前置知识
LLVM是C++编写的构架编译器的框架系统，可用于优化以任意程序语言编写的程序。
LLVM IR可以理解为LLVM平台的汇编语言，所以官方也是以语言参考手册(Language Reference Manual)的形式给出LLVM IR的文档说明。既然是汇编语言，那么就和传统的CUP类似，有特定的汇编指令集。但是它又与传统的特定平台相关的指令集(x86,ARM,RISC-V等)不一样，它定位为平台无关的汇编语言。也就是说，LLVM IR是一种相对于CUP指令集高级，但是又是一种低级的代码中间表示（比抽象语法树等高级表示更加低级)。
LLVM IR即代码的中间表示，有三种形式：
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.ll 格式：人类可以阅读的文本(汇编码) -->这个就是我们要学习的IR
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.bc 格式：适合机器存储的二进制文件
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内存表示
下面给出.ll格式和.bc格式生成及相互转换的常用指令清单：
.c -> .ll：clang -emit-llvm -S a.c -o a.ll.c -> .bc: clang -emit-llvm -c a.c -o a.bc.ll -> .bc: llvm-as a.ll -o a.bc.bc -> .ll: llvm-dis a.bc -o a.ll.bc -> .s: llc a.bc -o a.s
那么我们以一道CTF赛题来分析实验，学习LLVM IR
实验解析
题目附件直接给出了中间表示.II文件
打开查看一下汇编码，毕竟.II文件是人类可以阅读的文本，这边笔者使用的是Sublime Text（使用VScode查看即可）代码量不多，大概600行
题目初步分析
我们直接寻找一下main函数
我们可以看出题目经历了两次RC4，然后Base64，我们从上面可以看到密文，RC4_key,我们直接一把锁，cyberchef启动，会发现解不出来，那么程序应该做了其他的操作，最朴素的，我们可以想到把RC4魔改了，base64魔改等等。
So！继续学习研究ing
.II详细分析
所以本着学习的态度，我们这时候应该掏出LLVM Language Reference Manual(官方文档)来简单了解学习一些常见指令、符号标识以及特性。这边给出一些分析 .ll 中间文件的算法流程
@ - 全局变量% - 局部变量alloca - 在当前执行的函数的堆栈帧中分配内存，当该函数返回到其调用者时，将自动释放内存i32 - 32位4字节的整数align - 对齐load - 读出，store写入icmp - 两个整数值比较，返回布尔值br - 选择分支，根据条件来转向label，不根据条件跳转的话类型gotolabel - 代码标签call - 调用函数
首先看到一些全局变量，知道了RC4_key = llvmbitccipher = "TSzkWKgbMHszXaj@kLBmRrnTxsNtZsSOtZzqYikCw="
我们继续分析，重点分析各个function
b64encode
b64encode 魔改
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每三个字符，24位，切分成4断，每段6位。
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将6位对应的值 (value+ 59)&0xff 则是编码后的值。
%22 = getelementptr inbounds i8, i8* %19, i64 %21 // 取出当前处理字符 %23 = load i8, i8* %22, align 1 %24 = zext i8 %23 to i32 // 类型强制转化 %25 = ashr i32 %24, 2 // 算数右移两位 input[i]>>2 %26 = add nsw i32 %25, 59 // input[i]+59 %27 = trunc i32 %26 to i8 // 强制转化 相当于 &0xff %28 = load i8*, i8** %6, align 8 %29 = load i32, i32* %9, align 4 %30 = sext i32 %29 to i64 %31 = getelementptr inbounds i8, i8* %28, i64 %30 // 存储base64 编码串 store i8 %27, i8* %31, align 1 %32 = load i8*, i8** %4, align 8 %33 = load i32, i32* %7, align 4 %34 = sext i32 %33 to i64 %35 = getelementptr inbounds i8, i8* %32, i64 %34 %36 = load i8, i8* %35, align 1 %37 = zext i8 %36 to i32 %38 = and i32 %37, 3 // 获取第一个字符 低两位 %39 = shl i32 %38, 4 // 左移四位
RC4_init
RC4_init 正常，无魔改
define dso_local void @Rc4_Init(i8*, i32) #0 { //RC4_init function %3 = alloca i8*, align 8 %4 = alloca i32, align 4 %5 = alloca i32, align 4 %6 = alloca i32, align 4 store i8* %0, i8** %3, align 8 store i32 %1, i32* %4, align 4 //初始化S，T盒 call void @llvm.memset.p0i8.i64(i8* align 16 getelementptr inbounds ([256 x i8], [256 x i8]* @s, i64 0, i64 0), i8 0, i64 256, i1 false) call void @llvm.memset.p0i8.i64(i8* align 16 getelementptr inbounds ([256 x i8], [256 x i8]* @t, i64 0, i64 0), i8 0, i64 256, i1 false) store i32 0, i32* %5, align 4 br label %77: ; preds = %26, %2 %8 = load i32, i32* %5, align 4 %9 = icmp slt i32 %8, 256 br i1 %9, label %10, label %29 //如果 %9 为真（即 %8 小于 256），跳转到标签 %10；否则跳转到标签 %29，根据t打乱s盒10: ; preds = %7 %11 = load i32, i32* %5, align 4 %12 = trunc i32 %11 to i8 %13 = load i32, i32* %5, align 4 %14 = sext i32 %13 to i64 %15 = getelementptr inbounds [256 x i8], [256 x i8]* @s, i64 0, i64 %14 store i8 %12, i8* %15, align 1 %16 = load i8*, i8** %3, align 8 %17 = load i32, i32* %5, align 4 %18 = load i32, i32* %4, align 4 %19 = urem i32 %17, %18 %20 = zext i32 %19 to i64 %21 = getelementptr inbounds i8, i8* %16, i64 %20 %22 = load i8, i8* %21, align 1 %23 = load i32, i32* %5, align 4 %24 = sext i32 %23 to i64 %25 = getelementptr inbounds [256 x i8], [256 x i8]* @t, i64 0, i64 %24 store i8 %22, i8* %25, align 1 br label %2626: ; preds = %10 %27 = load i32, i32* %5, align 4 %28 = add nsw i32 %27, 1 store i32 %28, i32* %5, align 4 br label %729: ; preds = %7 store i32 0, i32* %6, align 4 store i32 0, i32* %5, align 4 br label %3030: ; preds = %54, %29 %31 = load i32, i32* %5, align 4 %32 = icmp slt i32 %31, 256 br i1 %32, label %33, label %5733: ; preds = %30 %34 = load i32, i32* %6, align 4 %35 = load i32, i32* %5, align 4 %36 = sext i32 %35 to i64 %37 = getelementptr inbounds [256 x i8], [256 x i8]* @s, i64 0, i64 %36 %38 = load i8, i8* %37, align 1 %39 = zext i8 %38 to i32 %40 = add nsw i32 %34, %39 %41 = load i32, i32* %5, align 4 %42 = sext i32 %41 to i64 %43 = getelementptr inbounds [256 x i8], [256 x i8]* @t, i64 0, i64 %42 %44 = load i8, i8* %43, align 1 %45 = zext i8 %44 to i32 %46 = add nsw i32 %40, %45 %47 = srem i32 %46, 256 store i32 %47, i32* %6, align 4 %48 = load i32, i32* %5, align 4 %49 = sext i32 %48 to i64 %50 = getelementptr inbounds [256 x i8], [256 x i8]* @s, i64 0, i64 %49 %51 = load i32, i32* %6, align 4 %52 = sext i32 %51 to i64 %53 = getelementptr inbounds [256 x i8], [256 x i8]* @s, i64 0, i64 %52 call void @swap(i8* %50, i8* %53) //call swap function br label %54
RC4_enc
RC4_enc 魔改 多了一层xor 89
define dso_local void @Rc4_Encrypt(i8*, i32) #0 { //RC4_enc function %3 = alloca i8*, align 8 %4 = alloca i32, align 4 %5 = alloca i8, align 1 %6 = alloca i8, align 1 %7 = alloca i8, align 1 %8 = alloca i8, align 1 store i8* %0, i8** %3, align 8 store i32 %1, i32* %4, align 4 store i8 0, i8* %6, align 1 store i8 0, i8* %7, align 1 store i8 0, i8* %8, align 1 br label %99: ; preds = %14, %2 %10 = load i8, i8* %8, align 1 %11 = zext i8 %10 to i32 %12 = load i32, i32* %4, align 4 %13 = icmp ult i32 %11, %12 br i1 %13, label %14, label %6414: ; preds = %9 %15 = load i8, i8* %6, align 1 %16 = zext i8 %15 to i32 %17 = add nsw i32 %16, 1 %18 = srem i32 %17, 256 %19 = trunc i32 %18 to i8 store i8 %19, i8* %6, align 1 %20 = load i8, i8* %7, align 1 %21 = zext i8 %20 to i32 %22 = load i8, i8* %6, align 1 %23 = zext i8 %22 to i64 %24 = getelementptr inbounds [256 x i8], [256 x i8]* @s, i64 0, i64 %23 //生成密钥流 %25 = load i8, i8* %24, align 1 %26 = zext i8 %25 to i32 %27 = add nsw i32 %21, %26 %28 = srem i32 %27, 256 %29 = trunc i32 %28 to i8 store i8 %29, i8* %7, align 1 %30 = load i8, i8* %6, align 1 %31 = zext i8 %30 to i64 %32 = getelementptr inbounds [256 x i8], [256 x i8]* @s, i64 0, i64 %31 %33 = load i8, i8* %7, align 1 %34 = zext i8 %33 to i64 %35 = getelementptr inbounds [256 x i8], [256 x i8]* @s, i64 0, i64 %34 //经典Swap了再加 call void @swap(i8* %32, i8* %35) %36 = load i8, i8* %6, align 1 %37 = zext i8 %36 to i64 %38 = getelementptr inbounds [256 x i8], [256 x i8]* @s, i64 0, i64 %37 %39 = load i8, i8* %38, align 1 %40 = zext i8 %39 to i32 %41 = load i8, i8* %7, align 1 %42 = zext i8 %41 to i64 %43 = getelementptr inbounds [256 x i8], [256 x i8]* @s, i64 0, i64 %42 %44 = load i8, i8* %43, align 1 %45 = zext i8 %44 to i32 %46 = add nsw i32 %40, %45 %47 = srem i32 %46, 256 %48 = sext i32 %47 to i64 %49 = getelementptr inbounds [256 x i8], [256 x i8]* @s, i64 0, i64 %48 %50 = load i8, i8* %49, align 1 store i8 %50, i8* %5, align 1 %51 = load i8, i8* %5, align 1 %52 = zext i8 %51 to i32 %53 = xor i32 %52, 89 //xor 89 %54 = load i8*, i8** %3, align 8 %55 = load i8, i8* %8, align 1 %56 = zext i8 %55 to i64 %57 = getelementptr inbounds i8, i8* %54, i64 %56 %58 = load i8, i8* %57, align 1 %59 = zext i8 %58 to i32 %60 = xor i32 %59, %53 //xor k %61 = trunc i32 %60 to i8 store i8 %61, i8* %57, align 1 %62 = load i8, i8* %8, align 1 %63 = add i8 %62, 1 store i8 %63, i8* %8, align 1 br label %964: ; preds = %9 ret void}
main
main函数逻辑cipher -->RC4_init-->RC4_enc-->RC4_enc-->b64encode需要注意一下在RC4_enc的参数中，传入的数据块长度是固定的16，所以说程序进行两次RC4_enc的原因也就确定了，是为了分两次对程序进行加密，也算是一点点小手段，总之，即使让你好好分析.II代码，考察对软件分析的细节，耐心，嘻嘻。
OK,理清楚逻辑，就可以试着敲代码解密啦。
解密
逆向分析过程明了之后，那么写代码就简单多了
#include<stdio.h>unsigned char s[300],t[300];void b64decode(unsigned char * enc,unsigned char* dec);void Rc4_dec1(int len, unsigned char *enc);void Rc4_Init(char *key,int len);void Rc4_dec2(int len, unsigned char *enc);int main() { unsigned char enc[50]="TSz`kWKgbMHszXaj`@kLBmRrnTxsNtZsSOtZzqYikCw="; unsigned char dec1[50]={0x00}; char key[10] ="llvmbitc"; unsigned char a[50]; int i=0; b64decode(enc,dec1); Rc4_Init(key,8); Rc4_dec1(16,&dec1[16]); for(i=0;i<16;i++) { dec1[i+16]^=dec1[i]; } Rc4_Init(key,8); Rc4_dec2(16,dec1); printf("%s",dec1); return 0;}void b64decode(unsigned char * enc,unsigned char* dec) { int i=0,j=0; for(i=0;i<40;i+=4) { dec[j] = ((enc[i]-59)<<2)&0xfc | (((enc[i+2]-59)>>4))&3; dec[j+1] = (((enc[i+2]-59)&0xf)<<4) | (((enc[i+1]-59)>>2)&0xf); dec[j+2] = (((enc[i+1]-59)&3)<<6) | ((enc[i+3]-59)&0x3f); j+=3; } dec[j] = ((enc[i]-59)<<2)&0xfc | (((enc[i+1]-59)>>4))&3; dec[j+1] = (((enc[i+2]-59)>>2)&0xf) | (((enc[i+1]-59)<<4)&0xf0); dec[j+2]=0;} void Rc4_Init(char *key,int len) { int i=0,v5=0; unsigned char temp; for(i=0;i<256;i++) { s[i] =i; t[i] = key[i%len]; } for(i=0;i<256;i++) { v5=(s[i]+t[i]+v5)%256; temp = s[i]; s[i]= s[v5]; s[v5]=temp; }}void Rc4_dec1(int len, unsigned char *enc) { int v3=0,v5=0,i,j; unsigned char temp; for(i=0;i<len;i++) { v3=(v3+1)%256; v5=(s[v3]+v5)%256; temp=s[v3]; s[v3]=s[v5]; s[v5]=temp; } v5=v3=0; for(i=0;i<len;i++) { v3=(v3+1)%256; v5 = (s[v3]+v5)%256; temp = s[v3]; s[v3]=s[v5]; s[v5]=temp; enc[i]^=s[(s[v5]+s[v3])%256]^0x59; }}void Rc4_dec2(int len, unsigned char *enc) { int v3=0,v5=0,i,j; unsigned char temp; v5=v3=0; for(i=0;i<len;i++) { v3=(v3+1)%256; v5 = (s[v3]+v5)%256; temp = s[v3]; s[v3]=s[v5]; s[v5]=temp; enc[i]^=s[(s[v5]+s[v3])%256]^0x59; }}
flag{Hacking_for_fun@reverser$!}
总结
通过这么一道CTF题目，深入学习LLVM IR的冰山一角，认真实验，细细分析，相信会对你有极大帮助。当然，如果单从解题来说，对于解决这道题有很多的办法，比如说将.II转化为可执行文件，然后IDA分析，但我们旨在学习LLVM IR，这里不再过多赘述。
