I wrote a very simple assembly program (x64, linux) which reads the contents from /etc/passwd (doesn't really matter from where) and writes it to stdout. I compile it with nasm (nasm -f elf64 foo.asm -o bar
). I receive the following error: ./bar: 24: Syntax error: EOF in backquote substitution
Here is my code:
global _start
section .data
fn: db '/etc/passwd',0
section .bss
section .text
_start:
; open the file
xor rax,rax
add al, 2
lea rdi, [fn]
xor rsi, rsi
syscall
; read the file, use some area
; in the stack to store the contents
mov rdi, rax
sub sp, 0xfff
lea rsi, [rsp]
xor rdx, rdx
mov dx, 0x200
xor rax, rax
syscall
; write to stdout
xor rdi, rdi
add dil, 1
mov rdx, rax
xor rax, rax
add al,1
syscall
; exit
xor rax,rax
add al, 60
syscall
Is is also possible to get more info about the error? the program compiles with no errors in nasm. Thanks :)!