From: nowhere@nevernet.at   
      
   On 23/10/2022 03:31, muta...@gmail.com wrote:   
   ...   
   > They do it so that the compiler generates   
   > the required code to add a large value   
   > to a far pointer, for example 80k,   
   > and the segment will be manipulated   
   > to get it to reach that memory address 80k   
   > further in absolute memory, as opposed   
   > to getting the added value truncated   
   > modulus 64k at most.   
      
   right, but...   
   you cannot have 80Kb in the 16 bit world :)   
   it needs two registers to hold that value,   
   UNREAL mode with 32-bit register usage   
   within 16 bit wasn't available before 80386.   
      
   why you insists on 5 bit shift-Add?   
   this would result in many unused bits.   
      
   Nibbles and bytes are the given opportunities.   
   __   
   wolfgang   
      
   --- SoupGate-Win32 v1.05   
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