汇编获取二进制

2025/4/22 18:48:01 来源：https://blog.csdn.net/wwq921220/article/details/147162863 浏览: 次关键词：汇编获取二进制

文章目录

AT&T
- asm
Intel

AT&T

mov_test.s

mov $0,%r8d

gcc -c mov_test.s 输出 mov_test.o,
objdump -D mov_test.o 查看 mov_test.o:


mov_test.o:     file format elf64-x86-64Disassembly of section .text:0000000000000000 <.text>:0:   41 b8 00 00 00 00       mov    $0x0,%r8d

mov_.S

mov %r8d,0
nop

执行命令： gcc -c mov_.S 会输出 mov_.o 文件：
objdump -D mov_.o :

mov_.o：     文件格式 elf64-x86-64Disassembly of section .text:0000000000000000 <.text>:0:   44 89 04 25 00 00 00    mov    %r8d,0x07:   00 8:   90                      nop

main.c:

void main() {
}

执行命令： gcc -S main.c 输出文件 main.s
main.s:

       .file   "main.c".text.globl  main.type   main, @function
main:
.LFB0:.cfi_startprocpushq   %rbp.cfi_def_cfa_offset 16.cfi_offset 6, -16movq    %rsp, %rbp.cfi_def_cfa_register 6noppopq    %rbp.cfi_def_cfa 7, 8ret.cfi_endproc
.LFE0:.size   main, .-main.ident  "GCC: (Uos 8.3.0.3-3+rebuild) 8.3.0".section        .note.GNU-stack,"",@progbits

doubao.s:

section .textglobal _start_start:; 将 0 存入 r8d 寄存器mov r8d, 0; 退出程序mov eax, 1xor ebx, ebxint 0x80

asm

asm_test.c:

#include <stdio.h>int main() {int num = 10;__asm__ ("movl %1, %%eax\n\t""movl %%eax, %0\n\t": "=m" (num): "r" (num): "%eax");printf("The value of num is: %d\n", num);return 0;
}