Difference between revisions of "Z80 Optimization"

Revision as of 06:43, 4 November 2009

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

;smaller, destroys a
loop:
  ld  a, d
  or  e
  jp  nz,loop
;faster, uses b, bigger
  dec  de
  ld  b, e
  inc  b
  inc  d
loop2:
  djnz loop2
  dec  d
  jp  nz,loop2
 

Size vs. Speed

;Unroll ldir to ldi's and make use of flag po
 

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.

Related topics

• MaxCodez topic
• ticalc docs
• doc by some TI programmer in Balley Z80

Acknowledgements

• fullmetalcoder and Galandros contributions