A86: Re: port 0
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A86: Re: port 0
That's because if you mask out the top bits of $76 you get $36, which is
$f600
>
>all documentation i've seen on port 0 has said that the byte sent to it is
>added to $c0, then multiplied by $0100 to become the pointer to the display
>memory.
>
>however, i've found that when i made a prog which sent a value greater than
>$3c to port 0, ($76) and used $fc00 as a double buffer, copying the entire
>contents of $fc00 to $f600 seemed to make it display. any ideas why?
>
>port 0 is REALLY confusing!