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[原创]第七题 声名远扬WP
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2021-12-2 11:09
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程序使用了兼容模式下32位代码与64位代码的混合执行的技术。
.text:00B2D88E 6A 00 push 0 .text:00B2D890 8B 5D 10 mov ebx, [ebp+arg_8] .text:00B2D893 53 push ebx .text:00B2D894 8B 4D 0C mov ecx, [ebp+arg_4] .text:00B2D897 51 push ecx .text:00B2D898 8B 4D 08 mov ecx, [ebp+arg_0] .text:00B2D89B 51 push ecx .text:00B2D89C 6A 00 push 0 .text:00B2D89E 8D 4D EC lea ecx, [ebp+var_14] .text:00B2D8A1 51 push ecx .text:00B2D8A2 FF 5D F4 call fword ptr [ebp+var_C] //进入IA-32e模式
当执行到.text:00B2D8A2 时，栈内存ebp+var_C如下：
Stack[000057AC]:012FEDB4 dd offset byte_B256F0 //远调用目标绝对地址 Stack[000057AC]:012FEDB8 dd 33h //远调用CS段选择子
当前CS段选择子为23h, 调用后的CS段选择子0x33(64位用户代码段).
0xB256F0所指的代码是64位代码，将其导出为二进制文件，使用IDA将其作为64位代码进行分析：
seg000:0000000000000000 sub_0 proc far seg000:0000000000000000 seg000:0000000000000000 arg_0 = qword ptr 10h seg000:0000000000000000 arg_8 = qword ptr 18h seg000:0000000000000000 arg_10 = qword ptr 20h seg000:0000000000000000 arg_18 = qword ptr 28h seg000:0000000000000000 arg_20 = qword ptr 30h seg000:0000000000000000 arg_28 = qword ptr 38h seg000:0000000000000000 seg000:0000000000000000 mov rdi, [rsp+arg_0] seg000:0000000000000005 mov rsi, [rsp+arg_8] seg000:000000000000000A mov r11, 1 seg000:0000000000000011 lea r10, [rsp+arg_10] seg000:0000000000000016 cmp rsi, 1 seg000:000000000000001A jb short loc_50 seg000:000000000000001C mov rcx, [rsp+arg_10] seg000:0000000000000021 cmp rsi, 2 seg000:0000000000000025 jb short loc_50 seg000:0000000000000027 mov rdx, [rsp+arg_18] seg000:000000000000002C cmp rsi, 3 seg000:0000000000000030 jb short loc_50 seg000:0000000000000032 mov r8, [rsp+arg_20] seg000:0000000000000037 cmp rsi, 4 seg000:000000000000003B jb short loc_50 seg000:000000000000003D mov r9, [rsp+arg_28] seg000:0000000000000042 mov rax, rsi seg000:0000000000000045 shl rax, 3 seg000:0000000000000049 add r10, rax seg000:000000000000004C sub r10, 8 seg000:0000000000000050 seg000:0000000000000050 loc_50: ; CODE XREF: sub_0+1A↑j seg000:0000000000000050 ; sub_0+25↑j ... seg000:0000000000000050 cmp rsi, r11 seg000:0000000000000053 jb short loc_62 seg000:0000000000000055 mov rax, [r10] seg000:0000000000000058 push rax seg000:0000000000000059 sub r10, 8 seg000:000000000000005D inc r11 seg000:0000000000000060 jmp short loc_50 seg000:0000000000000062 ; --------------------------------------------------------------------------- seg000:0000000000000062 seg000:0000000000000062 loc_62: ; CODE XREF: sub_0+53↑j seg000:0000000000000062 call rdi //实际验证函数 seg000:0000000000000064 shl rsi, 3 seg000:0000000000000068 add rsp, rsi seg000:000000000000006B mov rbx, [rsp+8] seg000:0000000000000070 mov [rbx], rax seg000:0000000000000073 retf //返回兼容模式
通过分析堆栈，可知验证函数为0xB255AC(同样为64位代码)，导出分析：
__int64 __fastcall sub_B255AC(char *seri, __int64 )
{
//...
//初始化
v27 = 44;
v4 = 0;
*(__m128i *)v28 = _mm_load_si128((const __m128i *)0xB25408);
*(__m128i *)&v28[32] = _mm_load_si128((const __m128i *)0xB253F8);
*(__m128i *)&v28[16] = _mm_load_si128((const __m128i *)0xB25418);
//序列号生成
do
{
v5 = v4 - 52;
v6 = v4++;
v28[v6] ^= v5;
}
while ( v4 < v27 );
v28[v27] = 0;
//序列号输入检查
v7 = 0;
chr = *seri;
if ( *seri )
{
v9 = seri;
do
{
if ( chr != v9[v28 - seri] )
break;
++v9;
++v7;
chr = *v9;
}
while ( *v9 );
}
v10 = v7;
i = 0;
v24 = -48;
if ( seri[v10] == v28[v10] )
{
//验证成功
}
else
{
//验证失败
}
return 0i64;
}算法较简单，静态分析即可，若想进行动态调试，可配置bochs调试（注意依赖数据也须导入到数据库），配置如下：
序列号算法：
//.idc
auto bytes = get_bytes(0xB25408, 0x10, 1);
bytes = bytes + get_bytes(0xB25418, 0x10, 1);
bytes = bytes + get_bytes(0xB253F8, 0x10, 1);
auto i = 0;
for(i = 0; i<44; i++){
auto v = ord(bytes[i]) ^ ((i - 52) & 0xff);
msg("%02X ", (v & 0xff));
bytes[i] = v;
}
bytes[i] = '\x00';
msg("\n%s\n", bytes);
//GYldGg-iIoJlPX9hPXpjPqfdEY21B01TBTzeGqfKNR!!输入的序列号需要进行类似base64的编码，编码函数位于.text:00B2E530，因此需要做对应的解码操作：
//.idc
auto seri = "GYldGg-iIoJlPX9hPXpjPqfdEY21B01TBTzeGqfKNR";
auto sseri = "";
auto chr = 0;
auto bits = 0;
auto result = 0;
auto tree = ECX; //调试拿到tree地址
auto i = 0;
auto j = 0;
for(i = 0; i < strlen(seri); i++){
auto found = -1;
for(j = 0; j <= 0x40; j++){
auto c = j;
auto sc = bsearch(tree, &c); //APPCALL .text:00B2E970 ; void *__thiscall bsearch(void *bin_tree, char *chr)
//msg("%02X ", byte(sc));
if(ord(seri[i]) == byte(sc)){
found = j;
break;
}
}
if(found == -1){
msg("\n###EXCEPTION %d %d %X\n", i, strlen(seri), ord(seri[i]));
break;
}
else{
bits = bits + 6;
chr = chr << 6;
chr = chr | found;
while(bits >= 8){
auto v = chr >> (bits - 8);
chr = chr - (v << (bits - 8));
bits = bits -8;
msg("%02X ", (v & 0xFF));
}
}
}
msg("\nDONE bits = %d, chr = %02X\n", bits, chr);
/*
66 6C 61 67 7B 32 30 32 31 2D 31 30 2D 30 34 2D 79 61 6E 67 79 61 6E 67 62 75 64 65 79 69 7D
DONE bits = 4, chr = 00
*/解码得到序列号为：
flag{2021-10-04-yangyangbudeyi}
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