Re: Question about the A20 line



Thanks for explaining :-)

could you explain how address 0xffff:0x10 and 0x0000:0x0000
are the same when a wrap around occurs? Shouldnt it be 0x0000:0x0010?
Or am i not understanding the logic correctly.


"Jean-François Michaud" <spamtrap@xxxxxxxxxx> wrote in message
news:1135881237.372100.254460@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
> It tests out to see if 2 memory zones overlap. When the A20 line isn't
> flipped, you only have 1Mb (20 bits) of addressable memory. Those
> memory addresses wrap around when you reach the end. so address 0:0 and
> address 0FFFFh:10h are actually the same! The thing is. If your A20
> line was flipped correctly, these 2 memory addresses would not overlap.
> If you put a byte in the first address and can't get the same value in
> the second address then your A20 flip worked!
>
> Regards
> Jean-François Michaud
>
>


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