6502 indirect addressing issues

i'm having some trouble with the following code for a 6502 machine:

C000 LDA #$00
C002 STA $FE
C004 LDA #$20
C006 STA $FF
C008 LDY #$08
C00A LDX #$00
C00F BEQ $C01B
C011 LDA ($FE),Y
C013 CMP #$2F
C015 BPL $C00C
C017 INX
C018 JMP $C00C

The exercise is to store the numbers 2, 1 and 4 starting from the address 2000 and say what are the values of A, X and Y.

I'm "running" my code with pen and paper but I got stuck at C011 for the following reason:
It loads in A the value stored at the memory address calculated this way:

  1. pick $FE value (that at first is 00)
  2. Add the value of Y (that at first iteration is 7)
  3. I now have 07
  4. Load A with the value stored at 07

Is this correct? Am I missing something? If I'm not, where do I use the values stored in 2000 2001 and 2002 ?

Thanks in advance..


  • No, you're not correct. You're missing the meaning of LDA ($FE),Y which uses the indirect indexed (as opposed to indexed indirect) addressing mode. Indirect means the value inside the parentheses is the address of a 16-bit pointer, low byte first. That's the $00 and $20 you set up earlier, so $2000.

    The indexing is done with Y, and your loop exit condition is based on Y, so you have that. The value of A is the last value read, so you have that too.

    But your comment & question on Weather Vane's answer is very relevant. The values in the other memory locations matter because of the CMP #$2F and subsequent BPL and INX. CMP acts like a subtract, and the N flag is set if the compared register < compared memory; see here.

    So the value of X depends on those other memory values.