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 Post subject: 6502 Assembly question
PostPosted: Sun Jul 05, 2020 3:46 pm 
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I need to state that I am very new to assembly programming.

I am reviewing some code and I am confused as to how it is functioning and looking for someone to explain.

The lines of code that I am confused about are:

lda ScreenRAMRowStartLow,Y ; load low address byte
sta ZeroPageLow

lda ScreenRAMRowStartHigh,Y ; load high address byte
sta ZeroPageHigh

My question is when LDA is performed what exactly is going on. What part does they Y play and how does the ScreenRAMRowStartLow used.

Below is the ScreenRAMRowStartLow definition.

ScreenRAMRowStartLow ; SCREENRAM + 40*0, 40*1, 40*2 ... 40*24
byte <SCREENRAM, <SCREENRAM+40, <SCREENRAM+80
byte <SCREENRAM+120, <SCREENRAM+160, <SCREENRAM+200
byte <SCREENRAM+240, <SCREENRAM+280, <SCREENRAM+320
byte <SCREENRAM+360, <SCREENRAM+400, <SCREENRAM+440
byte <SCREENRAM+480, <SCREENRAM+520, <SCREENRAM+560
byte <SCREENRAM+600, <SCREENRAM+640, <SCREENRAM+680
byte <SCREENRAM+720, <SCREENRAM+760, <SCREENRAM+800
byte <SCREENRAM+840, <SCREENRAM+880, <SCREENRAM+920
byte <SCREENRAM+960

ScreenRAMRowStartHigh ; SCREENRAM + 40*0, 40*1, 40*2 ... 40*24
byte >SCREENRAM, >SCREENRAM+40, >SCREENRAM+80
byte >SCREENRAM+120, >SCREENRAM+160, >SCREENRAM+200
byte >SCREENRAM+240, >SCREENRAM+280, >SCREENRAM+320
byte >SCREENRAM+360, >SCREENRAM+400, >SCREENRAM+440
byte >SCREENRAM+480, >SCREENRAM+520, >SCREENRAM+560
byte >SCREENRAM+600, >SCREENRAM+640, >SCREENRAM+680
byte >SCREENRAM+720, >SCREENRAM+760, >SCREENRAM+800
byte >SCREENRAM+840, >SCREENRAM+880, >SCREENRAM+920
byte >SCREENRAM+960

Secondary question is what do the > and < do in the definition of the byte?

Really looking to figure this out and would appriciate anyone taking the time to explain. If I did not explain the question right just so and I will try to rephrase.

Thanks
Mark


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PostPosted: Sun Jul 05, 2020 4:02 pm 
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Welcome!

The addressing mode base,Y will access memory at a base address plus the content of the Y register. The two-byte address 'base' from the program will have Y added to it to make the effective address. That is, 'base' can be thought of as an array, and accessing base,Y is like accessing base[y] in another language.

In this case you have two base addresses, so in effect you have two arrays of bytes.

I'll rewrite your code snippet with different variable names:

Code:
lda ScreenRAMRowStartLow,Y ; load low address byte
sta ZeroPageLow

lda ScreenRAMRowStartHigh,Y ; load high address byte
sta ZeroPageHigh


as
Code:
lda lowbytes,Y ; load low address byte
sta pointerlow

lda highbytes,Y ; load high address byte
sta pointerhigh


So we load the Yth byte from the first array and store it in zero page, then the Yth byte from the second array and store in zero page. We can think of these two arrays as a single array of two-byte values.

In all likelihood these two zero page locations are adjacent: we've just constructed a pointer in zero page, as one of N possible pointers from our array.

Having got an address (or a pointer) into zero page, it's now possible to access memory that's pointed to by that pointer, typically with a new index value but once again placed in the Y register:
LDA (zp),Y

Most likely the idea here is that the array contains pointers to start of each line, so Y in the first two accesses gets a pointer to screen memory at some line of the screen, and then the subsequent access (with a new value for Y) can be used to read or write the character in that position along the line.

Now you know the intended meaning of the two arrays, it should be relatively easy to puzzle out why (and how) it's being filled with the values relating to the location of screen memory.

Hope this helps.


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PostPosted: Sun Jul 05, 2020 4:04 pm 
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The lines of code get the start address for a line of characters in 40x25 video RAM area using lookup tables and store in two zero page locations (to create a pointer).

If you have a '<' character in from of an expression the assembler will calculate the low 8 bits of the value (e.g. value MOD 256). Similarly a '>' character calculate the high 8 bits of a value (e.g. value / 256).

The two tables have applied < and > to the same addresses so that the high bytes are in one and the low bytes are in the other.

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PostPosted: Sun Jul 05, 2020 5:16 pm 
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BigEd and BitWise,

Thank you so much for the quick response.

Both of your answers have been very helpful and I think I understand how this working now.

Mark


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PostPosted: Sun Jul 05, 2020 7:42 pm 
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Note that there are some design decisions in this arrangement that could be made otherwise. Splitting the high and low bytes of a 16-bit address into two separate tables means all you have to do is load the Y-register with the screen row you're interested in - you can use that directly as an index into the two tables to get the address of that row.

But that's not the only way to do it. You could decide to have only one table, like so:
Code:
word screenAddr, screenAddr+40, screenAddr+40*2, screenAddr_40*3, [etc]


Then you would have to do something like:

Code:
lda currRow
asl
tay


to double the current row in order to get the proper index into the table. The advantage is it's a lot easier to write this form in source code and less prone to mistakes, since in the split form you have to be very careful that the two tables match up. The disadvantage is that this form takes a little more code (two bytes) and little more time (four cycles) to access.


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PostPosted: Sun Jul 05, 2020 7:59 pm 
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teamtempest wrote:
The disadvantage is that this form takes a little more code (two bytes) and little more time (four cycles) to access.
It also maxes out with only 128 workable values for Y (whereas the other scheme would allow up to 256). But yeah, it's easier (and less error prone) setting up one table than two. :|

-- Jeff

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