Question

Getting opcode sldt to work

Asked by: MvanderKooij

I'm trying to get the Get_ldtr_base to work. I've copied the code from a C++ sample:

unsigned long
get_idt_base (void)
{
      unsigned char      idtr[6];
      unsigned long      idt      = 0;

      _asm sidt idtr
      idt = *((unsigned long *)&idtr[2]);
      
      return (idt);
}

unsigned long
get_ldtr_base (void)
{
      unsigned char   ldtr[5] = "\xef\xbe\xad\xde";
      unsigned long   ldt                  = 0;

      _asm sldt ldtr
      ldt = *((unsigned long *)&ldtr[0]);

      return (ldt);
}

but Delphi can not compile the "sldt ldtr" opcode. It turns out that Delphi wants to create a 16 bit register opcode. How can i fool delphi to get the sldt for me?

function Get_idt_base: Cardinal;
var
  idtr: array[0..5] of Byte;
begin
  asm
    sidt idtr
  end;
  Result := pCardinal(@idtr[2])^;
end;
 
function Get_ldtr_base: Cardinal;
var
  ldtr: array[0..4] of Byte;
begin
  ldtr[0] := $ef;
  ldtr[1] := $be;
  ldtr[2] := $ad;
  ldtr[3] := $de;
  ldtr[4] := $00;
 
  asm
    sldt ldtr
  end;
  Result := pCardinal(@ldtr[0])^;
end;

                                  
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Asked On
2009-09-30 at 02:14:01ID24772873
Tags

Delphi

,

assembly

,

sldt

Topics

Assembly Programming Language

,

Delphi Programming

Participating Experts
2
Points
500
Comments
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Answers

 

by: HypoPosted on 2009-09-30 at 04:37:07ID: 25457419

According to the Intel Instruction reference, the sldt instruction only returns it's result into a 16-bit register or memory location, and even if you would pass a 32-bit value, the high order 16 bits would only be set to zero.

Since the instruction only generates an output I don't see why you would initialize the ldtr values to $EF, $BE, $AD, $DE, but maybe i'm missing something here?

If not, then according to the reference manual, you should get the segment selector into that 16-bit result.

Is it perhaps the global descriptor table base you want to read, (instead of the local descriptor table?)

/Hypo

function Get_ldtr_base: Word; register;
asm
  SLDT AX
end;

                                              
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by: MvanderKooijPosted on 2009-09-30 at 04:57:08ID: 25457573

This is part of a VM detection scheme used by ScoopyNG (http://www.trapkit.de/)

After the '_asm sldt ldtr' call the resulting buffer is compared to $dead0000, the beef has gone there.
VM's do not properly clear those bits. So the test is using this behaviour.


 

by: epasquierPosted on 2009-09-30 at 05:36:10ID: 25457996

I have found this documentation

A.151 SGDT, SIDT, SLDT: Store Descriptor Table Pointers

SGDT mem                      ; 0F 01 /0             [286,PRIV]
SIDT mem                      ; 0F 01 /1             [286,PRIV]
SLDT r/m16                    ; 0F 00 /0             [286,PRIV]

SGDT and SIDT both take a 6-byte memory area as an operand: they store the contents of the GDTR (global descriptor table register) or IDTR (interrupt descriptor table register) into that area as a 32-bit linear address and a 16-bit size limit from that area (in that order). These are the only instructions which directly use linear addresses, rather than segment/offset pairs.

SLDT stores the segment selector corresponding to the LDT (local descriptor table) into the given operand.

The Local Descriptor Table Register (LDTR) is stored by sldt as indicated by the effective address operand. LDTR is stored into the two-byte register or the memory location.

 

by: epasquierPosted on 2009-09-30 at 05:37:18ID: 25458012

Ok, Hypo said it all while I was looking

 

by: HypoPosted on 2009-09-30 at 05:53:53ID: 25458232

Ok,
Then I get it... have heard about that test, but never seen it before... anyway your only problem then is that that the array is defined as a byte, and the operator wants a word, so you simply typecast the operand.

Try the sample below. :)

/Hypo

function Get_idt_base: Cardinal;
var
  idtr: array[0..5] of Byte;
begin
  asm
    sidt idtr
  end;
  Result := pCardinal(@idtr[2])^;
end;
 
function Get_ldtr_base: Cardinal;
var
  ldtr: array[0..4] of Byte;
begin
  ldtr[0] := $ef;
  ldtr[1] := $be;
  ldtr[2] := $ad;
  ldtr[3] := $de;
  ldtr[4] := $00;
 
  asm
    sldt Word(ldtr)
  end;
  Result := pCardinal(@ldtr[0])^;
end;

                                              
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by: MvanderKooijPosted on 2009-09-30 at 06:01:29ID: 31635271

Ah, should have thought of that, thanks, this works great (and the test itself also seems to work correctly)

 

by: epasquierPosted on 2009-09-30 at 06:03:34ID: 25458344

nope, I don't think this would do
ldtr is an ARRAY of byte, so in fact a pointer.
I'm not sure how it will work casting the pointer  to word

And I don't see the point in having array of byte anyway

function Get_ldtr_base: Cardinal;
var
  ldtr: Cardinal;
begin
  ldtr := $deadbeef; // reverse order : x86 is little-endian
  asm
    sldt ldtr
  end;
  Result := ldtr;
end;
 
it might even work with (depends on optimizations done by compiler):
 
function Get_ldtr_base: Cardinal;
begin
  Result := $deadbeef; // reverse order : x86 is little-endian
  asm
    sldt Result
  end;
end;

                                              
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by: MvanderKooijPosted on 2009-09-30 at 07:09:46ID: 25458996

Hi epasquier

That was almost right :-)

See attached code, a cast was needed...

function Get_ldtr_base: Cardinal;
begin
  Result := $deadbeef; // reverse order : x86 is little-endian
  asm
    sldt Word(Result)
  end;
end;

                                              
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by: epasquierPosted on 2009-09-30 at 07:26:02ID: 25459178

That is the beauty of Ex/Ex : everybody adds his little contribution to create the purest code possible.

Since this function does not exactly return ldtr base, and the purpose is to detect VM, then I propose

function Is_VirtualMachine: Boolean;
Var
 ldtr_base: Cardinal;
begin
 ldtr_base := $deadbeef; // reverse order : x86 is little-endian
 asm
  sldt Word(ldtr_base)
 end;
 Result:=(ldtr_base And $FFFF0000)=$dead0000;
end;

                                              
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by: MvanderKooijPosted on 2009-09-30 at 07:43:11ID: 25459423

Ehh almost :-) That and is not what you want here
This ione should do the trick

function Is_VirtualMachine: Boolean;
Var
 ldtr_base: Cardinal;
begin
 ldtr_base := $deadbeef; // reverse order : x86 is little-endian
 asm
  sldt Word(ldtr_base)
 end;
 Result:= ldtr_base <> $dead0000;
end;
                                              
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by: epasquierPosted on 2009-09-30 at 07:52:00ID: 25459543

erhhh... are you sure ? I might not have understood the trick principle

If I understand the SLDT opcode and the issue with VM :
* SLDT should but something in the 2 lower bytes (but not always 0000, or what would be the point ?)
* VM do not clear the 2 upper bytes, where other real machines does

correct ? if not can you precise ?

 

by: MvanderKooijPosted on 2009-10-01 at 23:48:13ID: 25475581

How it works, it works :-)

looks like when a memory adress is used a 32 bit value is set to it. It turns out to be working and no it is not logical that a 16 bit register instruction writes 32 bits of info.

20120131-EE-VQP-002

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