Divide by 3: Difference between revisions
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== 16-bit dividend, no remainder == | == 16-bit dividend, no remainder == | ||
< | <pre> | ||
; 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 48: | Line 48: | ||
rts | rts | ||
.endproc | .endproc | ||
</ | </pre> | ||
== 8-bit dividend, no remainder == | == 8-bit dividend, no remainder == | ||
< | <pre> | ||
; enter with number to be divided in A | ; enter with number to be divided in A | ||
; answer returned in A | ; answer returned in A | ||
| Line 73: | Line 73: | ||
rts | rts | ||
.endproc | .endproc | ||
</ | </pre> | ||
==References== | ==References== | ||
*[http://forums.nesdev.org/viewtopic.php?p=74242#p74242 bogax's post] | *[http://forums.nesdev.org/viewtopic.php?p=74242#p74242 bogax's post] | ||
Revision as of 04:17, 11 September 2014
A lot of times, you need to divide something by 3. One way is to multiply by $55.
16-bit dividend, no remainder
; 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
8-bit dividend, no remainder
; 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
