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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:

    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
   adc #$01
   lsr       ;shift right
   eor #$ff  ;make it negative again
   adc #$01
   jmp ++

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
   ror       ;shift right as before
   bpl +     ;if positive number then skip
   adc #$00  ;else round up
+  ...

Then we would need 7,5 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:

   lda divtable,x

When using illegal op-codes you might also want to use this version that gives you always a cleared carry afterwards:

   anc #$fe  ;copy N to C and mask out bit 0
   ror       ;rotate, carry will now be clear

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_divide_by_2_arithmetic_shift_right.txt · Last modified: 2015-04-17 04:33 by