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Lods byte ptr ds:[esi] ;i know that it loads a byte that is a pointer my question is " is the pointer ds with esi as an offset or is it the other way around or am i just plain wrong" ?

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Dani AI

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A short, practical clarification (building on and ):

ds:[esi] means “memory at the address formed by the DS segment base plus the contents of ESI.” The assembler/CPU treats ESI as an offset; the segment register supplies the segment base (so the effective address is segment-base + ESI) and the LODS family reads from that memory into AL/AX/EAX and then updates (E)SI automatically. See the Intel instruction description for LODS. (Intel SDM — LODS reference). (studizba.com)

How that becomes a runnable address depends on CPU mode: in 16-bit real mode the CPU forms a 20-bit address by shifting the 16-bit segment value left 4 bits and adding the offset; in protected mode the segment register is a selector and the segment descriptor supplies a 32/64-bit base that is added to the offset (and if paging is enabled that linear address is then translated to a physical address). (Intel SDM — segmentation and addressing). (manuals.plus)

Defaults and overrides you should watch for: most memory operands use DS by default for index registers like (E)SI; accesses that use BP/RBP or SP/RSP default to SS. You can supply segment-override prefixes (cs:, ds:, es:, fs:, gs:, ss:) to change the chosen segment on a particular instruction; the LODS documentation notes DS can be overridden when appropriate. Also note that in 64-bit (long) mode CS/DS/ES/SS are effectively flat (base = 0) and only FS/GS carry nonzero bases for things like TLS — so many segment overrides are ignored in long mode. (studylib.net)

Little, immediately useful example (behavior controlled by the direction flag DF):

; forward (lowest->highest addresses)
cld         ; clear DF
lodsb       ; AL := byte ptr [DS:ESI], ESI += 1

; backward (highest->lowest)
std         ; set DF
lodsb       ; AL := byte ptr [DS:ESI], ESI -= 1

For authoritative detail read the LODS and segmentation sections of the Intel SDM linked above; they state the DS:(E)SI convention, the DF behavior, and the rules for segment bases/overrides. (studizba.com)

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Member Avatar for Ancient Dragon

esi just contains an address that is within the ds segment. ds is not a pointer -- it contains the address of the data segment. Sometimes you might even see cs:[esi] or even ss:[esi] in which case esi is pointer to somewhere in the code segment or stack segment, respectively.

Member Avatar for avarionist

esi just contains an address that is within the ds segment. ds is not a pointer -- it contains the address of the data segment. Sometimes you might even see cs:[esi] or even ss:[esi] in which case esi is pointer to somewhere in the code segment or stack segment, respectively.

thanks AncientDragon :) you have all the answers :D

Member Avatar for wildgoose

The default for a [si] [esi] [rsi] access is the ds: Data segment / selector. But it can be overriden with an alternate such as AncientDragon indicated
cs: Code Segment/Selector,
es:Extra Segment/Selector,
ss: Stack Segment/Selector
fs, gs Selectors

Another item to note is LODS also increments the index by the size of the data referenced!

Assuming 32-bit Protected Mode

lodsb
   mov al,ds:[esi]
   add esi,1
lodsw
   mov ax,ds:[esi]
   add esi,2
lodsd
   mov eax,ds:[esi]
   add esi,4

You can use the Repeat instruction rep

rep lodsd
L1: mov eax,ds:[esi]
   add esi,4
   dec ecx
   jne L1
Member Avatar for avarionist

The default for a [si] [esi] [rsi] access is the ds: Data segment / selector. But it can be overriden with an alternate such as AncientDragon indicated
cs: Code Segment/Selector,
es:Extra Segment/Selector,
ss: Stack Segment/Selector
fs, gs Selectors

Another item to note is LODS also increments the index by the size of the data referenced!

Assuming 32-bit Protected Mode

lodsb
   mov al,ds:[esi]
   add esi,1
lodsw
   mov ax,ds:[esi]
   add esi,2
lodsd
   mov eax,ds:[esi]
   add esi,4

You can use the Repeat instruction rep

rep lodsd
L1: mov eax,ds:[esi]
   add esi,4
   dec ecx
   jne L1

Thanks for the info lods is an instruction i rarely see let alone use :)

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