From: muta...@gmail.com   
      
   On Friday, July 16, 2021 at 2:44:41 PM UTC+10, Joe Monk wrote:   
      
   > "For programs executed in protected mode, the D-bit in   
   > executable-segment descriptors determines the default   
   > attribute for both address size and operand size. These   
   > default attributes apply to the execution of all instructions   
   > in the segment. A value of zero in the D-bit sets the default   
   > address size and operand size to 16 bits; a value of one, to 32 bits."   
      
   And what are we left with with this issue from Rod?   
      
   > ;32-bit code   
   > 00000028 89FB mov ebx,edi   
   > 0000002A 8B4720 mov eax,[edi+32]   
   >   
   > ;16-bit code   
   > 00000028 89FB mov bx,di   
   > 0000002A 8B4720 mov ax,[bx+0x20]   
      
   Can I get the 16-bit flavor to work by just setting   
   the D-bit to 0?   
      
   Or is it still different in PM32?   
      
   And if the latter, then how important is that (second) instruction?   
   Can I just ensure that the C compiler doesn't emit that   
   instruction? Will I still have a Turing machine operating   
   close to full speed compared to if I was in RM16 instead?   
      
   Thanks. Paul.   
      
   --- SoupGate-Win32 v1.05   
    * Origin: you cannot sedate... all the things you hate (1:229/2)