From: muta...@gmail.com   
      
   On Saturday, July 17, 2021 at 1:06:38 AM UTC+10, muta...@gmail.com wrote:   
      
   > > Since I can only get 512 MiB for 16-bit programs anyway,   
   > > I may as well have the mandatory 6 selectors or whatever   
   > > to address the first 3.5 GiB of memory for 32-bit programs   
   > > and the rest of the selectors used for 16-bit programs in   
   > > the other 512 MiB of memory.   
      
   > I don't suppose there is a way to have, in addition to the   
   > above, of having 64-bit programs run in the 4 GiB to   
   > 32 GiB region?   
      
   How about a new mode for x64, called CM64,   
   active in PM32, where if the selector (cs/ds)   
   has a base of 0xffffffff and a length of 1 (or   
   something similar), then the x64 long mode   
   instruction set is active?   
      
   In fact, is there any reason why the x64 instruction   
   set can't always be active? If that's the case then   
   we are left with the only difference being the   
   default address size being 64 bits. So that's the   
   only thing that the dummy base above activates.   
   It would be similar to having a D "bit" of 2.   
      
   How many extra transistors would that require?   
   As a percentage.   
      
   Thanks. Paul.   
      
   --- SoupGate-Win32 v1.05   
    * Origin: you cannot sedate... all the things you hate (1:229/2)