Is there an offset writeback on x86?

Question

With ARM I can access a memory location like this:

LDR r0, [r1, #4]!

Meaning I want to load the value pointed to by r1, with an offset of 4 bytes, and the ! means that I want the value of r1 to be updated after this instruction with the new value (i.e., +4).

Is there an equivalent on x86 (AT&T syntax if possible)? For instance:

movl 4(%ebx), %eax

This will load the value pointed by %ebx with an offset of 4 bytes into %eax. How do I update the value of %ebx after the load though?

Thanks.

Answer 1

I think the instruction you're looking for is LODSD which means "load DWORD from string". You don't get to pick the registers though: The pointer is in ESI and the destination register is EAX.

After the byte, word, or doubleword is transferred from the memory location into the AL, AX, or EAX register, the (E)SI register is incremented or decremented automati- cally according to the setting of the DF flag in the EFLAGS register. (If the DF flag is 0, the (E)SI register is incremented; if the DF flag is 1, the ESI register is decre- mented.) The (E)SI register is incremented or decremented by 1 for byte operations, by 2 for word operations, or by 4 for doubleword operations.

Sorry, I've got Intex syntax ingrained in my head right now. Assuming flat 32-bit protected mode:

mov   esi, _source_data_       ; ESI points to source data
cld                            ; Clear the direction flag
                               ; (ESI will increment)
lodsd                          ; Essentially   mov eax, [ds:esi]
                               ;               add esi, 4