From: rick.c.hodgin@gmail.com   
      
   On Tuesday, December 4, 2018 at 3:16:31 PM UTC-5, R.Wieser wrote:   
   > Rick,   
   >   
   > > The Mod/Reg/RM byte is a traditional encoding byte in all   
   > > of x86 assembly, and it's not specific to FPU operations or   
   > > anything else specific.   
   >   
   > I know how it works for the x86 comands.    For the FPU commands ?  No so   
   > much.   
   >   
   > Even if I read the explanation to that "/{digit}" wrong and it doesn't   
   > implicitily exclude the "Mod" part from consideration, than I'm still left   
   > with the "Reg" part (which refers to a register), which has no meaning for   
   > most FPU commands (with the exclusion of one or two none use a register as   
   > source *or* target).   
   >   
   > If I use both the "Mod" as well as the "RM" parts the INT 0x35, 0x7E 0xFA   
   > sequence decodes to :   
   >   
   > FNSTENV [BP-0x06]   
   >   
   > , which looks sensible to me.   
   >   
   > But that stil leaves the "Reg" part, which for th 0x7E value refers to the   
   > DI register.  Which does not mean anything to me.   
   >   
   > So, what *does* it mean ?   
      
   0x7E = 0111 1110 in binary   
      
   Applying that to Mod/Reg/RM fields (2:3:3):   
      
       01 111 110   
      
   The value /6 you're seeing there is what you see in the RM field.   
   It doesn't refer to the di register, but is where the bits of the   
   /6 it's referring to go.   
      
   --   
   Rick C. Hodgin   
      
   --- SoupGate-Win32 v1.05   
    * Origin: you cannot sedate... all the things you hate (1:229/2)