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by lostlogin 4557 days ago
Thanks. This is close but off then: The drawing area is evenly divisible by 13 long words (13x32=416). This is exactly the number of registers that were available in the loop above for the MOVEM.L operation (It appears it could have used 14 registers, but I am guessing the extra 32 pixels would have made the drawing area more cramped by reducing the gray whitespace.
1 comments

MOVEM.L can address all 16 registers on the 68K but keeping the source and destination pointers in the register bank made the routine run much quicker. You also need SP to get back home, so 13 was all Hertzfeld could use.
> Hertzfeld

Atkinson

Ooops, you're right. My mistake.