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【网络安全】关于简单程序的简单逆向分析，零基础入门到精通教程！

news 2026/7/3 5:10:25

前言

本博文以两道题目为例，初步体验通过对程序的逆向分析来增强自身对代码的理解和运用，提高编译和汇编能力。

题一：

从网站中下载下来一个easyre.exe文件，查壳，发现有壳，脱壳，

将脱壳后的程序扔进ida，发现主要函数，

不难发现红框内的就是重点，通过for循环次数可知flag内容有12位字符串，

双击byte_402000，即可编写脚本，

# -*- coding:utf-8 -*- v4 = [42,70,39,34,78,44,34,40,73,63,43,64] //由于后半段字符串中包含双引号，为避免引发歧义，故后半段用单引号表示引用 model = r"}|{zyxwvutsrqponmlkjihgfedcba`_^]\[ZYXWVUTSRQPONMLKJIHGFEDCBA@?>=<;:9876543210/.-,+*)(" + chr(0x27) + r'&%$# !"' pos = [] for i in v4: pos.append(model.find(chr(i))+1) s = [chr(x + 1) for x in pos] flag = ''.join(s) print ('flag{'+flag+'}')

获得flag，flag{U9X_1S_W6@T?}。

题二：

从网站中下载来一个reverse_3.exe, 检查无壳，直接扔进ida进行反编译。

发现主函数_main_0，

int __cdecl main_0(int argc, const char **argv, const char **envp) { int v3; // eax const char *v4; // eax size_t v5; // eax signed int j; // [esp+DCh] [ebp-ACh] signed int i; // [esp+E8h] [ebp-A0h] signed int v9; // [esp+E8h] [ebp-A0h] char Dest[108]; // [esp+F4h] [ebp-94h] char Str; // [esp+160h] [ebp-28h] char v12; // [esp+17Ch] [ebp-Ch] for ( i = 0; i < 100; ++i ) { if ( (unsigned int)i >= 0x64 ) j____report_rangecheckfailure(); Dest[i] = 0; } sub_41132F("please enter the flag:"); sub_411375("%20s", &Str); v3 = j_strlen(&Str); v4 = (const char *)sub_4110BE((int)&Str, v3, (int)&v12); strncpy(Dest, v4, 0x28u); v9 = j_strlen(Dest); for ( j = 0; j < v9; ++j ) Dest[j] += j; v5 = j_strlen(Dest); if ( !strncmp(Dest, Str2, v5) ) sub_41132F("rigth flag!\n"); else sub_41132F("wrong flag!\n"); return 0; }

仔细一看还是蛮简单的，主要经过三个步骤：

先用str存储所输入的字符串，然后对str进行sub_4110BE加密。
接着使用一个for循环进行变换。
最后与str2比较，由此可知str2就是加密后的flag。

str2字符串如下，

.data:0041A034 ; char Str2[] .data:0041A034 Str2 db 'e3nifIH9b_C@n@dH',0 ; DATA XREF: _main_0+142↑o

sub_4110BE加密函数如下，

void *__cdecl sub_411AB0(char *a1, unsigned int a2, int *a3) { int v4; // STE0_4 int v5; // STE0_4 int v6; // STE0_4 int v7; // [esp+D4h] [ebp-38h] signed int i; // [esp+E0h] [ebp-2Ch] unsigned int v9; // [esp+ECh] [ebp-20h] int v10; // [esp+ECh] [ebp-20h] signed int v11; // [esp+ECh] [ebp-20h] void *Dst; // [esp+F8h] [ebp-14h] char *v13; // [esp+104h] [ebp-8h] if ( !a1 || !a2 ) return 0; v9 = a2 / 3; if ( (signed int)(a2 / 3) % 3 ) ++v9; v10 = 4 * v9; *a3 = v10; Dst = malloc(v10 + 1); if ( !Dst ) return 0; j_memset(Dst, 0, v10 + 1); v13 = a1; v11 = a2; v7 = 0; while ( v11 > 0 ) { byte_41A144[2] = 0; byte_41A144[1] = 0; byte_41A144[0] = 0; for ( i = 0; i < 3 && v11 >= 1; ++i ) { byte_41A144[i] = *v13; --v11; ++v13; } if ( !i ) break; switch ( i ) { case 1: *((_BYTE *)Dst + v7) = aAbcdefghijklmn[(signed int)(unsigned __int8)byte_41A144[0] >> 2]; v4 = v7 + 1; *((_BYTE *)Dst + v4++) = aAbcdefghijklmn[((byte_41A144[1] & 0xF0) >> 4) | 16 * (byte_41A144[0] & 3)]; *((_BYTE *)Dst + v4++) = aAbcdefghijklmn[64]; *((_BYTE *)Dst + v4) = aAbcdefghijklmn[64]; v7 = v4 + 1; break; case 2: *((_BYTE *)Dst + v7) = aAbcdefghijklmn[(signed int)(unsigned __int8)byte_41A144[0] >> 2]; v5 = v7 + 1; *((_BYTE *)Dst + v5++) = aAbcdefghijklmn[((byte_41A144[1] & 0xF0) >> 4) | 16 * (byte_41A144[0] & 3)]; *((_BYTE *)Dst + v5++) = aAbcdefghijklmn[((byte_41A144[2] & 0xC0) >> 6) | 4 * (byte_41A144[1] & 0xF)]; *((_BYTE *)Dst + v5) = aAbcdefghijklmn[64]; v7 = v5 + 1; break; case 3: *((_BYTE *)Dst + v7) = aAbcdefghijklmn[(signed int)(unsigned __int8)byte_41A144[0] >> 2]; v6 = v7 + 1; *((_BYTE *)Dst + v6++) = aAbcdefghijklmn[((byte_41A144[1] & 0xF0) >> 4) | 16 * (byte_41A144[0] & 3)]; *((_BYTE *)Dst + v6++) = aAbcdefghijklmn[((byte_41A144[2] & 0xC0) >> 6) | 4 * (byte_41A144[1] & 0xF)]; *((_BYTE *)Dst + v6) = aAbcdefghijklmn[byte_41A144[2] & 0x3F]; v7 = v6 + 1; break; } } *((_BYTE *)Dst + v7) = 0; return Dst; }

对aAbcdefghijklmn函数进行查看，

.rdata:00417B30 aAbcdefghijklmn db 'ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/=' .rdata:00417B30 ; DATA XREF: .text:004117E8↑o .rdata:00417B30 ; .text:00411827↑o ...

不难发现这是base64加密，因此只需解密即可。

编写脚本，

import base64 str1 = 'e3nifIH9b_C@n@dH' x = '' flag = '' for j in range(0, len(str1)): x += chr(ord(str1[j]) - j) flag = base64.b64decode(x) flag = flag.decode('ASCII') print(flag)

运行结果，