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base:signed_8bit_divied_by_2_arithmetic_shift_right

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 base:signed_8bit_divied_by_2_arithmetic_shift_right [2015-04-17 04:33] base:signed_8bit_divied_by_2_arithmetic_shift_right [2015-04-17 04:33] (current) Line 1: Line 1: + ====== 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: + + + 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 + + + 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: + + + 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 + ++ + + + This would need 13 cycles in average, but it can be done faster as well using the first method: + + + cmp #\$80  ;copy bit 7 to carry + bcc +     ;positive number? + adc #\$00  ;add one (carry still set) + sec       ;fix clobbered carry + + + ror       ;shift + + + 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: + + + tax + lda divtable,x + + + ===== 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.