6502 assembly shift bits

Question

I have a number 1f stored at address 00, and I have another number 1e stored at address 01. Now I want to make 1f into 1f00 and 1e into 001e, then add them up to get 1f1e. How do I do this? I know the instructions asl and lsr deal with this, but I'm a 6502 beginner, so I would really appreciate it someone can show me how it's done efficiently(least amount of cycles).

Since 6502 is a 8 bit machine, you can't work with 16 bit numbers directly. You can just manipulate the two halves separately, no shifts needed. — Jester
@Jester Do you mean that if I want to store something into 1f1e I need to use sta ($00),y? — Meteorite
Yes, but remember that 6502 is little endian, so to get 1f1e you need to swap the two bytes. You can of course also use sta ($00, x) assuming x is zero. — Jester
@Jester I'd promote that to an answer if I were you; it clearly answers the question and will be more prominent to people who find this question in the future. Failing that I guess somebody else could post it as community wiki. — Tommy

LawrenceC LawrenceC · Accepted Answer · 2014-10-30T17:02:48

If you are working with 16-bit quantities, you need to make sure your variables are 16 bits, or 2 bytes wide.

You probably want to use the 6502 little-endian order, especially if your variable could be used as an address. That means low-byte, high-byte order.

I have a number 1f stored at address 00, and I have another number 1e stored at address 01.

You really want to have $001F stored at address $00-$01 ($00 will equal #$1F and $01 will equal #$00), and $001E stored at address $02-$03 ($02 will equal #$1E and $03 will equal #$00). Then you probably want another variable to hold the result, let's say $04-$05.

But, if you really have two 8-bit variables and you want to "convert" them to 16-bit variables ...

Assume (using labels for clarity):

$00 - your first 8-bit variable (label VAR_A_8)

$01 - your second 8-bit variable (label VAR_B_8)

$02-$03 - where you want to place your first converted 8-bit variable (label VAR_A_16)

$04-$05 - where you want to place your second converted 8-bit variable (label VAR_B_16)

Are they unsigned? If so, just LDA #$00, STA VAR_A_16, STA VAR_B_16, LDA $00, >STA VAR_A_16, LDA $01, STA >VAR_B_16.

The > is a common assembler syntax to mean "low byte of" (It might actually be < - I always got confused on that)

Are they signed? If the 8-bit values are negative (bit 7 set), you want to store $FF in the new bytes, otherwise you want to store $00.

One way:

               LDX #$00
               LDA VAR_A_8       ;Loading .A will set N flag in .P if bit 7 is set
               BPL VAR_A_NOT_NEG
               DEX               ;.X will now be $FF
VAR_A_NOT_NEG: STX <VAR_A_16     ;store .X in high byte
               STA >VAR_A_16     ;copy low byte
               LDX #$00
               LDA VAR_B_8
               BPL VAR_B_NOT_NEG
               DEX
VAR_B_NOT_NEG: STX <VAR_B_16
               STA >VAR_B_16

6502 assembly shift bits

2 Answers