assembly code:understanding the lea command

Question

I found this code:

lea 0x10(%edi),%esi
mov %esi,0x4(%edi)

but I really don't understand this combination.

what is exactly happens on the stack on the lea-command.
is it not easier just to write: mov 0x10(%edi),0x4(edi%) ?

Carl Norum · Accepted Answer · 2013-06-11T16:57:50.993

2

Nothing happens to the stack.
It might be, but that's not a valid instruction. mov supports at most one memory operand. Anyway, your example appears to have different semantics (as mentioned by @zch below).

You can grab a copy of the Intel Software Developers Manuals and read all you want about it.

Edit: Regarding your questions "what value is written in %esi ? lea is calculation the offset? of which address?"

esi gets edi + 0x10; that's what that 0x10(%edi) means. lea stands for "load effective address". That is, it interprets edi as a pointer, and increments it by 0x10, storing the result in esi.

edited Jun 11 '13 at 16:57

answered Jun 11 '13 at 16:52

Carl Norum

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It would also mean different thing. Writing value, rather then address. – zch Jun 11 '13 at 16:53
2. I see, thanks. but 1. what value is written in %esi ? lea is calculation the offset? of which address? – Muten Roshi Jun 11 '13 at 16:53
Ah, wait. Is it just a tricky way to simply write a 0x10 to %esi? – Muten Roshi Jun 11 '13 at 16:56
1

Not quite. It writes `edi + 0x10` to `esi`. The documentation I linked to has everything you need to know (except maybe translating intel syntax to at&t syntax). – Carl Norum Jun 11 '13 at 16:58

score 0 · Answer 2 · answered Jun 14 '13 at 15:02

The use of lea vs. mov in x86 assembly is the same kind of thing as, in C/C++, saying:

char *ptr;
...
ptr = &val;

vs.

char *ptr;
...
*ptr = val;

lea calculates the address, mov (or other instructions with memory operands) dereferences (accesses) it.

So lea does in x86 assembly what's called "pointer arithmetics" in C/C++ - no memory access is involved.

assembly code:understanding the lea command

2 Answers2