question on assembler notation found in Yar's revenge

[+xucaen]

Hi, I was just looking at the disassembly of Yar's Revenge courtesy of Debro.

Around line 3248, there is a block of code that controls the swirl motion.

The code uses the character "|".

Now, I know in C and in C++ this is a bit operator for OR.

How is it being used here? I did searches in my 6502 documents and couldn't find any reference to it.

 

SwirlMotionTable

  IF COMPILE_VERSION = NTSC
  
  .byte HMOVE_0  | 10
  .byte HMOVE_L2 | 11
  .byte HMOVE_L3 | 13
  .byte HMOVE_L5 | 14
  .byte HMOVE_L6 | 0
  .byte HMOVE_L5 | 2
  .byte HMOVE_L3 | 3
  .byte HMOVE_L2 | 5
  .byte HMOVE_0  | 6
  
  ELSE

  .byte HMOVE_0  | 9
  .byte HMOVE_L3 | 10
  .byte HMOVE_L4 | 12
  .byte HMOVE_L6 | 13
  .byte HMOVE_L7 | 0
  .byte HMOVE_L6 | 3
  .byte HMOVE_L4 | 4
  .byte HMOVE_L3 | 6
  .byte HMOVE_0  | 7
  
  ENDIF

[djmips]

Yup, you are correct. It's bitwise OR.

 

 

For example. Referencing the source you are talking about

 

HMOVE_0 = $70

 

So .byte HMOVE_0 | 10 would generate $7A (since 10 = $0A)

Edited June 29, 2006 by djmips

[vdub_bobby]

I did searches in my 6502 documents and couldn't find any reference to it.

That's because it is pseudo-code for the assembler, not code for the 6502, so you'll want to look at the docs for the assembler - DASM, almost certainly.

 

I.e., this:

lda #($10|$01)

Will be assembled as:

lda #$11

 

For that reason, you can't bitwise OR two variables.

Edited June 29, 2006 by vdub_bobby