Subtracting registers with an LEA instruction?

Question

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!

@MikeCAT simply enough, you can make this as answers. Thank you! — user62453, Jun 05 '16 at 13:22
ARM addressing modes can add or subtract the index register, but [x86 can only add (with an optional scale factor)](http://stackoverflow.com/a/34058400/224132). — Peter Cordes, Jun 05 '16 at 16:50

score 12 · Accepted Answer · answered Jun 05 '16 at 13:24

12

The lea instruction doesn't care about the sign of the displacement. But you do need to always add the components together.

mov rax, -1
lea rsi, [rsp + rax]

Remember subtracting 1 is the same as adding -1.

answered Jun 05 '16 at 13:24

Sep Roland

1 Answers1