🚀 go-pugleaf

Hi!

(Hope this is not a FAQ or a stupid one):

What will happen to the upper 16 bit content of the registers EAX,EBX,...EBP,ESP
etc. when using 16 bit instructions?
Example:
MOV EAX HFFFFFFFF
Add ax,1
EAX=????????????

My 486DX seams to leave the upper 16 bits unchanged but the 
real question is: is this a documented (where) feature? What about
all these dozens of 386 and 486 clones, and the 486 part of the 
Pentium??

If the upper bits are unchanged, this would give you a second 
bank of 16 bit registers (switching between the "banks" with ROL E**,16). 

However, if for example XOR ax,ax is simply treated as XOR EAX,EAX
by the processor and the 16 upper bits are changed, this could lead
to *serious* problems when using DOS programs compiled with the -3 
option of BORLAND C++ 3.1: BC++3.11 code uses 32 bit registers to
handle "long int"s. MS-DOS interrupts use (and save) only the 
16-bit registers (the lower 16 bits of E**) 

I tried to look this up in various handbooks for assembly programming,
but none had the answer. I couldn`t find anything Intel-official on this, 
either.

Could someone please give me a hint or pointer to the right 
Intel-docu??

Please reply via e-mail.
Thanx in advance.

------------------------------------------------------------------------------
"I don't like signatures"
------------------------------------------------------------------------------