This may be dead obvious, but I'll point it out anyway.
You actually understand the 6 digit display routine.
This routine does a couple other things besides just a "simple" 6 digit display.
- Skips scanlines so text is displayed at correct vertical position (bottom of screen?)
- Sets the colours based on the current level
- Sets up the pointers for use in the 6 digit display based on a string-like structure
- Draws the 6 digit display
In effect, you're actually dealing with four distinct routines. They are:
Skip a certain number of scanlines (based on level?)
> ldy MsgDelayTab-1,x
> beq .skipWait
>.loopWait:
> sta WSYNC
> dey ; 2
> bne .loopWait ; 2³
>
>^ Simple Enough
>
>.skipWait:
Set colours:
> lda Planet
> and #~PLANETMASK
> bne .NTSCMsgCol
> lda MsgColTab_PAL,x ; 4
> bne .contMsgCol
> DEBUG_BRK
>.NTSCMsgCol:
> lda MsgColTab,x ; 4
>.contMsgCol:
> sta COLUP0 ; 3
> sta COLUP1 ; 3 = 10 <-- When I count, I get much more than 10?
Set up pointers:
ldy MsgTab,x ; 4
>^??????? Totally lost here |Load height of graphic?
>
> lda LetterPtrTab,y ; 4
> sta digitPtr ; 3
>^LetterPtrTab,Y = Address of first 8x8 Graphic?
>
> lda LetterPtrTab+1,y ; 4
>^Here is where I get lost? I don't understand how this gets compiled
>^If LetterPtrTab = $1200 and Y is $05, then this returns $1200 + 1 + 5? ($1206)???
>^Shouldn't this get the last byte of the second graphic? I'm confused how it does this?
>
> sta digitPtr+2 ; 3
>^digitPtr is in Zero Page... So the graphic data is loaded into RAM...
>
> lda LetterPtrTab+2,y ; 4
> sta digitPtr+4 ; 3
> lda LetterPtrTab+3,y ; 4
> sta digitPtr+6 ; 3
> lda LetterPtrTab+4,y ; 4
> sta digitPtr+8 ; 3
> lda LetterPtrTab+5,y ; 4
> sta digitPtr+10 ; 3 = 46
6 digit display:
> ldy #5 ; 2
>; ldy MsgHeightTab-1,x
> txa ; 2 don't change C!
> eor #MSG_C_2000 ; 2
> bne .msgLoop ; 2³
> ldy #7 ; 2 = 7/8
>.msgLoop:
>
>^ All graphics are 5 height, except Copyright which is 7...
>^ These are all wasted cycles for me... in order for this to work for me however,
>^ I have to get this exact cycle count?
>
> dey ; 2 61
> sty tmpVar ; 3 64
> lda (digitPtr+$8),y ; 5 69
> tax ; 2 71
> sta WSYNC ; 3 --- 76 ---
> lda (digitPtr),y ; 5 5
> sta GRP0 ; 3 8
> lda (digitPtr+$2),y ; 5 13
> sta GRP1 ; 3 16
> lda (digitPtr+$4),y ; 5 21
> sta GRP0 ; 3 24
> lda (digitPtr+$6),y ; 5 29
> sta tmpVar2 ; 3 32
> lda (digitPtr+$a),y ; 5 37
> ldy.w tmpVar2 ; 4 41
> sty GRP1 ; 3 44
> stx GRP0 ; 3 47
> sta GRP1 ; 3 50
> sta GRP0 ; 3 53
> ldy tmpVar ; 3 56
> bne .msgLoop ; 2³ 59
>
>^This all makes sense
>
> sty GRP0
> sty GRP1
> sty GRP0
>^Why is GRP0 cleared twice?
> rts
The part of the routine you seem to actually be having a problem with is the use of pointers, not really the display routine itself (Though I will also point out that the timing is only critical during the actual display part of the routine, the timing in other parts - particularly in the skip lines part (which you asked about) is not critical). And you didn't include the tables in the code you posted. MsgTab and LetterPtrTab are tables of offsets to the graphics data. So you need to consider and understand those tables if you're going to understand what's going on.
To simplify things, I'm just going to consider the first digit being displayed and ignore the other 5 digits.
To set up the first digit, we need to put a pointer to it's graphics data into digitPtr
If we want to setup digitPtr to point to the first letter in our alphabet (presumably a 0 or A but you didn't include the tables so I'm guessing and I'm not going to go hunting for the source..) we would:
lda #0*HeightOfLetters
sta digitPtr ;digitPtr+1 has been pre-set to point to the top of the graphics table at $FF00 or $FE00 or etc
If we want to setup digitPtr to point to the second letter in our alphabet (presumably a 1 or B but you didn't include the tables so I'm guessing and I'm not going to go hunting for the source..) we would:
lda #1*HeightOfLetters
sta digitPtr ;digitPtr+1 has been pre-set to point to the top of the graphics table at $FF00 or $FE00 or etc
And so on...
However, in this example Thomas has abstracted things a little to allow the routine to display different strings. So instead of loading the character values directly, he's using a pointer to some strings:
lda LetterPtrTab,y ; 4
sta digitPtr ; 3
LetterPtrTab is a table of strings, if each string is (probably) 6 letters wide, then to put the first letter of the first string into digitPtr
lda #0
lda LetterPtrTab,y ; 4
sta digitPtr ; 3
To put the first letter of the second string in digitPtr
lda #6
lda LetterPtrTab,y ; 4
sta digitPtr ; 3
But that's not particularly usefull unless you turn that #0 and #6 into a variable. Thomas could have used a simple variable like:
ldy MsgOffset
lda LetterPtrTab,y ; 4
sta digitPtr ; 3
Which would be fine if you only have a couple different messages to display, and they're all the same width. However, given many different messages to display it's convenient to put it all into a table:
ldy MsgTab,x ; 4
lda LetterPtrTab,y ; 4
sta digitPtr ; 3
So, to display the first message (string):
ldx #0
ldy MsgTab,x ; 4
lda LetterPtrTab,y ; 4
sta digitPtr ; 3
So, to display the second message (string):
ldx #1
ldy MsgTab,x ; 4
lda LetterPtrTab,y ; 4
sta digitPtr ; 3
If this still doesn't make any sense to you, it may not matter. If you just want to use the 6 digit display to display a score or a specific static text then you don't need all these extra abstractions that are desinged to handle a variety of different strings. All you really need is that 4th section of the code, and set up digitPtr et al to suit your own purposes.
Chris...
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