base:signed_8bit_divied_by_2_arithmetic_shift_right

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+ | ====== Arithmetic shift right ====== | ||

+ | By Bitbreaker | ||

+ | |||

+ | If we want to divide by the power of 2 we usually shift right. That is fine with unsigned numbers, but for signed numbers we would need a arithemtic shift right, that we have no opcode for. | ||

+ | So we need a trick to preserve bit 7 in another way: | ||

+ | |||

+ | <code> | ||

+ | cmp #$80 ;copy bit 7 to carry (i love that trick also for other situations where A should not be clobbered) | ||

+ | ror ;now rotate and we successfully preserved bit 7 | ||

+ | </code> | ||

+ | |||

+ | Easy like that, done in 4 cycles! However keep in mind that this way the result is rounded down and not up when dealing with negative numbers. So for more accuracy you might prefer: | ||

+ | |||

+ | <code> | ||

+ | bpl + ;positive number? | ||

+ | eor #$ff ;a = 0 - a to get a positive number | ||

+ | clc | ||

+ | adc #$01 | ||

+ | lsr ;shift right | ||

+ | eor #$ff ;make it negative again | ||

+ | clc | ||

+ | adc #$01 | ||

+ | jmp ++ | ||

+ | + | ||

+ | lsr | ||

+ | ++ | ||

+ | </code> | ||

+ | |||

+ | This would need 13 cycles in average, but it can be done faster as well using the first method: | ||

+ | |||

+ | <code> | ||

+ | cmp #$80 ;copy bit 7 to carry | ||

+ | bcc + ;positive number? | ||

+ | adc #$00 ;add one (carry still set) | ||

+ | sec ;fix clobbered carry | ||

+ | + | ||

+ | ror ;shift | ||

+ | </code> | ||

+ | |||

+ | Then we would need 8,5 cycles in average. Still pretty nice. If you need it even faster (6 cycles) and use a lot of div by 2, then a lookup table might also be an option: | ||

+ | |||

+ | <code> | ||

+ | tax | ||

+ | lda divtable,x | ||

+ | </code> | ||

+ | |||

+ | ===== And how's about shifting left? ===== | ||

+ | |||

+ | |||

+ | If you intend to multiply a signed number by two, just do a = a + a and you are fine. |

base/signed_8bit_divied_by_2_arithmetic_shift_right.txt ยท Last modified: 2015-04-17 04:33 (external edit)