Base 100: Difference between revisions

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m (Correct comment in 16-bit subtraction code.)
(Add code example for converting an unsigned byte to its base 100 encoding.)
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== Code examples ==
== Code examples ==
=== Byte to base 100 encoding ===
This takes in an unsigned 8-bit value in A and returns its base 100 encoding in X (high byte) and A (low byte):
<pre>
; Parameter: A = byte to encode
; Returns base 100 encoding as: X = high byte, A = low byte
ByteToBase100:
  ldx #0 ; Init high byte of result
  :
  cmp #100 ; 0~99 is valid
  bcc @ret
    inx
    sbc #100 ; Carry already set
    bcs :- ; Carry is guaranteed to be set
@ret:
  rts
</pre>
=== 16-bit increment ===
=== 16-bit increment ===
This increments a 16-bit base 100 number by 1, while preventing it from going over 9999:
This increments a 16-bit base 100 number by 1, while preventing it from going over 9999:

Revision as of 02:40, 13 April 2024

Unlike the regular 6502, the 2A03 does not have decimal mode. One workaround for this is to keep numbers in binary, and use a BCD conversion routine to convert to binary-coded decimal as needed. Base 100 is another workaround that simplifies displaying the numbers onscreen.

In base 100, a number consists of a series of bytes that range from 0-99 (or $00-$63 in hexadecimal). Unlike in regular 8-bit math, numbers wrap around at 100 instead of at 256, so $0063 + $0001 is $0100 instead of $0064. Given a base 100 number, you can use a 100-byte table to convert each byte to BCD, which is easy to display.

base_100_to_bcd:
  .byte $00, $01, $02, $03, $04, $05, $06, $07, $08, $09, $10, $11, $12, $13, $14, $15, $16, $17, $18, $19
  .byte $20, $21, $22, $23, $24, $25, $26, $27, $28, $29, $30, $31, $32, $33, $34, $35, $36, $37, $38, $39
  .byte $40, $41, $42, $43, $44, $45, $46, $47, $48, $49, $50, $51, $52, $53, $54, $55, $56, $57, $58, $59
  .byte $60, $61, $62, $63, $64, $65, $66, $67, $68, $69, $70, $71, $72, $73, $74, $75, $76, $77, $78, $79
  .byte $80, $81, $82, $83, $84, $85, $86, $87, $88, $89, $90, $91, $92, $93, $94, $95, $96, $97, $98, $99

Alternatively for even more speed you can have two tables that separately provide the ones digit and the tens digit of the resulting BCD number, letting you display each byte with something as simple as:

  ldx base_100_number
  lda base_100_tens,x
  sta $2007
  lda base_100_ones,x
  sta $2007

Base 100 is good for numbers that you want to display and do addition and subtraction on, but don't need for more complicated math. It can be good for things like a score, an amount of currency, or a timer or countdown.

Code examples

Byte to base 100 encoding

This takes in an unsigned 8-bit value in A and returns its base 100 encoding in X (high byte) and A (low byte):

; Parameter: A = byte to encode
; Returns base 100 encoding as: X = high byte, A = low byte
ByteToBase100:
  ldx #0 ; Init high byte of result
  :
  cmp #100 ; 0~99 is valid
  bcc @ret
    inx
    sbc #100 ; Carry already set
    bcs :- ; Carry is guaranteed to be set
@ret:
  rts

16-bit increment

This increments a 16-bit base 100 number by 1, while preventing it from going over 9999:

AddOneCoin:
  lda #99 ; Check for cap of 9999
  cmp Money+1
  bne :+
  cmp Money+0
  bne :+
    rts ; Exit if cap already reached
  :
  ; Otherwise increment the amount
  inc Money
  lda Money
  cmp #100
  bcc :+
    lda #0 ; Base 100 overflow
    sta Money
    inc Money+1
  :
  rts

16-bit decrement

This decrements a 16-bit base 100 number by 1, while preventing it from going under 0:

SubOneCoin:
  lda Money ; Check if amount is 0
  ora Money+1
  beq :+
  ; Otherwise decrement the amount
  lda #99
  dec Money
  bpl :+
    sta Money ; Base 100 underflow
    dec Money+1
  :
  rts 

16-bit addition

Here's an example that demonstrates base 100 addition, using a pair of 16-bit numbers. The result is clamped to 9999:

AddPrizeMoney:
  lda Money
  clc
  adc Prize
  cmp #100
  bcc :+
    ; Carry is set if the code ends up in here
    sbc #100 ; Guaranteed to set carry
  :
  sta Money
  ; Carry will correctly reflect base 100 overflow here
  lda Money+1
  adc Prize+1
  cmp #100
  bcc :+
    lda #99 ; Cap amount at 9999
    sta Money
  :
  ; Write the new amount
  sta Money+1
  rts

16-bit subtraction

Here's an example that demonstrates base 100 subtraction, using a pair of 16-bit numbers:

PurchaseItem:
  lda Money
  sec
  sbc Price
  bpl :+
    ; Carry is clear if the code ends up in here
    adc #100
    clc
  :
  sta Temp
  ; Carry will correctly reflect base 100 underflow here
  lda Money+1
  sbc Price+1
  bmi NotEnoughFunds
  ; Write the new amount
  sta Money+1
  lda Temp
  sta Money
  rts