此文以记录分析路程
实在是懒。比赛的时候分析到一半就不想分析了。耐心不够啊！感觉质量还是很不错！
Mainfest文件中定义在Android6.0SDK，在Android4.4.4中运行程序闪退。折腾了几个小时下AVD在Android6.0中跑还是闪退，只能硬着头皮看逻辑了。
Java层只有loadlibrary("cook")，分析so中JNI_ONLOAD，发现字符串全部被处理，使用的时候调用sub_1034解密，直接动态调试，顺便看下为什么会闪退。(自己编写一个Loader加载so调试会比较方便)。发现mkdir会失败。patch下(其实patch目的还是看后面字符串解密，写IDA脚本发现没法整理好栈中数据顺序，还是功力不够啊！)
由于看到ARM下堆栈传递的参数是乱序的直接试用x86的进行分析，脚本如下
def dec_string(enc_data):
ret_len = len(enc_data)
ret = ""
chr1 = None
chr2 = None
"""
*result =
~(( ~ cur_int&0xff | ((unsigned __int16)(cur_int & 0xFF00) >> 8)) & (~((unsigned __int16)(cur_int & 0xFF00) >> 8) | cur_int));
result[1] = HIBYTE(cur_int) ^ ((cur_int & 0xFF0000) >> 16);
"""
for i in range(ret_len):
cur_int = enc_data
chr1 = (~(((~cur_int & 0xff) | ((cur_int & 0xff00) >> 8)) & ((~(cur_int & 0xff00) >> 8) | cur_int))) & 0xff
chr2 = ((cur_int & 0xff000000) >> 24) ^ ((cur_int & 0xff0000) >> 16)
ret += chr(chr1)
ret += chr(chr2)
return ret
pDecode = 0x680
for x in XrefsTo(pDecode,flags = 0):
#MakeCode((x.frm & 0xFFFFFFFE));
addr = x.frm
print hex(addr)
while True:
addr = PrevHead(addr)
if "esp" in GetOpnd(addr,0):
if "x" not in GetOpnd(addr,1):
index = GetOperandValue(addr,1)
break
dict = []
count = 0
while count < index:
addr = PrevHead(addr)
if "esp" in GetOpnd(addr,0):
dict.append(GetOperandValue(addr,1))
count += 1
MakeComm(x.frm,dec_string(dict))
解密出来可以看到大概的逻辑了
[i]
发现写入 /data/data/com.google.ctf.food/files/d.dex
导出后Jeb分析，发现com.google.ctf.food.F.cc类被抽空了。那么后面肯定还有读写内存的操作，继续分析下去，发现sub_1098中就有这种操作。
[i]
![5$N)E%X_Z[P3UH(8O1][T.png](https://attach.52pojie.cn/forum/201706/23/172705gii1cjjyy2cy4ciy.png "5$N)E%X_Z[P3UH(8O1][T.png")
从/proc/self/maps中读取d.dex基址，匹配''dex\n0'，在偏移：0x720处写入数据，数据被异或解密，如图
修改dex文件，填充修改的部分
a = [0x49, 0x5E, 0x52, 0x5A, 0x79, 0x1B, 0x7B, 0x5A, 0x7C,
0x5B, 0x66, 0x5A, 0x5A, 0x5A, 0x48, 0x5A, 0x6F, 0x1A,
0x55, 0x5A, 0x12, 0x58, 0x5B, 0x5A, 0xE, 9, 0x5F, 0x5A,
0x12, 0x59, 0x59, 0x5A, 0xED, 0x68, 0xD7, 0x78, 0x15,
0x58, 0x5B, 0x5A, 0x82, 0x5A, 0x5A, 0x5B, 0x72, 0xA8,
0x78, 0x5A, 0x45, 0x5A, 0x2A, 0x7A, 0x7E, 0x5A, 0x4A,
0x5A, 0x40, 0x5B, 0x5A, 0x5A, 0x34, 0x7A, 0x7F, 0x5A,
0x4A, 0x5A, 0x50, 0x5A, 0x63, 0x5A, 0x47, 0x5A, 0xE,
0xA, 0x58, 0x5A, 0x34, 0x4A, 0x5B, 0x5A, 0x5A, 0x5A,
0x56, 0x5A, 0x78, 0x5B, 0x45, 0x5A, 0x38, 0x58, 0x5E,
0x5A, 0xE, 9, 0x5F, 0x5A, 0x2B, 0x7A, 0x78, 0x5A, 0x68,
0x5A, 0x56, 0x58, 0x2A, 0x7A, 0x7E, 0x5A, 0x7B, 0x5A,
0x48, 0x48, 0x2B, 0x6A, 0x4F, 0x5A, 0x4A, 0x58, 0x56,
0x5A, 0x34, 0x4A, 0x4C, 0x5A, 0x5A, 0x5A, 0x54, 0x5A,
0x5A, 0x59, 0x5B, 0x5A, 0x52, 0x5A, 0x5A, 0x5A, 0x40,
0x41, 0x44, 0x5E, 0x4F, 0x58, 0x48, 0x5D]
b = []
for i in a:
b.append(i^0x5A)
f = open("d.dex",'rb')
file = f.read()
buf = list(file)
for i in range(0x90):
buf[0x720+i] = chr(b[i])
f.close()
f = open("f.dex",'wb')
f.write("".join(buf))
f.close()
修复好了之后，分析dex，这里直接贴代码，java逻辑层就很清楚了
import java.util.Arrays;
public class test {
public static void main(String[] argv){
byte[] v1 = new byte[]{26, 27, 30, 4, 21, 2, 18, 7};
byte[] v2 = new byte[]{0x13, 0x11, 0x13, 3, 4, 3, 1, 5};
for(int i=0;i<v1.length;i++){
v1 ^= v2;
}
byte[] flag = new byte[]{-19, 116, 58, 108, -1, 33, 9, 61, -61, -37, 108, -123, 3, 35, 97, -10, -15, 15, -85, -66, -31, -65, 17, 79, 31, 25, -39, 95, 93, 1, -110, -103, -118, -38, -57, -58, -51, -79};
System.out.println(new String(a(flag,v1)));
}
public static byte[] a(byte[] arg8, byte[] arg9) {
int v7 = 256;
byte[] v3 = new byte[v7];
byte[] v4 = new byte[v7];
int v0 = 0;
int v1;
for(v1 = 0; v1 != v7; ++v1) {
v3[v1] = ((byte)v1);
v4[v1] = arg9[v1 % arg9.length];
}
int v2 = v1 ^ v1;
v1 = 0;
while(v2 != v7) {
v1 = v1 + v3[v2] + v4[v2] & 255;
v3[v1] = ((byte)(v3[v1] ^ v3[v2]));
v3[v2] = ((byte)(v3[v2] ^ v3[v1]));
v3[v1] = ((byte)(v3[v1] ^ v3[v2]));
++v2;
}
v4 = new byte[arg8.length];
v2 ^= v2;
v1 ^= v1;
while(v0 != arg8.length) {
v2 = v2 + 1 & 255;
v1 = v1 + v3[v2] & 255;
v3[v1] = ((byte)(v3[v1] ^ v3[v2]));
v3[v2] = ((byte)(v3[v2] ^ v3[v1]));
v3[v1] = ((byte)(v3[v1] ^ v3[v2]));
v4[v0] = ((byte)(arg8[v0] ^ v3[v3[v2] + v3[v1] & 255]));
++v0;
}
return v4;
}
}```
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发表于 2017-6-23 17:27
发表于 2017-6-26 18:38
|
发表于 2017-6-23 21:26
！
很难用！