实战逆向RUST语言程序
实战为主,近日2024年羊城杯出了一道Rust编写的题目,这里将会以此题目为例,演示Rust逆向该如何去做。
题目名称:sedRust_happyVm
题目内容:unhappy rust, happy vm
关于Rust逆向,其实就是看汇编,考验选手的基础逆向能力。在汇编代码面前,任何干扰都会成为摆设。
1、初步分析
64为程序,使用IDA 64打开
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通过字符串定位分析点
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现在我们知道 inputflag的长度大于 0x15
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接下来在汇编层面下一个断点,输入假flag,去观察相关寄存器的值
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好像并没有什么内容
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继续单步 步过,直到发现下一个要注意的地方!
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字符串长度:0x28
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我们继续单步步过跟踪
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开辟空间的时候,说明快到真正函数处理过程了。
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2、分析加密流程
2.1 base64分割模块
这里简单将 3 字节变成4字节的操作,称之为 base64分割模块
这里举个例子
输入的:"111"
->二进制字符串 001100010011000100110001
经过base64分割模块
->001100 010011 000100 110001
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发现程序执行完后正好是这样的结果
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2.2 组合
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举个例子:
假如分割之后的4字节为:
0xC、0x13、0x4、0x31
那么组合后的字符串
rax = 0xC
rcx = 0x1300
edx = 0xB1130C18
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2.3 VM处理模块
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发现func3 非常乱
并且频繁调用sub_40A800()
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发现这是一道VM类型的题,那么VM的题加密应该会很简单,基本是异或之类。
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在 sub_40A800 里面找到 异或,下断点
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这个al每经过两次就是秘钥
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解题脚本
int main() {
//提取的密文
unsigned char s1[] = { 0x00,0x82,0x11,0x92,0xa8,0x39,0x82,0x28,0x9a,0x61,0x58,0x8b,0xa2,0x43,0x68,0x89,0x4,0x8f,0xb0,0x43,0x49,0x3a,0x18,0x39,0x72,0xc,0xba,0x76,0x98,0x13,0x8b,0x46,0x33,0x2b,0x25,0xa2,0x8b,0x27,0xb7,0x61,0x7c,0x3f,0x58 };
//提取的秘钥
unsigned char s2[] = { 0x18,0xb1,0x9,0xa4,0xa6,0x2a,0x9e,0x1b,0x96,0x57,0x5d,0xad,0xae,0x75,0x65,0xac,0x9,0x8c,0xa0,0x76,0x47,0x2c,0x10,0x1,0x7c,0xf,0xba,0x47,0x95,0x30,0x9b,0x74,0x3f,0x2d,0x2d,0x9a,0x87,0x31,0xba,0x43,0x70,0x2c,0x4c };
unsigned char s3[128] = { 0 };
for (int i = 0; i < 43; i++) {
s3[i] = s1[i] ^ s2[i];
}
//还原base64分割模块
char s4[128] = { 0 };
int j = 0;
for (int i = 0; i < 44; i += 4, j += 3) {
s4[j] = (s3[i] << 2) | (s3[i + 1] >> 4);
s4[j+1] = (s3[i+1] << 4) | (s3[i + 2] >> 2);
s4[j+2] = (s3[i+2] << 6) | s3[i + 3];
}
printf("%s", s4);
return 0;
}
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