在Linux的64位程序的命令行 [英] Process command line in Linux 64 bit
问题描述
我在访问来自Linux的64位汇编程序过程中的命令行的问题。要重现此以最小的code,我做它打印的程序名的前5个字符这32位程序:
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
此程序工作。当我把它翻译为64位和Linux的64运行,它不打印任何东西:
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
在哪里是我的错?
Where is my mistake?
推荐答案
您加载正确的地址为%RCX
。
INT 0x80的
然后调用32位的系统调用接口。截断的地址为32位,这使得它不正确。 (如果你使用调试器,只是之后的第一个 INT 0x80的
,你会看到,它在-14 返回%EAX 设置断点code>,这是
-EFAULT
)
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
.)
第二个系统调用,退出
,工程确定,因为截断到32位没有做任何伤害在这种情况下。
The second syscall, exit
, works OK because the truncation to 32 bits doesn't do any harm in that case.
如果你想传递一个64位的地址,系统调用,你将不得不使用64位系统调用接口:
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
, notint 0x80
; - different registers are used: see here;
- the system call numbers are different as well: see here.
下面是你的code的工作版本:
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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