to be divided by 3 ; Y containing the lo byte of the number to be divided by 3 ; the hi byte of the partial product is kept in A or saved ; on the stack when neccessary ; the product (N/3 quotient) is returned hi byte in A, ; lo byte in Y
.proc div3_ay
; save the number in lo_temp, hi_temp
sty lo_temp sty lo_product sta hi_temp
ldy #$09 clc bcc ENTER
; each pass through loop adds the number in ; lo_temp, hi_temp to the partial product and ; then divides the partial product by 4
LOOP:
pha lda lo_product adc lo_temp sta lo_product pla adc hi_temp
ENTER:
ror ror lo_product lsr ror lo_product dey bne LOOP ldy lo_product rts
.endproc </source>
8-bit dividend, no remainder
<source lang="6502">
; enter with number to be divided in A ; answer returned in A
.proc div3_a
sta temp lsr lsr adc temp ror lsr adc temp ror lsr adc temp ror lsr adc temp ror lsr rts
.endproc </source>
