Z80 Optimization

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Introduction

Sometimes it is needed some extra speed in ASM or make your game smaller to fit on the calculator. Examples: mapping, grayscale and 3D graphics.

Registers and Memory

General algorithm improvements and correct use of registers.

General use of registers:

a accumulator
b counter

hl 16-bit accumulator/pointer to memory
de pointer of destination in memory

Stack

When you run out of registers, stack may offer an interesting alternative to fixed RAM location for temporary storage.

Allocation

You can either allocate stack space with repeated push, which allows to initialize the data but restricts the allocated space to multiples of 2. An alternate way is to allocate uninitialized stack space (hl may be replaced with an index register) :

 ; allocates 7 bytes of stack space : 5 bytes, 27 T-states instead of 4 bytes, 44 T-states with 4 push which would have forced the alloc of 8 bytes
 ld hl, -7
 add hl, sp
 ld sp, hl

Access

The most common way of accessing data allocated on stack is to use an index register since all allocated "variables" can be accessed without having to use inc/dec but this is obviously not a strict requirement. Beware though, using stack space is not always optimal in terms of speed, depending (among other things) on your register allocation strategy :

 ; 4 bytes, 19 T-states
 ld c, (ix + n)   ; n is an immediate value in -128..127
 
 ; 4 bytes, 17 T-states, destroys a
 ld a, (somelocation)
 ld c, a

If your needs go beyond simple load/store however, this method start to show its real power since it vastly simplify some operations that are complicated to do with fixed storage location (and generally screw up register in the process).

 ; 3 bytes, 19 T-states
 cp (ix + n)

 sub (ix + n)
 sbc a, (ix + n)
 add a, (ix + n)
 adc a, (ix + n)

 inc (ix + n)
 dec (ix + n)

 and (ix + n)
 or (ix + n)
 xor (ix + n)

 ; 4 bytes, 23 T-states
 rl (ix + n)
 rr (ix + n)
 rlc (ix + n)
 rrc (ix + n)
 sla (ix + n)
 sra (ix + n)
 sll (ix + n)
 srl (ix + n)
 bit k, (ix + n)   ; k is an immediate value in 0..7
 set k, (ix + n)
 res k, (ix + n)

Again, choose wisely between hl and an index register depending on the structure of your data the smallest/fastest allocation solution may vary (hl equivalent instructions are generally 2 bytes smaller and 12 T-states faster but do not allow indexing so may require intermediate inc/dec).

Deallocation

If you want need to pop an entry from the stack but need to preserve all registers remember that sp can be incremented/decremented like any 16bit register :

 ; drops the top stack entry : waste 1 byte and 2 T-states but may enable better register allocation...
 inc sp
 inc sp

If you have a large amount of stack space to drop and a spare 16 bit register (hl, index, or de that you can easily swap with hl) :

 ; drop 16 bytes of stack space : 5 bytes, 27 T-states instead of 8 bytes, 80 T-states for 8 pop
 ld hl, 16
 add hl, sp
 ld sp, hl

The larger the space to drop the more T-states you will save, and at some point you'll start saving space as well (beyond 8 bytes)

Shadow registers

In some rare cases, when you run out of registers and cannot to either refactor your algorithm(s) or to rely on RAM storage you may want to use the shadow registers : af', bc', de' and hl'

These registers behave like their "standard" counterparts (af, bc, de, hl) and you can swap the two register sets at using the following instructions :

 ex af, af'  ; swaps af and af' as the mnemonic indicates

 exx         ; swaps bc, de, hl and bc', de', hl'

Shadow registers can be of a great help but they come with two drawbacks :

they cannot coexist with the "standard" registers : you cannot use ld to assign from a standard to a shadow or vice-versa. Instead you must use nasty constructs such as :

 ; loads hl' with the contents of hl
 push hl
 exx
 pop hl

they require interrupts to be disabled since they are originally intended for use in Interrupt Service Routine. There are situations where it is affordable and others where it isn't. Regardless, it is generally a good policy to restore the previous interrupt status (enabled/disabled) upon return instead of letting it up to the caller. Hopefully it s relatively easy to do (though it does add 4 bytes and 29/33 T-states to the routine) :

  ld a, i  ; this is the core of the trick, it sets P/V to the value of IFF so P/V is set iff interrupts were enabled at that point
  push af  ; save flags
  di       ; disable interrupts
  
  ; do something with shadow registers here

  pop af   ; get back flags
  ret po   ; po = P/V reset so in this case it means interrupts were disabled before the routine was called
  ei       ; re-enable interrupts
  ret

Small Tricks

Note that the following tricks act much like a peephole optimizer and are the last optimization step : remember to first optimize your algorithm and register allocation before applying any of the following if you really want the fastest speed and the smallest code.

Optimize size and speed

Loading stuff

;Instead of:
 cp 0
;Use
 or a
; -> save 1 byte and 3 T-states

;Instead of:
 ld a,0
;Try this:
 xor a    ;disadvantages: changes flags
;or
 sub a    ;disadvantages: changes flags
; -> save 1 byte and 3 T-states

;Instead of
	ld b,$20
	ld c,$30
;try this
	ld bc,$2030
;or this
	ld bc,(b_num * 256) + c_num
; -> save 1 byte and 4 T-states

;Instead of
	xor a
	ld (data1),a
	ld (data2),a
	ld (data3),a
	ld (data4),a
	ld (data5),a ;if data1 to data5 are one after the other
;try this
	ld hl,data1
	ld de,data1+1
	xor a
	ld (hl),a
	ld bc,4
	ldir
; -> save 3 bytes for every ld (dataX),a

;Instead of
	ld a,(var)
	inc a
	ld (var),a
;try this ;if hl is not tied up and all you do is check flags, use indirection:
	ld hl,var
	inc (hl)
	ld a,(hl)

; Instead of :
 ld a, (hl)
 ld (de), a
 inc hl
 inc de
; Use :
 ldi
 inc bc
; -> save 1 byte and 4 T-states

Math tricks

;Instead of
    ld de,-767
    add hl,de
;try this
    dec h  ; -256
    dec h  ; -512
    dec h  ; -768
    inc hl  ; -767

;Instead of
	srl a
	srl a
	srl a
;try this
	rrca
	rrca
	rrca
	and %00011111

;Instead of
loop:
 ld a,2
;code1
 ld a,0
;code2

;try this
 ld a,2
loop:
;code1
 xor $01
;code2

Looping with 16 bit counter

There are two ways to make loops with a 16bit counter :

the naive one, which results in smaller code but increased loop overhead (24 * n T-states) and destroys a

  ld bc, ...
loop:
  ; loop body here
 
  dec bc
  ld  a, b
  or  c
  jp  nz,loop

the slightly trickier one, which takes a couple more bytes but has a much lower overhead (12 * n + 14 * (n / 16) T-states)

  dec  de
  ld  b, e
  inc  b
  inc  d
loop2:
  ; loop body here
  
  djnz loop2
  dec  d
  jp  nz,loop2

The rationale behind the second method is to reduce the overhead of the "inner" loop as much as possible and to use the fact that when b gets down to zero it will be treated as 256 by djnz.

You can therefore use the following macros for setting proper values of 8bit loop counters given a 16bit counter in case you want to do the conversion at compile time :

  #define inner_counter8(counter16) (((counter16) - 1) & 0xff) + 1
  #define outer_counter8(counter16) (((counter16) - 1) >> 8) + 1

Size vs. Speed

The classical problem of optimization in computer programming, Z80 is no exception. In ASM most frequently size is what matters because generally ASM is fast enough and it is nice to give a user a smaller program that doesn't use up most RAM. Speed can also be needed...

General Unrolling You can unroll some loop several times instead of looping, this is used frequently on math routines of multiplication.

Unroll commands

; "Classic" way : ~21 T-states per byte copied
 ld hl,src
 ld de,dest
 ld bc,size
 ldir

; Unrolled : (16 * size + 10) / n -> ~18 T-states per byte copied when unrolling 8 times
 ld hl,src
 ld de,dest
 ld bc,size  ; if the size is not a multiple of the number of unrolled ldi then a small trick must be used to jump appropriately inside the loop for the first iteration
loopldi:
 ldi
 ldi
 ldi
 ldi
 ldi
 ldi
 ldi
 ldi
 jp pe, loopldi    ; jp used as it is faster and in the case of a loop unrolling we assume speed matters more than size

This unroll of ldi also works with outi and ldr.

Setting flags

In some occasion you might want to selectively set/reset a flag.

Here are the most common uses :

; set Carry flag
scf

; reset Carry flag (alters Sign and Zero flags as defined)
or a

; alternate reset Carry flag (alters Sign and Zero flags as defined)
and a

; set Zero flag (resets Carry flag, alters Sign flag as defined)
cp a

; reset Zero flag (alters a, reset Carry flag, alters Sign flag as defined)
or 1

; set Sign flag (negative) (alters a, reset Zero and Carry flags)
or $80

; reset Sign flag (positive) (set a to zero, set Zero flag, reset Carry flag)
xor a

Other possible uses (much rarer) :

; Set parity/overflow (even):
xor a

Reset parity/overflow (odd):
sub a

; set half carry (hardly ever useful but still...)
and a

; reset half carry (hardly ever useful but still...)
or a

As you can see these are extremely simple, small and fast ways to alter flags which make them interesting as output of routines to indicate error/success or other status bits that do not require a full register.

Were you to use this, remember that these flag (re)setting tricks frequently overlap so if you need a special combination of flags it might require slightly more elaborate tricks. As a rule of a thumb, always alter the carry last in such cases because the scf and ccf instructions do not have side effects.

Acknowledgements

fullmetalcoder
Galandros

Z80 Optimization

Contents

Introduction

Registers and Memory

Stack

Allocation

Access

Deallocation

Shadow registers

Small Tricks

Optimize size and speed

Loading stuff

Math tricks

Looping with 16 bit counter

Size vs. Speed

Setting flags

Related topics

Acknowledgements

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