在 Linux 64 位中处理命令行 [英] Process command line in Linux 64 bit

问题描述

我在从 Linux 64 位汇编程序访问进程命令行时遇到问题.为了用最少的代码重现这一点，我制作了这个 32 位程序，它打印程序名称的前 5 个字符:

<前>.section .text.globl _start_开始:movl %esp, %ebpmovl $4, %eax # 写movl $1, %ebx # 标准输出movl 4(%ebp), %ecx #程序名地址(argv[0])movl $5, %edx # 硬编码长度整数 $0x80movl $1, %eaxmovl $0, %ebx整数 $0x80

这个程序正在运行.当我将其转换为 64 位并在 Linux 64 中运行时，它不打印任何内容:

<前>.section .text.globl _start_开始:movq %rsp, %rbpmovq $4，%raxmovq $1, %rbxmovq 8(%rbp), %rcx # 程序名称地址?movq $5，%rdx整数 $0x80movq $1, %raxmovq $0, %rbx整数 $0x80

我的错误在哪里?

解决方案

您正在将正确的地址加载到 %rcx 中.

int 0x80 然后调用 32 位系统调用接口.这会将地址截断为 32 位，从而使其不正确.(如果您使用调试器并在第一个 int 0x80 之后设置断点，您将看到它在 %eax 中返回 -14，即 -EFAULT.)

第二个系统调用 exit 工作正常，因为在这种情况下截断为 32 位不会造成任何伤害.

<小时>

如果要将 64 位地址传递给系统调用，则必须使用 64 位系统调用接口:

• 使用syscall，而不是int 0x80;
• 使用了不同的寄存器:参见这里;
• 系统调用号也不同:参见这里.

这是您的代码的工作版本:

.section .text.globl _start_开始:movq %rsp, %rbpmovq $1, %raxmovq $1, %rdimovq 8(%rbp), %rsi # 程序名称地址?movq $5，%rdx系统调用movq 60 美元，%raxmovq $0, %rdi系统调用

I have problems accessing the process command line from Linux 64 bit Assembly program. To reproduce this with minimal code, I made this 32-bit program which prints first 5 characters of the program name:

.section .text

.globl _start
_start:
 movl  %esp, %ebp

 movl $4, %eax        # write
 movl $1, %ebx        # stdout
 movl 4(%ebp), %ecx   # program name address (argv[0])
 movl $5, %edx        # hard-coded length
 int  $0x80

 movl $1, %eax
 movl $0, %ebx
 int  $0x80

This program is working. When I translate it to 64 bit and run in Linux 64, it doesn't print anything:

.section .text

.globl _start
_start:
 movq  %rsp, %rbp

 movq $4, %rax
 movq $1, %rbx
 movq 8(%rbp), %rcx       # program name address ?
 movq $5, %rdx
 int  $0x80

 movq $1, %rax
 movq $0, %rbx
 int  $0x80

Where is my mistake?

解决方案

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

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