Extended Memory in 6.0
TERRY MICHAEL AUSPITZ
that the second byte of the pointer is always 0. It seems as if changing this
value may allow the user to access memory over 1mb. Is this so? If not, where
is it pointing when I set that pointer (via variant records) to something other
than 0? All I know for sure is that the second byte is *NOT* ignored.
--
_______________________________________________________________________________
_/ _/ _/_/_/ | __o Terry M. Auspitz at Lehigh University
_/ _/ _/ _/ | _ \ <,_ (tm...@lehigh.edu)
_/_/_/_/ _/ | (_)/ (_) Would anyone else share these opinions?
JFK
>Hi... I was recently researching the way pointers worked in TP6, and I
>noticed
>that the second byte of the pointer is always 0. It seems as if
>changing this
>value may allow the user to access memory over 1mb. Is this so? If not,
>where
>is it pointing when I set that pointer (via variant
>records) to something other
>than 0? All I know for sure is that the second
>byte is *NOT* ignored.
If you're running under DOS, then due to the infinite wisdom of Intel you
have a segment and offset part of the pointer. To decipher this jibberish
into an absolute memory location, take the segment part of the pointer,
multiply by sixteen, and add the offset. This gives you a 20-bit address
location from a "32-bit" (16:16 bit) pointer.
So no, if you change this value, you'll be relocating the pointer into yet
another spot in the 1Mb range. Running under protected mode, however, Intel
realized their folley and used the 32-bit pointers as true 32-bit address
pointers (with additional complications only Intel could come up with in
the name of "compatibility").
Please! I'm not trying to start an anti-/pro- Intel flame war! I just calls
'em as I sees 'em.
-------
James Knowles /|
aristo...@cc.usu.edu /_|MIGA provides creative solutions to problems
sl...@cc.usu.edu / | created by other computers