Does the LEA instruction support negative displacement?
mov rax, 1
lea rsi, [rsp - rax]
When I use the above code in my asm file I got the error:
$ nasm -f macho64 test.asm
$ error: invalid effective address
I Know that we can do pointer arithmetic like this in C:
void foo(char *a, size_t b) {
*(a - b) = 1;
}
then I assume that:
lea rsi, [rsp - rax]
will work.
And I also try to see what the GCC compiler do by using:
$ gcc -S foo.c // foo.c has the function foo(above) in it
but my asm knowleage is not enough for me the understand the asm output from the GCC compiler.
Can anyone explain why:
lea rsi, [rsp - rax] ;; invalid effective address
does not work. And I'm using these to achieve the samething:
;; assume rax has some positive number
neg rax
lea rsi, [rsp + rax]
neg rax
or
sub rsp, rax
mov rsi, rsp
add rsp, rax
What is a more standard way of doing it?
I'm using NASM version 2.11.08 compiled on Nov 26 2015 on MAC OSX 10.11
Thank you in advance for your help!
The lea instruction doesn't care about the sign of the displacement (integer constant part; it's encoded as 8-bit or 32-bit and sign-extended).
But you do need to always add the components together.
mov rax, -1
lea rsi, [rsp + rax] ; negative index in a register
lea rsi, [rsp + -1] ; negative displacement
lea rsi, [rsp - 1] ; assemblers let you write it this way
lea rdi, [rsp + rax*4 - 8] ; With all 3 components, and even scaling the idx
Remember subtracting 1 is the same as adding -1.
For example, [rax - 1] is actually encoded as [rax + (-1)]
with a disp8 of 0xFF in the machine code.
(The scale factor is encoded as a 2-bit left-shift count, so it can't be negative either.)
See also Referencing the contents of a memory location. (x86 addressing modes) for details on how x86 addressing modes work.