Process command line in Linux 64 bit

 

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  • What happens if you use the 32-bit int 0x80 Linux ABI in 64-bit code? 1 answer

    • You are loading the correct address into %rcx .

    int 0x80 then invokes the 32-bit syscall interface. That truncates the address to 32 bits, which makes it incorrect. (If you use a debugger and set a breakpoint just after the first int 0x80 , you will see that it returns with -14 in %eax , which is -EFAULT .)

    The second syscall, exit , works OK because the truncation to 32 bits doesn't do any harm in that case.

    If you want to pass a 64-bit address to a system call, you will have to use the 64-bit syscall interface:

  • use syscall , not int 0x80 ;
  • different registers are used: see here;
  • the system call numbers are different as well: see here.

    • Here is a working version of your code: 

    .section .text

.globl _start
_start:
 movq  %rsp, %rbp

 movq $1, %rax
 movq $1, %rdi
 movq 8(%rbp), %rsi       # program name address ?
 movq $5, %rdx
 syscall

 movq $60, %rax
 movq $0, %rdi
 syscall

    As stated in the X86_64 ABI: Use the syscall instruction instead of int $0x80 . The Kernel uses different registers in 64 Bit as syscall arguments, and the number assigned for a syscall function varies between i386 and x86_64, too.

    An example - in german, sorry - can be found here:
    http://zygentoma.de/codez/linux_assembler.php

    链接地址: http://www.djcxy.com/p/72446.html

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