LEA confusing me

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Teehee

Joined: 05 Aug 2009
Posts: 570
Location: Brazil

Teehee 31 May 2011, 19:39

Code:

mov eax,1
lea eax,[eax+4]
; eax = 5

if eax has 1, + 4, so eax = 5, ok. But loading as [xxx] it seams you are returning a value pointed for an address, and not the result itself, like in:

Code:

mov eax,1
mov eax,[eax+4]
; eax = any value at address 5

for me this is more readable:

Code:

mov eax,1
lea eax,eax+4
; eax = 5

get my point? Razz

what do you think?

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Sorry if bad english.

31 May 2011, 19:39

edfed

Joined: 20 Feb 2006
Posts: 4334
Location: Now

edfed 31 May 2011, 19:52

i think the reason you think that is because you still not pointed that lea means load, and not move.

every load instructions in x86 uses a pointer to an adress as an operand.

for example, lgdt loads the content of gdtr pseudo descriptor in a ram location.

Code:

lgdt  [ebx+eax*8+sys.gdt]

for example, will use the conventionnal adress formula as an operand,

there are lea, lgdt, lidt, lldt, ltr, lsl, etc... all these instructions uses a memory operand.

and lea is just the Load Effective Adress, means exactlly what it does, it loads the effective adress using the 32 bits adress operand.

Code:

[segment : offset + pointer + index * scale ]
         : effective adress

the segment part is not included in the result of the adress calculation because the result will be relative to a segment.

31 May 2011, 19:52

ouadji

Joined: 24 Dec 2008
Posts: 1081
Location: Belgium

ouadji 31 May 2011, 21:20

most often, I use "lea" to compute a "result",
and not to compute an address.

Code:

        lea     ebx,[(ecx*2) - 6]
   mul     ebx

I like this opcode, very convenient
and much more powerful than it appears at the first glance.

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31 May 2011, 21:20

Teehee

Joined: 05 Aug 2009
Posts: 570
Location: Brazil

Teehee 31 May 2011, 22:12

now it makes more sense Smile

thanks

31 May 2011, 22:12

MazeGen

Joined: 06 Oct 2003
Posts: 977
Location: Czechoslovakia

MazeGen 01 Jun 2011, 14:04

LEA is one of those weird instructions in x86 ISA. It uses source operand that has nothing to do with memory, it is just encoded as a memory operand.

01 Jun 2011, 14:04

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