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If there is an address of a char array in eax, what's the difference between 0(%eax) and (%eax)?Or are they the same, both refer to the first element in the array?

melpomene
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vingin
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    when you assembled then disassembled them did you see a difference? – old_timer May 17 '17 at 21:06
  • Think about what `0(%eax)` means. It means that there is an offset of 0 from the value in `eax`. Now, what does `(%eax)` mean? It means the value in `eax`. So as long as `x == x + 0`, no, there is obviously no difference. – Cody Gray - on strike May 19 '17 at 08:21
  • If you're seeing something like `0(%eax)` in disassembler output (as opposed to assembly source code) then the 0 is probably the target of a relocation that would change it some other value when the object file is linked. – Ross Ridge May 19 '17 at 18:51

2 Answers2

2

They are exactly the same. You could prove this easily by assembling and then disassembling the two instructions:

movl (%eax),%edx
movl 0(%eax),%edx

Disassembly of section .text:

0000000000000000 <.text>:
   0:   67 8b 10                mov    (%eax),%edx
   3:   67 8b 10                mov    (%eax),%edx

Notice that they have exactly the same encoding in bytes.

Cody Gray - on strike
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old_timer
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0

They are the same.

Asuming that you had eax set correctly to point to the address of the first element in the array, would could use movb to put a byte of data in the array using it.

lostbard
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  • What does `MOVB` have to do with anything here? If it's just a random example, it's particularly confusing because you cannot use `MOVB` with `EAX`, as shown in the question; a BYTE-sized register is required. Perhaps a better example would have been `MOVZBL`? Or just `MOVL` to store a DWORD-sized value. – Cody Gray - on strike May 19 '17 at 08:24
  • The question asked about char array in eax, so accessing elements of the array would require a movb. ie: movb $32, 0(%eax) – lostbard May 19 '17 at 10:14