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A question about using brackets in assembly language?
Sorry for the very unspecific title. I don't know what it is called when you do this. I know that in Intel syntax, putting brackets around a memory address means "the value at that address". So, if we wrote:
mov eax, [ebx]
It would take the value at the address stored in ebx and put it into eax. But what if we do something like this:
mov eax, [ebx-24]
Would this take the value at address ebx-24 and store it into eax (like you would think), or would it simply take the value of ebx-24 and store it into eax?
Also, what would this do?
lea eax, [ebx-24]
How is this different from
mov eax, [ebx-24]
1 Answer
- tLv 57 years agoFavorite Answer
Yes, `mov eax, [ebx-24]` takes the value at address `ebx - 24`. The `lea` instruction computes an address the same way as `mov`, but instead of loading the value stored at that address it loads the address itself. So `lea eax, [ebx-24]` does the same thing as (pseudocode) `eax := ebx - 24`.
