[stella] Question about Circus Atari and BCD addition

[bob.montgomery at thomson.com]

Hi,

this is how Stella does BCD adds:

  // Compute the BCD lookup table
  uInt16 t;
  for(t = 0; t < 256; ++t)
  {
    ourBCDTable[0][t] = ((t >> 4) * 10) + (t & 0x0f);
    ourBCDTable[1][t] = (((t % 100) / 10) << 4) | (t % 10);
  }

define(M6502_ADC, `{
  uInt8 oldA = A;
  {
    Int16 sum = ourBCDTable[0][A] + ourBCDTable[0][operand] + (C ? 1 :
0);

    C = (sum > 99);
    A = ourBCDTable[1][sum & 0xff];
    notZ = A;
    N = A & 0x80;
    V = ((oldA ^ A) & 0x80) && ((A ^ operand) & 0x80);
  }
}')

If that's correct, then it looks like the 6507 first converts both
operands to BCD, treating each nibble like a decimal digit (whether or
not the nibble is <= 9), so $F0 = 150 and $FF = 165.
Then it adds the operands together, throws away the hundreds digit and
that's your result.

So $FF + $09 = 165 + 9 = 174 = $74

The real trickiness is when the result of adding the two operands is >
255.  Then it wraps.  So $FF + $FF = 165 + 165 = 74 (i.e., 330 - 256) =
$74

So apparently adding $0A is exactly the same as adding $10.

All assuming that Stella is accurate.

-Bob