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We have already seen why there are 3 variants of '2600 units - these variations (PAL, NTSC, SECAM) exist because of the differences in TV standards in various countries. Specifically, the colour information is encoded in different ways into the analogue TV signal for each system, and the '2600 hardware is responsible for inserting that colour information in the data sent to the TV.
Not only do these different '2600 systems write the colour information in different ways, they also write totally different colours! What is one colour on a NTSC system is probably NOT the same colour on PAL, and almost certainly not the same colour on SECAM! Here's some wonderful colour charts from B. Watson to show the colours used by each of the systems...
http://www.urchlay.c...colorchart.html
Colours are represented on the '2600 by numbers. How else could it be? The colour to number correspondence is essentially an arbitrary association - so, for example on a NTSC machine the value $1A is yellowish, on PAL the same colour is grey, and on SECAM it is aqua (!). If the same colour values were used on a game converted between a NTSC and PAL system, then everything would look very weird indeed! To read the colour charts on the above URL, form a 2-digit hex number from the hue and the lum values (ie: hue 2, lum 5 -> $25 value -> brown(ish) on NTSC and as it happens a very similar brown(ish) on PAL.
We've already played with colours in our first kernel! In the picture section (the 192 scanlines) we had the following code...
; 192 scanlines of picture... ldx #0 REPEAT 192; scanlines inx stx COLUBK sta WSYNC REPEND
We should know by now what that "sta WSYNC" does - and now it's time to understand the rest of it. Remember the picture that the kernal shows? A very pretty rainbow effect, with colour stripes across the screen. It's the TIA producing those colours, but it's our kernal telling the TIA what colour to show on each line. And it's done with the "stx COLUBK" line.
Remember how the TIA maps to memory in locations 0 - $7F, and that WSYNC is a label representing the memory location of the TIA register (which happens, of course, to be called WSYNC). In similar fashion, COLUBK is a label which corresponds to the TIA register of the same name. This particular register allows us to set the colour of the background that the TIA sends to the TV!
A quick peek at the symbol table shows...
COLUBK 0009 (R )
In fact, the very best place to look is in the Stella Programmer's guide - for here you will be able to see the exact location and usage of this TIA register. This is a pretty simple one, though - all we do is write a number representing the colour we want (selected from the colour charts linked to, above) and the TIA will display this colour as the background.
Don't forget that it also depends on what system we're running on! If we're doing a PAL kernel, then we will see a different colour than if we're doing a NTSC or SECAM kernel. The bizarre consequence of this is that if we change the number of scanlines our kernel generates, the COLOURS of everything also change. That's because (if we are running on an emulator or plug a ROM into a console) we are essentially switching between PAL/NTSC/SECAM systems, and as noted these systems send different colour information to the TV! It's weird, but the bottom line is that when you choose colours, you choose them for the particular TV standard you are writing your ROM to run on. If you change to a different TV system, then you will also need to rework all the colours of all the objects in your game.
Let's go back to our kernel and have a bit of a look at what it's doing to achieve that rainbow effect. There's remarkably little code in there for such a pretty effect.
As we've learned, the 6502 has just three "registers". These are named A, X and Y - and allow us to shift bytes to and from memory - and perform some simple modifications to these bytes. In particular, the X and Y registers are known as "index registers", and these have very little capability (they can be loaded, saved, incremented and decremented). The accumulator (A) is our workhorse register, and it is this register used to do just about all the grunt-work like addition, subtraction, and bit manipulation.
Our simple kernel, though, uses the X register to step a colour value from 0 (at the start), writing the colour value to the TIA background colour register (COLUBK), incrementing X by one each scanline. First (outside the repeat) we have "ldx #0". This instruction moves the numeric value 0 into the X register. ld is an abbreviation for "load", and we have lda, ldx, ldy. st is the similar abbreviation for store, and we have stx sty sta. Inside our repeat structure, we have "stx COLUBK". As noted, this will copy the current contents of the x register into the memory location 9 (which is, of course, the TIA register COLUBK). The TIA will then *immediately* use the value we wrote as the background colour sent to the TV. Next we have an instruction "inx". This increments the current value of the X register by one. Likewise, we have an "iny" instruction, which increments the y register. But, alas, we don't have an "ina" instruction to increment the accumulator (!). We are also able to decrement (by 1) the x and y registers with "dex" and "dey".
The operation of our kernel should be pretty obvious, now. The X register is initialised with 0, and every scanline it is written to the background colour register, and incremented. So the background colour shows, scanline by scanline, the colour range that the '2600 is capable of. In actual fact, you could throw another "inx" in there and see what happens. Or even change the "inx" to "dex" - what do you think will happen? As an aside, it was actually possible to blow up one early home computer by playing around with registers like this (I kid you not!) - but you can't possibly damage your '2600 (or emulator!) doing this. Have fun, experiment.
Since we're only doing 192 lines, the X register will increment from 0 to 192 by the time we get to the end of our block of code. But what if we'd put two "inx" lines in? We'd have incremented the X register by 192 x 2 = 384 times. What would its value be? 384? No - because the X register is only an 8-bit register, and you would need 9 bits to hold 384 (binary %110000000). When any register overflows - or is incremented or decremented past its maximum capability, it simply "wraps around". For example, if our register had %11111111 in it (255, the maximum 8-bit number) and it was incremented, then it would simply become %00000000 (which is the low 8-bits of %100000000). Likewise, decrementing from 0 would leave %11111111 in the register. This may seem a bit confusing right now, but when we get used to binary arithmetic, it will seem quite natural. Hang in there, I'll avoid throwing the need to know this sort of stuff at you for a while.
Now you've had a little introduction to the COLUBK register, I'd just like to touch briefly on the difference apparent between the WSYNC register and the COLUBK register. The former (WSYNC) was a strobe - you could simply "touch" it (by writing any value) and it would instantly halt the 6502. Didn't matter what value you wrote, the effect was the same. The latter register (COLUBK) was used to send an actual VALUE to the TIA (in this case, the value for the colour for the background) - and the value written was very much important. In fact, this value is stored internally by the TIA and it keeps using the value it has internally as the background colour until it changes.
If you think about the consequences of this, then, the TIA has at least one internal memory location which is in an uknown (at least by us) state when the machine first powers on. We'd probably see black - which happens to be value 0 on all machines), but you never know. I believe it is wise to initialise the TIA registers to known-states when your kernel first starts - so there are no surprises on weird machines or emulators. We have done nothing, so far, to initialise the TIA - or the 6502, for that matter - and I think we'll probably have a brief look at system startup code in a session real-soon-now.
Until then, have a play with the picture-drawing section, and see what happens when you write different values to the COLUBK register. You might even like to change it several times in succession and see what happens. Here's somtehing to try (with a bit of headscratching, you should be able to figure all this out by now)...
; 192 scanlines of picture... ldx #0 ldy #0 REPEAT 192; scanlines inx stx COLUBK nop nop nop dey sty COLUBK sta WSYNC REPEND
Try inserting more "nop" lines (what does nop do, again?) - can you see how the timing of the 6502 and where you do changes to the TIA is reflected directly onscreen because of the synchronisation between the 6052 and the TIA which is drawing the lines on-the-fly?
Have a good play with this, because once you've cottoned-on to what's happening here, you will have no problems programming anything on the '2600.
See you next time!
NOP = waste 1 cycle
So my biggest misunderstanding at the moment is... how do I know where to begin? Do I control the tv or do I try to keep up with it? For instance, if I move #2 into VSYNC in the middle of drawing the playfield, does the electron beam return to the top of the screen? I just did a test and that's what it looks like is happening. And is it the same for VBLANK and RSYNC as well?
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