>The TIA reference says there are 45 reserved hardware registers. These are
>mapped into the 128 bytes, right?
>
>So that means a game like Video Chess, which uses 64 bytes for board
>position, only has 19 free bytes for the program and the stack, right?
No. The 128 bytes of RAM are in addition to the register space.
>I have a question about the 76 cycles available during the horizontal
>scanline. It appears as if this ignores the leftover right overscan area.
>
>Is the horizontal blank (beam return) forced at that point or can the
>programmer allow the scanline to draw through the overscan or something?
>
>Doesn't the background color do that (extend through right overscan) anyway?
No and no. From a programming standpoint (which you're unfamiliar
with), the scanline ends immediately at the right border of the visible
area... the right overscan area is for all intents and purposes part of
the horizontal blank region for the next scanline.
>I have overheard discussions of not calling WSYNC and using up all
>available processor time, so I was wondering whether that was time left
>during the active portion of the screen or after the 76 cycle mark in the
>overscan.
No. That just means that your kernel uses all 76 cycles exactly, and is
therefore already synched with the beginning of the next scanline
without needing to Wait for SYNC.
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