标签:enc goto log size 等于 eof amp 异或 告诉
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交叉引用的地方
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signed __int64 sub_4009C6()
{
__int64 v0; // rax
signed __int64 result; // rax
unsigned __int64 v2; // rax
__int64 v3; // rax
const __m128i *v4; // ST10_8
const __m128i *v5; // ST18_8
const __m128i *v6; // ST20_8
const __m128i *v7; // ST28_8
const __m128i *v8; // ST30_8
const __m128i *v9; // ST38_8
const __m128i *v10; // ST40_8
const __m128i *v11; // ST48_8
const __m128i *v12; // ST50_8
__int64 v13; // ST58_8
int i; // [rsp+Ch] [rbp-114h]
char v15; // [rsp+60h] [rbp-C0h]
char v16; // [rsp+61h] [rbp-BFh]
char v17; // [rsp+62h] [rbp-BEh]
char v18; // [rsp+63h] [rbp-BDh]
char v19; // [rsp+64h] [rbp-BCh]
char v20; // [rsp+65h] [rbp-BBh]
char v21; // [rsp+66h] [rbp-BAh]
char v22; // [rsp+67h] [rbp-B9h]
char v23; // [rsp+68h] [rbp-B8h]
char v24; // [rsp+69h] [rbp-B7h]
char v25; // [rsp+6Ah] [rbp-B6h]
char v26; // [rsp+6Bh] [rbp-B5h]
char v27; // [rsp+6Ch] [rbp-B4h]
char v28; // [rsp+6Dh] [rbp-B3h]
char v29; // [rsp+6Eh] [rbp-B2h]
char v30; // [rsp+6Fh] [rbp-B1h]
char v31; // [rsp+70h] [rbp-B0h]
char v32; // [rsp+71h] [rbp-AFh]
char v33; // [rsp+72h] [rbp-AEh]
char v34; // [rsp+73h] [rbp-ADh]
char v35; // [rsp+74h] [rbp-ACh]
char v36; // [rsp+75h] [rbp-ABh]
char v37; // [rsp+76h] [rbp-AAh]
char v38; // [rsp+77h] [rbp-A9h]
char v39; // [rsp+78h] [rbp-A8h]
char v40; // [rsp+79h] [rbp-A7h]
char v41; // [rsp+7Ah] [rbp-A6h]
char v42; // [rsp+7Bh] [rbp-A5h]
char v43; // [rsp+7Ch] [rbp-A4h]
char v44; // [rsp+7Dh] [rbp-A3h]
char v45; // [rsp+7Eh] [rbp-A2h]
char v46; // [rsp+7Fh] [rbp-A1h]
char v47; // [rsp+80h] [rbp-A0h]
char v48; // [rsp+81h] [rbp-9Fh]
char v49; // [rsp+82h] [rbp-9Eh]
char v50; // [rsp+83h] [rbp-9Dh]
char v51[32]; // [rsp+90h] [rbp-90h]
int v52; // [rsp+B0h] [rbp-70h]
char v53; // [rsp+B4h] [rbp-6Ch]
char v54; // [rsp+C0h] [rbp-60h]
char v55; // [rsp+E7h] [rbp-39h]
char v56; // [rsp+100h] [rbp-20h]
unsigned __int64 v57; // [rsp+108h] [rbp-18h]
v57 = __readfsqword(0x28u);
v15 = 73;
v16 = 111;
v17 = 100;
v18 = 108;
v19 = 62;
v20 = 81;
v21 = 110;
v22 = 98;
v23 = 40;
v24 = 111;
v25 = 99;
v26 = 121;
v27 = 127;
v28 = 121;
v29 = 46;
v30 = 105;
v31 = 127;
v32 = 100;
v33 = 96;
v34 = 51;
v35 = 119;
v36 = 125;
v37 = 119;
v38 = 101;
v39 = 107;
v40 = 57;
v41 = 123;
v42 = 105;
v43 = 121;
v44 = 61;
v45 = 126;
v46 = 121;
v47 = 76;
v48 = 64;
v49 = 69;
v50 = 67; //v15-v50这是一个字符串数组
memset(v51, 0, sizeof(v51));
v52 = 0;
v53 = 0;
sub_4406E0(0LL, (__int64)v51);
v53 = 0;
LODWORD(v0) = sub_424BA0((const __m128i *)v51);
if ( v0 == 36 )
{
for ( i = 0; ; ++i )
{
LODWORD(v2) = sub_424BA0((const __m128i *)v51);
if ( i >= v2 )
break;
if ( (unsigned __int8)(v51[i] ^ i) != *(&v15 + i) ) //这里做了一次数组的异或
{
result = 0xFFFFFFFELL;
goto LABEL_13;
}
}
sub_410CC0((const __m128i *)"continue!");
memset(&v54, 0, 0x40uLL);
v56 = 0;
sub_4406E0(0LL, (__int64)&v54); //v54是又一次输入
v55 = 0;
LODWORD(v3) = sub_424BA0((const __m128i *)&v54);
if ( v3 == 39 )
{
v4 = (const __m128i *)sub_400E44((const __m128i *)&v54);//这里反复套用做了十次的base64加密
v5 = (const __m128i *)sub_400E44(v4);
v6 = (const __m128i *)sub_400E44(v5);
v7 = (const __m128i *)sub_400E44(v6);
v8 = (const __m128i *)sub_400E44(v7);
v9 = (const __m128i *)sub_400E44(v8);
v10 = (const __m128i *)sub_400E44(v9);
v11 = (const __m128i *)sub_400E44(v10);
v12 = (const __m128i *)sub_400E44(v11);
v13 = sub_400E44(v12);
if ( !(unsigned int)sub_400360(v13, (__int64)off_6CC090) )//这里是判断v13这个结果base64处理之后的值和off_6CC090是不是相等
{
sub_410CC0((const __m128i *)"You found me!!!");
sub_410CC0((const __m128i *)"bye bye~");
}
result = 0LL;
}
else
{
result = 0xFFFFFFFDLL;
}
}
else
{
result = 0xFFFFFFFFLL;
}
LABEL_13:
if ( __readfsqword(0x28u) != v57 )
sub_444020();
return result;
}
/* Orphan comments:
Vm0wd2VHUXhTWGhpUm1SWVYwZDRWVll3Wkc5WFJ
*/
sub_400E44:这表明做了Base64加密 这个数组就是v54的这个数组
off_6CC090的值是:
把off_6CC090的值解密10次base64之后,得到的内容是网址:https://bbs.pediy.com/thread-254172.htm
这个网址看起来没东西,flag就不是在这里。不对劲
回到这里 发现 有两组数据 没见过的 Ctrl+X
它们用的是同一段的函数
再次交叉引用 可以看到 出现了_fini. 这样的字样
.fini段的解释是:此节区包含了可执行的指令,是进程终止代码的一部分。程序正常退出时,系统将安排执行这里的代码。
这一段里面有数据
我们分析该段
unsigned __int64 sub_400D35()
{
unsigned __int64 result; // rax
unsigned __int64 v1; // rt1
unsigned int v2; // [rsp+Ch] [rbp-24h]
signed int i; // [rsp+10h] [rbp-20h]
signed int j; // [rsp+14h] [rbp-1Ch]
unsigned int v5; // [rsp+24h] [rbp-Ch]
unsigned __int64 v6; // [rsp+28h] [rbp-8h]
v6 = __readfsqword(0x28u);
v2 = sub_43FD20(0LL) - qword_6CEE38;
for ( i = 0; i <= 1233; ++i )
{
sub_40F790(v2);
sub_40FE60();
sub_40FE60();
v2 = (unsigned __int64)sub_40FE60() ^ 0x98765432;
}
v5 = v2;
if ( ((unsigned __int8)v2 ^ byte_6CC0A0[0]) == ‘f’ && (HIBYTE(v5) ^ (unsigned __int8)byte_6CC0A3) == ‘g‘) //通过这里反向推出v2和v5的值
{
for ( j = 0; j <= 24; ++j )
sub_410E90((unsigned __int8)(byte_6CC0A0[j] ^ *((_BYTE *)&v5 + j % 4)));
}
v1 = __readfsqword(0x28u);
result = v1 ^ v6;
if ( v1 != v6 )
sub_444020();
return result;
}
Byte_6CC0A0[]=[0x40,0x35,0x20]
Byte_6CC0A3[]=[0x56,0x5D,0x18,0x22,0x45,0x17,0x2F,0x24,0x6E,0x62,0x3C,0x27,0x54,0x48,0x6C,0x24,0x6E,0x72,0x3C,0x32,0x45,0x5B]
我们可以看到就是v2和Byte_6CC0A0[0]异或后等于‘f‘ v5和Byte_6CC0A3异或后等于‘g‘
合理推测v2v3v4v5 四个应该是flag 那么它们对应的应该是Byte_6CC0A0[0]-[3]和Byte_6CC0A3[0]
那么flag的值就应该在Byte_6CC0A3[]里面一共25个字符
后面这段数据咋找的思路:
一般出题人不会无缘无故给一段没用的数据
这段代码是会执行的,.fini段,是在程序的结束之后执行。了解了init,fini段后基本上都会检查这两个段的。
[V&N2020 公开赛]strangeCpp这道题,也是把数据藏起来了没有引用,需要找。
写程序的时候一般不会添加无用数据,所以发现有意义的数据看看交叉引用啥的……
key?=?[73,111,100,108,62,81,110,98,40,111,99,121,127,121,46,105,127,100,96,51,119,125,119,101,107,57,123,105,121,61,126,121,76,64,69,67]
flag?=?‘‘
for?i?in?range(len(key)):
?????flag+=chr(key[i]^i)
print(flag)
结果是:Info:The first four chars are flag
这告诉我们该flag开头是flag
v5=‘‘
enc=‘flag‘
key1=[0x40,0x35,0x20,0x56,0x5D,0x18,0x22,0x45,0x17,0x2F,0x24,0x6E,0x62,0x3C,0x27,0x54,0x48,0x6C,0x24,0x6E,0x72,0x3C,0x32,0x45,0x5B]
for?i?in?range(4):
????????v5+=chr(key1[i]^ord(enc[i]))
print(v5)
v5的值是:&YA1
flag2=‘‘
for?i?in?range(25):
????flag2+=chr(key1[i]^ord(v5[i%4]))
print(flag2)
flag{Act1ve_Defen5e_Test}
标签:enc goto log size 等于 eof amp 异或 告诉
原文地址:https://www.cnblogs.com/Nickyl07/p/13344953.html