Synthetic instructions: Difference between revisions

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(Added sign extension, gave unnamed labels names)
(Change wording on second bit counting snippet from "cleared" to "shifted out")
 
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  sec
  sec
  adc #0
  adc #0
Or if we know carry is clear coming in:
; A = -A
eor #$FF
adc #1


== Reverse subtraction ==
== Reverse subtraction ==
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  neg:
  neg:


Constant-time version:
        ; calculates upper byte of sign-extension of A, alternate version
        and #$80
        bpl pos
        lda #$ff
pos:
 
This constant-time version (7 bytes, 8 cycles) destroys the carry, so don't try using it in the middle of a multi-byte addition:
 
  asl a    ;  cpx #$80 or cpy #$80 is also possible
  lda #$00
  adc #$FF  ; C is unchanged and A = $00 if C was set or $FF if C was clear
  eor #$FF  ; now all bits of A are set to what bit 7 was
 
If you're just trying to add an 8-bit delta to a 16-bit value, you could try subtracting 256 from the value by decrementing the high byte if the value is negative and then adding as if it were unsigned.


        asl a
  lda delta_value
        lda #0
  bpl notneg
        adc #$FF
  dec value_hi
        eor #$FF
notneg:
  clc
  adc value_lo
  sta value_lo
  lda #0
  adc value_hi
  sta value_hi
 
[http://forums.nesdev.org/viewtopic.php?p=124949#p124949 This forum thread] describes addition of signed deltas with clamping.


== Arithmetic shift right ==
== Arithmetic shift right ==
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A could be saved and restored using other methods, like TAX and TXA, etc.
A could be saved and restored using other methods, like TAX and TXA, etc.
Branching can also be used:
    ; 8-bit rotate right A
    lsr a
    bcc skip
    adc #$80-1 ; carry is set, so will add extra 1
skip:


If the operand is in memory:
If the operand is in memory:
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  lsr a
  lsr a
  ror Value
  ror Value
To rotate A left twice:<ref name="wilson">http://www.6502.org/source/general/SWN.html</ref>
; double 8-bit rotate left A
asl a
adc #$80
rol a
== Nybble swap ==
The nybbles of A can be swapped (e.g. $1F becomes $F1) in 8 bytes and 12 cycles by doing the "double 8-bit rotate left A" trick above twice:<ref name="wilson"/>
; swap nybbles of A
asl a
adc #$80
rol a
asl a
adc #$80
rol a


== 16-bit increment and decrement ==
== 16-bit increment and decrement ==
Line 117: Line 171:
  nodec:  dec Word
  nodec:  dec Word


16-bit increment shows even more advantage when used to control a loop, because the 16-bit decrement conveniently leaves the zero flag set at the end ''only'' if the entire 16-bit value is zero.
16-bit increment shows even more advantage when used to control a loop, because the 16-bit increment conveniently leaves the zero flag set at the end ''only'' if the entire 16-bit value is zero.


== X/Y as Operand ==
== X/Y as Operand ==
''Main article: [[Identity table]]''


Normally X and Y cannot be used as an operand to an instruction operating on A. For example, CMP X isn't possible. Where X or Y needs to be used in such a way, they are usually saved to a temporary variable:
Normally X and Y cannot be used as an operand to an instruction operating on A. For example, CMP X isn't possible. Where X or Y needs to be used in such a way, they are usually saved to a temporary variable:
Line 156: Line 212:


See [[Jump table]] for further explanation and alternate approaches.
See [[Jump table]] for further explanation and alternate approaches.
== Toggle carry ==
Invert the carry flag without affecting A (N and Z flags are destroyed, however):<ref>[https://archive.org/details/eor6502 EOR #$FF - 6502 Ponderables and Befuddlements], puzzle $08</ref>
rol a
eor #$01
ror a
== Set overflow flag ==
The overflow flag can be set by using the BIT instruction with any byte that has the second-most significant bit set. The N and Z flags are destroyed, however.<ref>[https://archive.org/details/eor6502 EOR #$FF - 6502 Ponderables and Befuddlements], puzzle $09</ref>
    bit label
    ...
label:
    rts        ; opcode $60 (%01100000)
== Test whether A is in range ==
Test whether A (unsigned) is between constants ''min'' and ''max'' (inclusive). A is destroyed.<ref>[https://archive.org/details/eor6502 EOR #$FF - 6502 Ponderables and Befuddlements], puzzle $20</ref>
Set carry flag if A is in range, otherwise clear carry:
clc
adc #$ff-max
adc #max-min+1
Clear carry flag if A is in range, otherwise set carry:
sec
sbc #min
sbc #max-min+1
== Count bits set ==
Count the number of bits set in ''myVal''. The result will be in X. ''myVal'' is destroyed. (On each round of the loop, the least significant set bit is cleared.)<ref>[https://archive.org/details/eor6502 EOR #$FF - 6502 Ponderables and Befuddlements], puzzle $29</ref>
    ldx #0
    lda myVal
    beq end
loop:
    inx
    dec myVal
    and myVal
    sta myVal
    bne loop
end:
Count the number of bits set in A. The result will be in X and A. (On each round of the loop, the most significant bit is shifted out.)<ref>[http://forum.6502.org/viewtopic.php?t=1206 6502.org Forum - Bit Counting]</ref>
    ldx #$ff  ; Set count to -1
incr:
    inx        ; Add one to count
loop:
    asl        ; Shift by one bit
    bcs incr  ; Count one bits
    bne loop  ; Repeat till zero
    txa        ; Move count to A
== See also ==
* [[6502 assembly optimisations]]


== External links ==
== External links ==
*[http://www.cs.umd.edu/class/spring2003/cmsc311/Notes/Mips/pseudo.html Pseudoinstructions in MIPS]
*[http://www.cs.umd.edu/class/spring2003/cmsc311/Notes/Mips/pseudo.html Pseudoinstructions in MIPS]
== References ==
<references/>

Latest revision as of 09:13, 1 July 2024

There are several additional instructions that would be nice to have on the NES. Even though not present, they can be synthesized using existing instructions, resulting in synthetic instructions or pseudoinstructions. If turned into assembler macros, they can be used almost as if they were natively supported. Being able to think of them as native instructions lightens the mental load when programming, because instructions are an important tool for abstraction. Even without making the following into macros, after reading them you will be more likely to think of one of these while coding, saying "I need a subtract-from instruction here".

Negate A

Many processors have a native negate instruction, which subtracts the value from zero. Here we must manually calculate the two's complement of A, which involves a one's complement and increment:

; A = -A
eor #$FF
sec
adc #0

Or if we know carry is clear coming in:

; A = -A
eor #$FF
adc #1

Reverse subtraction

Using SBC, a value can be subtracted from A, but there's no direct way to subtract A from some value. The ARM instruction set includes instructions RSB (reverse subtract) and RSC (reverse subtract with carry) that negate the register and add the value. The 6502 can do this too:

; A = Value - A
eor #$FF
sec
adc Value

As a special case, if we want to subtract A from 255, we can just do

; A = 255 - A
eor #$FF

This also shows another way of understanding the general subtract from; we first subtract A from 255, then add one, so it's as if we subtracted from 256, which is the same as subtracting from zero, since A is only 8 bits.

Sign-extend

When increasing the number of bits in a signed value, the new high bits are filled with copies of the sign bit. This is called sign extension. For example, sign-extending the 8-bit value $80 (-128) to 16 bits sets the new bits, resulting in $FF80; sign-extending $7F (+127) to 16 bits results in $007F. The following sequences calculate the upper byte of the sign-extended value.

        ; calculates upper byte of sign-extension of A
        ora #$7F
        bmi neg
        lda #0
neg:
        ; calculates upper byte of sign-extension of A, alternate version
        and #$80
        bpl pos
        lda #$ff
pos:

This constant-time version (7 bytes, 8 cycles) destroys the carry, so don't try using it in the middle of a multi-byte addition:

  asl a     ;  cpx #$80 or cpy #$80 is also possible
  lda #$00
  adc #$FF  ; C is unchanged and A = $00 if C was set or $FF if C was clear
  eor #$FF  ; now all bits of A are set to what bit 7 was

If you're just trying to add an 8-bit delta to a 16-bit value, you could try subtracting 256 from the value by decrementing the high byte if the value is negative and then adding as if it were unsigned.

  lda delta_value
  bpl notneg
  dec value_hi
notneg:
  clc
  adc value_lo
  sta value_lo
  lda #0
  adc value_hi
  sta value_hi

This forum thread describes addition of signed deltas with clamping.

Arithmetic shift right

The ARM instruction set has an arithmetic right shift, which doesn't alter the sign (top) bit. This shift is used to divide a signed value by two. But the 6502 lacks this instruction; LSR doesn't work because it shifts the sign bit to the right, then clears it.

To implement this, we need carry set to the sign bit, then we can use ROR. CMP #$80 performs this task; if the value is less than $80, carry is cleared, otherwise it's set:

; Arithmetic shift right A
cmp #$80
ror a

If the operand is in memory, we just use ASL to move the sign bit into carry:

; Arithmetic shift right Value
lda Value
asl a
ror Value

8-bit rotate

The 65xx series rotate instructions are all 9-bit, not 8-bit as often imagined. If they really were 8-bit, then eight ROR or ROL instructions in a row would leave A with its original value. In actuality, nine are required to do so, since the carry acts as a ninth bit of A.

Similar to arithmetic right shift, we must set carry to the top or bottom bit in advance of the rotate. For 8-bit rotate left, it's simple:

; 8-bit rotate left A
cmp #$80
rol a

; alternate method
asl a
adc #0

For 8-bit rotate right, we must save and restore A:

; 8-bit rotate right A
pha
lsr a
pla
ror a

A could be saved and restored using other methods, like TAX and TXA, etc.

Branching can also be used:

    ; 8-bit rotate right A
    lsr a
    bcc skip
    adc #$80-1 ; carry is set, so will add extra 1
skip:

If the operand is in memory:

; 8-bit rotate left Value
lda Value
asl a
rol Value

; 8-bit rotate right Value
lda Value
lsr a
ror Value

To rotate A left twice:[1]

; double 8-bit rotate left A
asl a
adc #$80
rol a

Nybble swap

The nybbles of A can be swapped (e.g. $1F becomes $F1) in 8 bytes and 12 cycles by doing the "double 8-bit rotate left A" trick above twice:[1]

; swap nybbles of A
asl a
adc #$80
rol a
asl a
adc #$80
rol a

16-bit increment and decrement

Incrementing/decrementing a 16-bit value involves first adjusting the low byte, then adjusting the high byte if necessary. Increment is simpler, since the high byte is adjusted when the low byte wraps around to zero; for decrement, the high byte is adjusted when the low byte wraps around to $FF.

        ; 16-bit increment Word
        inc Word
        bne noinc
        inc Word+1
noinc:

        ; 16-bit decrement Word
        lda Word
        bne nodec
        dec Word+1
nodec:  dec Word

16-bit increment shows even more advantage when used to control a loop, because the 16-bit increment conveniently leaves the zero flag set at the end only if the entire 16-bit value is zero.

X/Y as Operand

Main article: Identity table

Normally X and Y cannot be used as an operand to an instruction operating on A. For example, CMP X isn't possible. Where X or Y needs to be used in such a way, they are usually saved to a temporary variable:

; Compare A with X
stx temp
cmp temp

By putting a 256-byte table in memory with each entry simply having the value of its index, X and Y can be used as operands:

table:  .byte $00,$01,$02,$03,$04,$05,$06,$07,$08,$09,$0A,$0B,$0C,$0D,$0E,$0F
        .byte $10,$11,$12,$13,$14,$15,$16,$17,$18,$19,$1A,$1B,$1C,$1D,$1E,$1F
        ...
        .byte $F0,$F1,$F2,$F3,$F4,$F5,$F6,$F7,$F8,$F9,$FA,$FB,$FC,$FD,$FE,$FF
cmp table,x      ; CMP X

eor table,x      ; EOR X

clc
adc table,y      ; ADC Y

JMP (addr,X)

The JMP (addr,X) instruction is present in later 65xx processors. It behaves like JMP (addr), except it fetches the 16-bit value from addr+X. The least-problematic way to implement this is using RTI:

; Jump to address stored at addr+X
lda addr+1,x
pha
lda addr,x
pha
php
rti

See Jump table for further explanation and alternate approaches.

Toggle carry

Invert the carry flag without affecting A (N and Z flags are destroyed, however):[2]

rol a
eor #$01
ror a

Set overflow flag

The overflow flag can be set by using the BIT instruction with any byte that has the second-most significant bit set. The N and Z flags are destroyed, however.[3]

   bit label
   ...
label:
   rts        ; opcode $60 (%01100000)

Test whether A is in range

Test whether A (unsigned) is between constants min and max (inclusive). A is destroyed.[4]

Set carry flag if A is in range, otherwise clear carry:

clc
adc #$ff-max
adc #max-min+1

Clear carry flag if A is in range, otherwise set carry:

sec
sbc #min
sbc #max-min+1

Count bits set

Count the number of bits set in myVal. The result will be in X. myVal is destroyed. (On each round of the loop, the least significant set bit is cleared.)[5]

   ldx #0
   lda myVal
   beq end
loop:
   inx
   dec myVal
   and myVal
   sta myVal
   bne loop
end:

Count the number of bits set in A. The result will be in X and A. (On each round of the loop, the most significant bit is shifted out.)[6]

   ldx #$ff   ; Set count to -1
incr:
   inx        ; Add one to count
loop:
   asl        ; Shift by one bit
   bcs incr   ; Count one bits
   bne loop   ; Repeat till zero
   txa        ; Move count to A

See also

External links

References