Addressing mode

[buli]

<Base indexed addressing>

Code: [Select]

.MODEL SMALL
.STACK
.DATA
  T  BYTE ' ABCDEFGHIJKLMNOPQRSTUVWXYZ$'
  .CODE
MAIN:
      MOV AX,@DATA
    MOV DS,AX
    MOV BX,01H
            MOV SI,01H
            MOV DL,[BX][SI]       
    MOV AH,02H
    INT 21H
    MOV AH,4CH
    INT 21H

   END main
Result==> L
<Register indirect addressing>

Code: [Select]

.MODEL SMALL
.STACK
.DATA
        T  BYTE 'ABCDEFGHIJKLMNOPQRSTUVWXYZ$'
.CODE
MAIN:
      MOV AX,@DATA
    MOV DS,AX
    MOV BX,02H         
            MOV DL,[BX]         
    MOV AH,02H
    INT 21H
    MOV AH,4CH
    INT 21H

   END main
Result==> A
WHY they get different results ?
In my point,i think that both results are same because they addressing same address.

[Adamanteus]

Address of variable is missed :

Code: [Select]

MOV DL, T[BX][SI]
MOV DL, T[BX]


[buli]

Thanks sir
But  if i don't interpose T,Are their address different!?

[FORTRANS]

Hi,

But  if i don't interpose T,Are their address different!?

   Without the T to specify the variable address, an address of
zero will be used.  If the T variable is the first variable declared
in the data segment, it will have an address of zero.  If anything
is specified before T, T will not have an address of zero.  The
variable's address is its offset from the beginning of the segment.

HTH,

Steve N.

[nidud]

deleted

[nidud]

deleted

[FORTRANS]

Hi,

The segments are calculated at a 16-byte boundary so T may not be the same as DATA. There is also a SPACE in front of A in the first sample but not the second.

   Good catch, thanks.   The DATA area is called a segment (using
the SEGMENT directive) but it can be byte-aligned.  The segment
registers are limited to a 16 byte resolution.  According to a
textbook, the default alignment of DATA is a paragraph (16 bytes)
for earlier processors, and DWORD (8 bytes) when using a  .80386
directive.  And presumably then, for all of the 32-bit processors.
Old textbook.  And other alignment types can be specified.

   So an address of a variable is its offset from the start of the DATA
area, plus the DATA area's offset from a 16-byte alignment.

Regards,

Steve

P.S.  While I was typing, you added a WORD aligned example.  It
is not that the CODE and DATA areas overlap as such, but that the
DATA area is not on a 16-byte boundary.  The addressing of the
segment registers "overlap".

P.P.S.  My code is usually set up to have the DATA area on a 16-byte
boundary.  So I forgot that it need not be aligned.

SRN
SRN

[nidud]

deleted

[buli]

Thanks FORTRANS  and Nidud a lot
How patient you are!!
I think that i forgot a useful tool----"Debug"
Well,there some point i still can't understand(Occupy 30%). But i expected that as i finish the course,i will realize them.