攻防世界----＞Windows_Reverse1（补）

做题笔记。

做题回顾。

假设，我们不知道地址随机怎么办？不能动调，只能静态分析。

下载查壳

upx脱壳。

32ida打开。

动调报错。

重新打开，静态分析。

跟进关键函数。

不明白可以反汇编和汇编一起看。

溯源。

*decode 取值等于 byte_ [xxx] 是否说明了byte_ 是一张解密表？

可是没内容？

我们对 byte_ [xxx] 进行hex跟踪。

“ ？ “ 表示为不可见字符。用0代替就好。

而这，又是一个整体

进行整理：

编写思路：

脚本：

#include <stdio.h>
#include <string.h>
#include <stdlib.h>int main()
{unsigned char encode_table[] ={0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x4E,0xE6,0x40,0xBB,0xB1,0x19,0xBF,0x44,0xFF,0xFF,0xFF,0xFF,0xFF,0xFF,0xFF,0xFF,0xFE,0xFF,0xFF,0xFF,0x01,0x00,0x00,0x00,0x7E,0x7D,0x7C,0x7B,0x7A,0x79,0x78,0x77,0x76,0x75,0x74,0x73,0x72,0x71,0x70,0x6F,0x6E,0x6D,0x6C,0x6B,0x6A,0x69,0x68,0x67,0x66,0x65,0x64,0x63,0x62,0x61,0x60,0x5F,0x5E,0x5D,0x5C,0x5B,0x5A,0x59,0x58,0x57,0x56,0x55,0x54,0x53,0x52,0x51,0x50,0x4F,0x4E,0x4D,0x4C,0x4B,0x4A,0x49,0x48,0x47,0x46,0x45,0x44,0x43,0x42,0x41,0x40,0x3F,0x3E,0x3D,0x3C,0x3B,0x3A,0x39,0x38,0x37,0x36,0x35,0x34,0x33,0x32,0x31,0x30,0x2F,0x2E,0x2D,0x2C,0x2B,0x2A,0x29,0x28,0x27,0x26,0x25,0x24,0x23,0x22,0x21,0x20,0x00};unsigned char encode[] = "DDCTF{reverseME}";char flag[50] = " ";for (int i = 0; i < strlen(encode); i++){flag[i] = encode_table[encode[i]];}printf("flag{%s}\n", flag);system("pause");return 0;
}