Question: lea rsi, [rsp] in this context equivalent to mov rsi, rsp?

[sy2002]

Thank you for writing the blog article https://gaultier.github.io/blog/x11_x64.html

Was a fun read! :-)

I do have a question about this very piece of code:

%define SYSCALL_WRITE 1
%define STDOUT 1

print_hello:
  push rbp ; Save rbp on the stack to be able to restore it at the end of the function.
  mov rbp, rsp ; Set rbp to rsp

  sub rsp, 5 ; Reserve 5 bytes of space on the stack.
  mov BYTE [rsp + 0], 'h' ; Set each byte on the stack to a string character.
  mov BYTE [rsp + 1], 'e'
  mov BYTE [rsp + 2], 'l'
  mov BYTE [rsp + 3], 'l'
  mov BYTE [rsp + 4], 'o'

  ; Make the write syscall
  mov rax, SYSCALL_WRITE
  mov rdi, STDOUT ; Write to stdout.
  lea rsi, [rsp] ; Address on the stack of the string.
  mov rdx, 5 ; Pass the length of the string which is 5.
  syscall

  add rsp, 5 ; Restore the stack to its original value.

  pop rbp ; Restore rbp
  ret

You write lea rsi, [rsp] .

But if you would omit the brackets [ and ] : Wouldn't then a

mov rsi, rsp equivalent and sufficient in this situation?

I am not super fit in X86 assembly, so I might be totally wrong - just curious because I thought the brackets are "de-referencing" the pointer (i.e. deliver the value at the address) so what if you are not de-referencing the pointer in the first place and just move the actual stack pointer?

[gaultier]

Hi!
In this context, yes. mov & lea overlap in some cases, here they are the same, with one caveat.

On macos, lea is preferred: https://stackoverflow.com/a/47301555

I have not tested this code on macos, but I expect mov to fail and lea to succeed (see the stack overflow link).
On other Unices, changing lea to mov here should be perfectly fine.

[sy2002]

Thank you for taking the time and for your very helpful answer: You boosted my understanding of x86 assembly :-)