2 replies to this topic

[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 by djmips, Thu Jun 29, 2006 11:36 AM.

[vdub_bobby]

Quote

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 by vdub_bobby, Thu Jun 29, 2006 11:42 AM.