6502 assembly optimisations

This page is about optimisations that are possible in assembly language, or various things one programmer has to keep in mind to make his code as optimal as possible.

There is two major kind of optimisations: Optimisation for speed (code executes in fewer cycles) and optimisation for size (the code takes fewer bytes).

There is also some other kinds of optimisations, such as constant-executing-time optimisation (code execute in a constant number of cycle no matter what it has to do), or RAM usage optimisation (use as few variables as possible). Because those optimisations have more to do with the algorithm than with its implementation in assembly, only speed and size optimisations will be discussed in this article.

Optimise both speed and size of the code

Avoid a jsr + rts chain

A tail call occurs when a subroutine finishes by calling another subroutine. This can be optimised into a JMP instruction:

MySubroutine
  lda Foo
  sta Bar
  jsr SomeRandomRoutine
  rts

becomes :

MySubroutine
  lda Foo
  sta Bar
  jmp SomeRandomRoutine

Savings : 9 cycles, 1 byte

Split word tables in high and low components

This optimisation is not human friendly, makes the source code much bigger, but still makes the compiled size smaller and faster:

Example
  lda FooBar
  asl A
  tax
  lda PointerTable,X
  sta Temp
  lda PointerTable+1,X
  sta Temp+1
  ....

PointerTable
  .dw Pointer1, Pointer2, ....

Becomes :

Example
  ldx FooBar
  lda PointerTableL,X
  sta Temp
  lda PointerTableH,X
  sta Temp+1
  ....

PointerTableL
  .byt <Pointer1, <Pointer2, ....

PointerTableH
  .byt >Pointer1, >Pointer2, ....

Some assemblers may have a way to implement a macro to automatically make the table coded like this (Unofficial MagicKit Assembler is one such program).

Savings : 2 bytes, 4 cycles

Use Jump tables with RTS instruction instead of JMP indirect instruction

The so-called RTS Trick is a method of implementing jump tables by pushing a subroutine's entry point to the stack.

; This:

  ldx JumpEntry
  lda PointerTableH,X
  sta Temp+1
  lda PointerTableL,X
  sta Temp
  jmp [Temp]

; becomes this:

  ldx JumpEntry
  lda PointerTableH,X
  pha
  lda PointerTableL,X
  pha
  rts

Note that PointerTable entries must point to one byte before the intended target when the RTS trick is used, because RTS will add 1 to the offset.

If Temp is outside zero page, this saves 6 bytes and 1 cycle. If Temp is within zero page, this saves 4 bytes and costs 1 cycle.

In either case, it frees up the RAM occupied by Temp for other uses so long as the use of the RTS Trick does not occur at peak stack depth. In addition, it's reentrant, which means the NMI and IRQ handlers do not need dedicated 2-byte RAM allocations for their own Temp variables.

Combining this with the tail call optimization squeezes 1 more byte and 9 more cycles:

; This:

  jsr SomeOtherFunction  ; MUST NOT modify JumpEntry
  ldx JumpEntry
  lda PointerTableH,X
  pha
  lda PointerTableL,X
  pha
  rts

; Becomes this:

  ldx JumpEntry
  lda PointerTableH,X
  pha
  lda PointerTableL,X
  pha
  jmp SomeOtherFunction

Here, the CPU runs SomeOtherFunction, then returns to the function from the jump table, then returns to what called this dispatcher. One example is a SomeOtherFunction that mixes player input into the random number generator's entropy pool before calling the routine for a particular game state.

Use a macro instead of a subroutine which is only called once

What is the point to call a subroutine if you only call it at a single place? It would be more optimal to just insert the code where the subroutine is called. However this makes the code less structured and harder to understand. Inline expansion of a subroutine into another subroutine can be done with a macro. One drawback is that if the subroutine is called in a loop, it may require some JMPing to work around the 128-byte limit on branch length.

How macros are used depends on the assembler so no code examples will be placed here to avoid further confusion. In C, the static inline keyword acts as a hint to expand a function as a macro.

Savings : 4 bytes, 12 cycles.

Arithmetic shift right

Compact way to divide a variable by 2 but keep its sign:

   cmp #$80
   ror A

Easily test 2 upper bits of a variable

    lda FooBar
    asl A         ;C = b7, N = b6

Alternative:

    bit Foobar    ;N = b7, V = b6, regardless of the value of A.

This can be e.g. used to poll the sprite-0-hit flag in $2002.

Negating a value without temporaries

   eor #$FF
   clc
   adc #1

Avoiding the need for CLC/SEC with ADC/SBC

If you are using ADC #imm, and you know your carry is already cleared, you do not need to do CLC. However, if you know that carry is set (for example, your code is located in a branch that is only ever entered with a BCS instruction), you can still avoid using CLC by just doing ADC #(value-1). The PLOT subroutine in the Apple II Monitor uses this.

Similarly for SBC #imm: If you know know that carry is clear, you can still avoid using SEC by just doing SBC #(value+1).

Test bits in decreasing order

   lda foobar 
   bmi bit7_set 
   cmp #$40  ; we know that bit 7 wasn't set 
   bcs bit6_set 
   cmp #$20 
   bcs bit5_set 
             ; and so on

Or if you do not need to preserve the value of A:

   lda foobar 
   asl
   bcs bit7_set 
   asl
   bcs bit6_set 
   asl
   bcs bit5_set 
             ; and so on

This saves one byte per comparison, but 2 cycles more are used because of the extra ASL.

Test bits in increasing order

   lda foobar 
   lsr
   bcs bit0_set
   lsr
   bcs bit1_set
   lsr
   bcs bit2_set
             ; and so on

Note: This does not preserve the value of A.

Test bits without destroying the accumulator

The AND instruction can be used to test bits, but this destroy the value in the accumulator. The BIT can do this but it has no immediate adressing mode. A way to do it is to look for an opcode that has the bits you want to test, and use bit $xxxx on this opcode.

Example
   lda foobar
   and #$30
   beq bits_clear
   lda foobar
   ....

bits_clear
   lda foobar
   .....

becomes :

Example
   lda foobar
   bit _bmi_instruction ;equivalent to and #$30 but preserves A
   beq bits_clear
   ....

bits_clear
   .....

anywhere_in_the_code
    ....
_bmi_instruction    ;The BMI opcode = $30
    bmi somewhere

Savings : 2 cycles, 3 bytes

Use opposite rotate instead of a great number of shifts

To retrieve the 3 highest bits of a value in the low positions, you might be tempted to do 5 LSRs in a row. However, if you do not need the 5 top bits to be cleared, this is more efficient:

  lda value   ; got: 76543210 c
  rol         ; got: 6543210c 7
  rol         ; got: 543210c7 6 
  rol         ; got: 43210c76 5
  rol         ; got: 3210c765 4
  ; Only care about these ^^^

It works the same for replacing 5 ASLs with 4 RORs.

To replace 6 ASLs you can use 3 RORs:

  lda value   ; got: 76453210 c
  ror         ; got: c7654321 0
  ror         ; got: 0c765432 1
  ror         ; got: 10c76543 2
  and #$C0    ; got: 10------

Optimise speed at the expense of size

Those optimisations will make code faster to execute, but use more ROM. Therefore, it is useful in NMI routines and other things that need to run fast.

Use identity look-up table instead of temp variable

Example
    ldx Foo
    lda Bar
    stx Temp
    clc
    adc Temp    ;A = Foo + Bar

becomes :

Example
    ldx Foo
    lda Bar
    clc
    adc Identity,X    ;A = Foo + Bar

Identity
    .byt $00, $01, $02, $03, .....

If your program is very large, it is possible that this way uses slightly less ROM; also, it might save one byte of RAM in some circumstances.

Savings : 2 cycles

Use look-up table to shift left 4 times

Provided that the high nibble is already cleared, you can shift left by 4 by making a multiplication look-up table.

Example:
  lda rownum
  asl A
  asl A
  asl A
  asl A
  rts

becomes

Example:
  ldx rownum
  lda times_sixteen,x
  rts

times_sixteen:
  .byt $00, $10, $20, $30, $40, $50, $60, $70
  .byt $80, $90, $A0, $B0, $C0, $D0, $E0, $F0

In very large programs, this might save some ROM space. However, it will use up the X register, so it might not be ideal.

Savings: 4 cycles

Optimise code size at the expense of cycles

Those optimisations will produce code that is smaller but takes more cycles to execute. Therefore, it can be used in the parts of the program that do not have to be fast.

Use the stack instead of a temp variable

Example
   lda Foo
   sta Temp
   lda Bar
   ....
   ....
   lda Temp   ;Restores Foo
   .....

becomes:

Example
   lda Foo
   pha
   lda Bar
   ....
   ....
   pla   ;Restores Foo
   .....

Savings : 2 bytes.

Use an "intelligent" argument system

Each time a routine needs multiple bytes of arguments (>3) it's hard to code it without wasting a lot of bytes.

Example
   lda Argument1
   sta Temp
   lda Argument2
   ldx Argument3
   ldy Argument4
   jsr RoutineWhichNeeds4Args
   .....

Becomes something like:

Example
   jsr PassArguments
   .dw RoutineWhichNeeds4Args
   .db Argument1, Argument2, Argument3, Argument4
   .db $00
   ....

PassArguments
   pla 
   tay 
   pla 
   pha                    ; put the high byte back 
   sta pointer+1 
   ldx #$00 
   beq SKIP 
LOOP 
   sta parameters,x 
   inx 
SKIP 
   iny                    ; pointing one short first pass here fixes that 
   lda (pointer),y 
   bne LOOP      
   iny 
   lda (pointer),y 
   beq LOOP

   dey                    ; fix the return address guess we can't return to a 
                         ;  break        
   tya 
   pha 
   jmp (parameters)

Syscalls in Apple ProDOS and FDS BIOS work this way.

Savings : Complicated to estimate - only saves bytes if the trick is used fairly often across the program, in order to compensate for the size of the PassArguments routine.

Using relative branch instruction instead of absolute

If the state of one of the processor flags is already known at this point and the branch target is not too far away, then you can use a condition branch instruction.

Savings : 1 byte.

See also

• Synthetic instructions