🚀 go-pugleaf

In article <10344@saturn.ucsc.edu> ted@helios.ucsc.edu (Ted Cantrall) writes:
$How would things have worked out if IBM had put this 384k block at the bottom
$of the memory (0-384). That would have left no constrictions on upward
$expansion except the 8088. And that problem would have been remedied by
$the 80286.		-ted-

   It would have given us another 65 520 (64K-16) bytes when the 80286 came
out.  Don't forget that in order to use the 286's 16M address space, you must
be in protected mode.  In protected mode, segments have different meanings.
Instead of the meaning we're used to on the 8086/8088, they actually point
to a descriptor in memory giving information about the segment (such as
its address, length, privilege level, etc.)  So code written to run under
the 8086 processor will run fine in 80286 real mode, but not in protected
mode.
-- 
Stephen M. Dunn                               cs4g6ag@maccs.dcss.mcmaster.ca
          <std_disclaimer.h> = "\nI'm only an undergraduate!!!\n";
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       "I want to look at life - In the available light" - Neil Peart