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I'm reversing an x86 binary and I have an instruction that using the at&t syntax is

movl 0xc(%ebp), %eax

while using the intel syntax is

mov eax, dword [ebp+arg_3]

I'm using radare2, and at the start of the function it prints

; arg int arg_0_1      @ ebp+0x1
; arg int arg_3        @ ebp+0xc

Why [ebp+arg_3] instead of a simple substitution like [arg_3]?

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    your question has a typo - the '-' should be '+'. – peter ferrie Jul 29 '16 at 15:36
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The reference to ebp makes explicit which register is being used to access the variable, and also the fact that the reference is relative, not absolute. A compiler might have produced an instruction using esp instead, if no stack frame exists, or any other register if the code were written by hand.

Further, to show only [arg_3] implies that the access is equivalent to another memory location such as [401000h], but with a name instead of a number.

  • Could you add a generic expression using the syntax used in the OP? – Giuseppe Crinò Jul 29 '16 at 15:48
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    @giuscri it's simply a convention chosen by authors of a disassembler you're using. They seem to simply equate arg3 to 0xc and not bother with details :) If your question is about "; arg int arg_3 @ ebp+0xc" - then yeah I agree it's a bit illogical but hey, ask them to fix it :) – Vitaly Osipov Aug 3 '16 at 8:04
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    hi I am one of the authors, the reason why we made it like this is that many of the arguments/variables manipulating commands requires prior knowledge of that base register you will use, so think of the [ebp+arg_3] as if I was trying to say arg_3 is offsetted with respect to ebp instead of regular + operator – Ahmed Abd El Mawgood Aug 7 '16 at 8:28

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