本文介绍了如何使它在x86-64的装配工作?的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
最近,我正在学习组装,现在我有一些困惑。我从学会了。
Recently I'm learning assembly and now i have some confusion. I learned it from Professional Assembly language.
我的系统的拱门:
#uname -m
x86_64
这是我的code:
.section .data
output:
.asciz "This is section %d\n"
.section .text
.globl _start
_start:
pushq $1
pushq $output
call printf
addq $8, %rsp
call overhere
pushq $3
pushq $output
call printf
addq $8, %rsp
pushq $0
call exit
overhere:
pushq %rbp
movq %rsp, %rbp
pushq $2
pushq $output
call printf
addq $8, %rsp
movq %rbp, %rsp
popq %rbp
ret
我组装,链接,像这样运行它,得到所示的错误消息:
I assemble, link and run it like this, getting the error message shown:
#as -o calltest.o calltest.s
#ld -dynamic-linker /lib64/ld-linux-x86-64.so.2 -lc -o calltest calltest.o
#./calltest
Segmentation fault
我如何工作的?
推荐答案
x86_64的
有另外一种传递参数,请参阅:的
x86_64
has another kind of passing arguments, see: http://en.wikipedia.org/wiki/X86_calling_conventions#System_V_AMD64_ABI
这就是你们的榜样将如何工作:
This is how your example would work:
.section .data
output:
.asciz "This is section %d\n"
.section .text
.globl _start
_start:
movq $output, %rdi # 1st argument
movq $1, %rsi # 2nd argument
xorl %eax, %eax # no floating point arguments
call printf
call overhere
movq $output, %rdi # 1st argument
movq $3, %rsi # 2nd argument
xorl %eax, %eax # no floating point arguments
call printf
xor %edi, %edi
call exit
overhere:
pushq %rbp
movq %rsp, %rbp
movq $output, %rdi # 1st argument
movq $2, %rsi # 2nd argument
xorl %eax, %eax # no floating point arguments
call printf
movq %rbp, %rsp
popq %rbp
ret
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