hgame2023 vm
vm逆向
hgame2023 vm
简单翻阅一下发现，sub_140001000里面是vm_init，sub_1400010B0是主要的vm部分
查看vm_init部分，发现只知道cpu结构的大小以及初始值和24-32字节的结构，前24字节的结构未知，暂时还无法构建cpu的结构，需要更多的信息。
分析下面两图的函数可以得知，opcode总共有8个，byte_140005360的取值很多是0-7，故byte_140005360为opcode数组。看result类型可知opcode为4字节。(unsigned int *)(a1+24)在操作这个这个数组，因此a1+24为4字节的eip。sub_1400010B0中返回a1+32这个一字节的数据，他应该是某种标志位，某个标志寄存器。在sub_140001940中result为8字节，因此handle为8字节。在分析中，要确定cpu数组a1中各个成员的大小，以便创建cpu结构。
取v2 = opcode[a1[6] + 1]; 印证了a1[6]即a1+24为eip，此处为eip+1
此处出现了a1[7]，而且a1[7]先自增再存入内容
下一个指令是a1[7]--，而且是先减后用，搭配上数组140005D40仅有上下两图中的指令操作，得出a1[7]应为esp，操作的数组为栈空间。
定义各种运算
此处涉及到a1+32的功能，根据a1和a1+4处的内容为a1+32号位置赋值，相等为0不等为1。联想到cmp指令操作的寄存器，猜测a1+32应该是ZF标志位。
至此，可以猜测cpu的结构。前24字节为6个通用寄存器，每个4字节。还有4字节eip，4字节esp和1字节ZF。创建结构体
typedef struct
{
unsigned int R[6];
unsigned int eip;
unsigned int esp;
char zf;
vm_opcode op_list[OPCODE_N];
}vm_cpu;
typedef struct
{
unsigned int opcode;
void (*handle)(void*);
}vm_opcode;
修改类型让其更易读
可看到指令对应的函数已经被修复
修复后的指令功能识别
其他指令分析同理。写脚本打印一下执行流程。
#include<stdio.h>
#include <stdlib.h>
int main(){
unsigned char opcode[]={
0, 3, 2, 0, 3, 0, 2, 3, 0, 0, 0, 0, 0, 2, 1, 0, 0, 3, 2, 0x32, 3, 0, 2, 3, 0, 0, 0, 0,
3, 0, 1, 0, 0, 3, 2, 0x64, 3, 0, 2, 3, 0, 0, 0, 0, 3, 3, 1, 0, 0, 3, 0, 8, 0, 2, 2, 1,
3, 4, 1, 0, 3, 5, 2, 0, 3, 0, 1, 2, 0, 2, 0, 1, 1, 0, 0, 3, 0, 1, 3, 0, 3, 0, 0, 2, 0,
3, 0, 3, 1, 0x28, 4, 6, 0x5f, 5, 0, 0, 3, 3, 0, 2, 1, 0, 3, 2, 0x96, 3, 0, 2, 3, 0, 0,
0, 0, 4, 7, 0x88, 0, 3, 0, 1, 3, 0, 3, 0, 0, 2, 0, 3, 0, 3, 1, 0x28, 4, 7, 0x63, 0xff, 0xff};
unsigned int r[6]={0, 0, 0, 0, 0, 0};
int zf=0;
unsigned int esp=0,eip=0;
unsigned int input_and_mem[]={
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0x9B,0x0A8,0x2, 0xBC,0x0AC,0x9C,0x0CE,
0x0FA,0x2, 0xB9,0x0FF,0x3A,0x74,0x48,0x19,0x69,0x0E8,0x3, 0xCB,0x0C9,0x0FF,0x0FC,0x80,0x0D6,0x8D,
0x0D7,0x72,0x0, 0xA7,0x1D,0x3D,0x99,0x88,0x99,0x0BF,0x0E8,0x96,0x2E,0x5D,0x57,0x0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0xC9,0x0A9,0x0BD,0x8B,0x17,0x0C2,0x6E,0x0F8,0x0F5,0x6E,0x63,0x63,0x0D5,0x46,0x5D,0x16,
0x98,0x38,0x30,0x73,0x38,0x0C1,0x5E,0x0ED,0x0B0,0x29,0x5A,0x18,0x40,0x0A7,0x0FD,0x0A,0x1E,0x78,0x8B,
0x62,0x0DB,0x0F,0x8F,0x9C,0x0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0x4800,0x0F100,0x4000,0x2100,0x3501,0x6400,
0x7801,0x0F900,0x1801,0x5200,0x2500,0x5D01,0x4700,0x0FD00,0x6901,0x5C00,0x0AF01,0x0B200,0x0EC01,0x5201,
0x4F01,0x1A01,0x5000,0x8501,0x0CD00,0x2300,0x0F800,0x0C00,0x0CF00,0x3D01,0x4501,0x8200,0x0D201,0x2901,
0x0D501,0x601,0x0A201,0x0DE00,0x0A601,0x0CA01,0x0, 0, 0, 0, 0, 0, 0, 0, 0, 0};
unsigned int stack[80]={0};
int v2;
FILE *fp = fopen("log.txt", "a+");
while(opcode[eip]!=0xff){
switch (opcode[eip])
{
case 0:
v2=opcode[eip+1];
if(v2){
switch (v2)
{
case 1:
input_and_mem[r[2]]=r[0];
fprintf(fp,"input_and_mem[%d]=%d\n",r[2],r[0]);
break;
case 2:
*(r+opcode[eip+2])=*(r+opcode[eip+3]);
fprintf(fp,"r[%d]=r[%d]\n",opcode[eip+2],opcode[eip+3]);
break;
case 3:
*(r+opcode[eip+2])=opcode[eip+3];
fprintf(fp,"r[%d]=%d\n",opcode[eip+2],opcode[eip+3]);
break;
default:
break;
}
}else{
r[0]=input_and_mem[r[2]];
fprintf(fp,"r[0]=input_and_mem[%d] %d\n",r[2],r[0]);
}
eip=eip+4;
break;
case 1:
v2=opcode[eip+1];
if(v2){
switch (v2)
{
case 1:
stack[++esp]=r[0];
fprintf(fp,"stack[%d]=r[0] %d\n",esp,r[0]);
break;
case 2:
stack[++esp]=r[2];
fprintf(fp,"stack[%d]=r[2] %d\n",esp ,r[2]);
break;
case 3:
stack[++esp]=r[3];
fprintf(fp,"stack[%d]=r[3] %d\n",esp ,r[3]);
break;
}
}else{
stack[++esp]=r[0];
fprintf(fp,"stack[%d]=r[0] %d\n",esp ,r[0] );
}
eip=eip+2;
break;
case 2:
v2=opcode[eip+1];
if(v2){
switch (v2)
{
case 1:
r[1]=stack[esp--];
fprintf(fp,"r[1]=stack[%d] %d\n",esp+1,r[1]);
break;
case 2:
r[2]=stack[esp--];
fprintf(fp,"r[2]=stack[%d] %d\n",esp+1,r[2]);
break;
case 3:
r[3]=stack[esp--];
fprintf(fp,"r[3]=stack[%d] %d\n",esp+1 ,r[3]);
break;
}
}else{
r[0]=stack[esp--];
fprintf(fp,"r[0]=stack[%d] %d\n",esp+1 ,r[0]);
}
eip=eip+2;
break;
case 3:
switch (opcode[eip+1])
{
case 0:
*(r+opcode[eip+2])+=*(r+opcode[eip+3]);
fprintf(fp,"r[%d]=r[%d]+r[%d] %d\n",opcode[eip+2],opcode[eip+2],opcode[eip+3],*(r+opcode[eip+2]));
break;
case 1:
*(r+opcode[eip+2])-=*(r+opcode[eip+3]);
fprintf(fp,"r[%d]=r[%d]-r[%d] %d\n",opcode[eip+2],opcode[eip+2],opcode[eip+3] ,*(r+opcode[eip+2]));
break;
case 2:
*(r+opcode[eip+2])*=*(r+opcode[eip+3]);
fprintf(fp,"r[%d]=r[%d]*r[%d] %d\n",opcode[eip+2],opcode[eip+2],opcode[eip+3] ,*(r+opcode[eip+2]));
break;
case 3:
*(r+opcode[eip+2])^=*(r+opcode[eip+3]);
fprintf(fp,"r[%d]=r[%d]^r[%d] %d\n",opcode[eip+2],opcode[eip+2],opcode[eip+3] ,*(r+opcode[eip+2]) );
break;
case 4:
*(r+opcode[eip+2])<<=*(r+opcode[eip+3]);
fprintf(fp,"r[%d]=r[%d]<<r[%d] %d\n",opcode[eip+2],opcode[eip+2],opcode[eip+3],*(r+opcode[eip+2]));
*(r+opcode[eip+2])&=0xff00u;
fprintf(fp,"r[%d]=r[%d]&0xff00u %d\n",opcode[eip+2],opcode[eip+2],*(r+opcode[eip+2]));
break;
case 5:
*(r+opcode[eip+2])=*(r+opcode[eip+2]) >> *(r+opcode[eip+3]);
fprintf(fp,"r[%d]=r[%d]>>r[%d] %d\n",opcode[eip+2],opcode[eip+2],opcode[eip+3],*(r+opcode[eip+2]));
break;
default:
break;
}
eip=eip+4;
break;
case 4:
if(r[0]==r[1]){
zf=0;
fprintf(fp,"cmp r0 r1 r0==r1 zf=0\n");
}else{
zf=1;
fprintf(fp,"cmp r0 r1 r0!=r1 zf=1\n");
}
eip=eip+1;
break;
case 5:
eip=opcode[eip+1];
fprintf(fp,"jmp %d\n",eip);
break;
case 6:
if(zf){
eip=eip+2;
fprintf(fp,"jnz %d\n",eip);
}else{
eip=opcode[eip+1];
fprintf(fp,"jnz %d\n",eip);
}
break;
case 7:
if(zf){
eip=opcode[eip+1];
fprintf(fp,"jz %d\n",eip);
}else{
eip=eip+2;
fprintf(fp,"jz %d\n",eip);
}
break;
default:
break;
}
}
return 0;
}
编写代码时一定仔细检查
log:
r[2]=0
r[2]=r[2]+r[3] 0
r[0]=input_and_mem[0] 0
r[1]=r[0]
r[2]=50
r[2]=r[2]+r[3] 50
r[0]=input_and_mem[50] 155
r[1]=r[1]+r[0] 155
r[2]=100
r[2]=r[2]+r[3] 100
r[0]=input_and_mem[100] 201
r[1]=r[1]^r[0] 82
r[0]=8
r[2]=r[1]
r[1]=r[1]<<r[0] 20992
r[1]=r[1]&0xff00u 20992
r[2]=r[2]>>r[0] 0
r[1]=r[1]+r[2] 20992
r[0]=r[1]
stack[1]=r[0] 20992
r[0]=1
r[3]=r[3]+r[0] 1
r[0]=r[3]
r[1]=40
cmp r0 r1 r0!=r1 zf=1
jnz 93
jmp 0
r[2]=0
r[2]=r[2]+r[3] 1
r[0]=input_and_mem[1] 0
r[1]=r[0]
r[2]=50
r[2]=r[2]+r[3] 51
r[0]=input_and_mem[51] 168
r[1]=r[1]+r[0] 168
r[2]=100
r[2]=r[2]+r[3] 101
r[0]=input_and_mem[101] 169
r[1]=r[1]^r[0] 1
r[0]=8
r[2]=r[1]
r[1]=r[1]<<r[0] 256
r[1]=r[1]&0xff00u 256
r[2]=r[2]>>r[0] 0
r[1]=r[1]+r[2] 256
r[0]=r[1]
stack[2]=r[0] 256
r[0]=1
r[3]=r[3]+r[0] 2
r[0]=r[3]
r[1]=40
cmp r0 r1 r0!=r1 zf=1
jnz 93
jmp 0
....
r[2]=0
r[2]=r[2]+r[3] 37
r[0]=input_and_mem[37] 0
r[1]=r[0]
r[2]=50
r[2]=r[2]+r[3] 87
r[0]=input_and_mem[87] 46
r[1]=r[1]+r[0] 46
r[2]=100
r[2]=r[2]+r[3] 137
r[0]=input_and_mem[137] 15
r[1]=r[1]^r[0] 33
r[0]=8
r[2]=r[1]
r[1]=r[1]<<r[0] 8448
r[1]=r[1]&0xff00u 8448
r[2]=r[2]>>r[0] 0
r[1]=r[1]+r[2] 8448
r[0]=r[1]
stack[38]=r[0] 8448
r[0]=1
r[3]=r[3]+r[0] 38
r[0]=r[3]
r[1]=40
cmp r0 r1 r0!=r1 zf=1
jnz 93
jmp 0
r[2]=0
r[2]=r[2]+r[3] 38
r[0]=input_and_mem[38] 0
r[1]=r[0]
r[2]=50
r[2]=r[2]+r[3] 88
r[0]=input_and_mem[88] 93
r[1]=r[1]+r[0] 93
r[2]=100
r[2]=r[2]+r[3] 138
r[0]=input_and_mem[138] 143
r[1]=r[1]^r[0] 210
r[0]=8
r[2]=r[1]
r[1]=r[1]<<r[0] 53760
r[1]=r[1]&0xff00u 53760
r[2]=r[2]>>r[0] 0
r[1]=r[1]+r[2] 53760
r[0]=r[1]
stack[39]=r[0] 53760
r[0]=1
r[3]=r[3]+r[0] 39
r[0]=r[3]
r[1]=40
cmp r0 r1 r0!=r1 zf=1
jnz 93
jmp 0
r[2]=0
r[2]=r[2]+r[3] 39
r[0]=input_and_mem[39] 0
r[1]=r[0]
r[2]=50
r[2]=r[2]+r[3] 89
r[0]=input_and_mem[89] 87
r[1]=r[1]+r[0] 87
r[2]=100
r[2]=r[2]+r[3] 139
r[0]=input_and_mem[139] 156
r[1]=r[1]^r[0] 203
r[0]=8
r[2]=r[1]
r[1]=r[1]<<r[0] 51968
r[1]=r[1]&0xff00u 51968
r[2]=r[2]>>r[0] 0
r[1]=r[1]+r[2] 51968
r[0]=r[1]
stack[40]=r[0] 51968
r[0]=1
r[3]=r[3]+r[0] 40
r[0]=r[3]
r[1]=40
cmp r0 r1 r0==r1 zf=0
jnz 95
r[3]=0
r[1]=stack[40] 51968
r[2]=150
r[2]=r[2]+r[3] 150
r[0]=input_and_mem[150] 18432
cmp r0 r1 r0!=r1 zf=1
jz 136
分析可以发现，虽然步骤很多，但是通过jmp 0 指令分隔可分成40个部分，也就是40轮循环，每轮循环都类似。以第一轮为例
r[2]=0
r[2]=r[2]+r[3] 0
r[0]=input_and_mem[0] 0
r[1]=r[0] //r1=input_and_mem[0]
r[2]=50
r[2]=r[2]+r[3] 50
r[0]=input_and_mem[50] 155 //
r[1]=r[1]+r[0] 155 //r1=r1+input_and_mem[50]
r[2]=100
r[2]=r[2]+r[3] 100
r[0]=input_and_mem[100] 201
r[1]=r[1]^r[0] 82 //r1=r1^input_and_mem[100]
r[0]=8
r[2]=r[1] //r2=r1
r[1]=r[1]<<r[0] 20992 //r1=r1<<8
r[1]=r[1]&0xff00u 20992 //r1=r1&0xff
r[2]=r[2]>>r[0] 0 //r2=r2>>8
r[1]=r[1]+r[2] 20992 //r1=r1+r2 也就是交换高低字节
r[0]=r[1] //r0=r1
stack[1]=r[0] 20992 //push r0
r[0]=1
r[3]=r[3]+r[0] 1
r[0]=r[3]
r[1]=40
cmp r0 r1 r0!=r1 zf=1
jnz 93
jmp 0
也就是input_and_mem里前40位是输入，50-90位是加运算的数据，100-140位是异或的数据。在最后一次跳转中使用pop弹出了栈顶的数据，和input_and_mem里的150号数据进行比较，可以猜测150-190为加密后的数据，栈中的数据为加密后的数据，依次弹出和密文比较。还可推断是逐个比较，只要不正确即终止运行，因此此处只比较了一次。代码示意和脚本如下。
push (((input_and_mem[i]+input_and_mem[i+50])^input_and_mem[i+100])<<8)&0xff00u+(((input_and_mem[i]+input_and_mem[i+50])^input_and_mem[i+100])>>8)
#include<stdio.h>
#include <stdlib.h>
int main(){
unsigned int cipher[40]=
{
0x4800,0x0F100,0x4000,0x2100,0x3501,0x6400,0x7801,0x0F900,0x1801,0x5200,0x2500,0x5D01,
0x4700,0x0FD00,0x6901,0x5C00,0x0AF01,0x0B200,0x0EC01,0x5201,0x4F01,0x1A01,0x5000,0x8501,
0x0CD00,0x2300,0x0F800,0x0C00,0x0CF00,0x3D01,0x4501,0x8200,0x0D201,0x2901,0x0D501,0x601,
0x0A201,0x0DE00,0x0A601,0x0CA01
};
unsigned int key_add[40]={
0x9B,0x0A8,0x2, 0xBC,0x0AC,0x9C,0x0CE,
0x0FA,0x2, 0xB9,0x0FF,0x3A,0x74,0x48,0x19,0x69,0x0E8,0x3, 0xCB,0x0C9,0x0FF,0x0FC,0x80,0x0D6,0x8D,
0x0D7,0x72,0x0, 0xA7,0x1D,0x3D,0x99,0x88,0x99,0x0BF,0x0E8,0x96,0x2E,0x5D,0x57
};
unsigned int key_xor[40]={
0xC9,0x0A9,0x0BD,0x8B,0x17,0x0C2,0x6E,0x0F8,0x0F5,0x6E,0x63,0x63,0x0D5,0x46,0x5D,0x16,
0x98,0x38,0x30,0x73,0x38,0x0C1,0x5E,0x0ED,0x0B0,0x29,0x5A,0x18,0x40,0x0A7,0x0FD,0x0A,0x1E,0x78,0x8B,
0x62,0x0DB,0x0F,0x8F,0x9C
};
unsigned int temp;
for(int i=0;i<40;i++){
temp=cipher[39-i];
temp=((temp<<8)&0xff00u)+(temp>>8);
temp=temp^key_xor[i];
temp=temp-key_add[i];
printf("%c",temp);
}
return 0;
}
//hgame{y0ur_rever5e_sk1ll_i5_very_g0od!!}
参考：
