1、题目需求
读汇编是逆向基本功。
给出的文件是func函数的汇编
main函数如下
输出的结果即为flag,格式为flag{**********}
,请连flag{}一起提交
编译环境为linux gcc x86-64
调用约定为System V AMD64 ABI
请不要利用汇编器,IDA等工具。。这里考的就是读汇编与推算汇编结果的能力
int main(int argc, char const *argv[])
{
char input[] = {0x0, 0x67, 0x6e, 0x62, 0x63, 0x7e, 0x74, 0x62, 0x69, 0x6d,
0x55, 0x6a, 0x7f, 0x60, 0x51, 0x66, 0x63, 0x4e, 0x66, 0x7b,
0x71, 0x4a, 0x74, 0x76, 0x6b, 0x70, 0x79, 0x66 , 0x1c};
func(input, 28);
printf("%s\n",input+1);
return 0;
}
2、下载的汇编代码Func如下:
00000000004004e6 <func>:
4004e6: 55 push rbp
4004e7: 48 89 e5 mov rbp,rsp
4004ea: 48 89 7d e8 mov QWORD PTR [rbp-0x18],rdi //缓冲区也就是Input
4004ee: 89 75 e4 mov DWORD PTR [rbp-0x1c],esi //长度,也就是28
4004f1: c7 45 fc 01 00 00 00 mov DWORD PTR [rbp-0x4],0x1 //i=1
4004f8: eb 28 jmp 400522 <func+0x3c>
4004fa: 8b 45 fc mov eax,DWORD PTR [rbp-0x4]
4004fd: 48 63 d0 movsxd rdx,eax
400500: 48 8b 45 e8 mov rax,QWORD PTR [rbp-0x18]
400504: 48 01 d0 add rax,rdx
400507: 8b 55 fc mov edx,DWORD PTR [rbp-0x4]
40050a: 48 63 ca movsxd rcx,edx //rcx=i
40050d: 48 8b 55 e8 mov rdx,QWORD PTR [rbp-0x18]
400511: 48 01 ca add rdx,rcx //定位到Input+i地址
400514: 0f b6 0a movzx ecx,BYTE PTR [rdx] //取地址内容:Input[i]
400517: 8b 55 fc mov edx,DWORD PTR [rbp-0x4]
40051a: 31 ca xor edx,ecx //Input[i]=Input[i]^i
40051c: 88 10 mov BYTE PTR [rax],dl //保存结果
40051e: 83 45 fc 01 add DWORD PTR [rbp-0x4],0x1 //i++
400522: 8b 45 fc mov eax,DWORD PTR [rbp-0x4]
400525: 3b 45 e4 cmp eax,DWORD PTR [rbp-0x1c] //判断循环是否结束
400528: 7e d0 jle 4004fa <func+0x14>
40052a: 90 nop
40052b: 5d pop rbp
40052c: c3 ret
3、编写C语言代码解密
#include "stdafx.h"
#include <windows.h>
int main()
{
char input[] = { 0x0, 0x67, 0x6e, 0x62, 0x63, 0x7e, 0x74, 0x62, 0x69, 0x6d,
0x55, 0x6a, 0x7f, 0x60, 0x51, 0x66, 0x63, 0x4e, 0x66, 0x7b,
0x71, 0x4a, 0x74, 0x76, 0x6b, 0x70, 0x79, 0x66 , 0x1c };
for (int i = 1; i <= 28; i++)
input[i] = input[i] ^ i;
printf("%s\n", input + 1);
return 0;
}