[ASM] Code explanation required

[Nick]

Can you guys explain in detail what this code is doing besides shifting bytes to the left 3 times?

      asm
        push    edx
        mov     dl, [btch]
        rol     dl, 3
        mov     [btch], dl
        pop     edx
      end;

 

[username1]

Rotate left three bits.

Nothing else can be inferred, this is taken out of context.

 

[Nick]

Thats all its doing nothing more?


{ asm
push edx
mov dl, [btch]
rol dl, 3
mov [btch], dl
pop edx
end; }
IS this the quivelent to this: btch := btch shl 3;

edit: it would be btch := btch shl 3 + 3;

 

[username1]

sage

No.

Rotate, not shift. See the Intel IA-32 programmer's reference manual volume II for more information.

 

[Nick]

Re: sage

No.

Rotate, not shift. See the Intel IA-32 programmer's reference manual volume II for more information.

Ok changed function to rotating but it kind of out puts the same.. any idea why the numbers off by 3 compared to delphi. Using vb6.

 

[username1]

Re: sage

compared to delphi. Using vb6.

Wait, what?

 

[Nick]

Re: sage

Wait, what?

Sorry, the asm is from a delphi function.. and I can't seem to figure out the appropriate function to cope for the asm in vb6.

function rol(Value : integer ; N : integer) : integer ;
var
val1:integer;
begin
val1:=8*sizeof(value);
Result := (Value shl N) or (Value shr (val1-N))
end ;

is what I'm using to ROL in delphi.. gives 64 for h and the asm function gives 67 any idea why?

 

[Negata]

Possibly it's that you're using a shift, which doesn't carry the overflowed bits back from the low-end like ROL does.

 

[Nick]

I figured it out.. was due to me not using a byte to return the value. stupid me.

 

[username1]

sage

Sorry, the asm is from a delphi function.. and I can't seem to figure out the appropriate function to cope for the asm in vb6.

Ever thought of the fact that all languages are not created equal?

Delphi > VB.