From: joemonk64@gmail.com   
      
   On Sunday, October 23, 2022 at 7:26:51 PM UTC-5, Joe Monk wrote:   
   > > Since the memory to be managed is more than 64k, the OS    
   > > needs to effectively using huge pointers, and as such the    
   > > huge memory model OS needs to generate code that is    
   > > aware of the segment shift value.    
   > >   
   > Straight from the intel 8086 book:    
   >    
   > "In order to be dynamically relocatable, a program must not load or alter   
   its segment registers and must not transfer directly to a location outside the   
   current code segment. In other words, all offsets in the program must be   
   relative to fixed values    
   contained in the segment registers. This allows the program to be moved   
   anywhere in memory as long as the segment registers are updated to point to   
   the new base addresses."    
   >    
   Also straight from the book:   
      
   "Whenever the BIU accesses memory-to fetch an instruction or to obtain or   
   store a variable-it generates a physical address from a logical address. This   
   is done by shifting the segment base value four bit positions and adding the   
   offset..."   
      
   Joe   
      
   --- SoupGate-Win32 v1.05   
    * Origin: you cannot sedate... all the things you hate (1:229/2)