app检测root怎么逆向(包含脱下来的java代码不正确如何解决)

本篇文章仅供交流学习，侵权删(满满的求生欲)

本次app样本:5LqM5LiJ6YeMNy4wLjM=

逆向部分

1.绕过root检测

首先现象就是打开app，会提示手机已经root，然后闪退。

当然是使用面具的隐藏功能了

我使用的版本是23.0，主界面右上角设置里面有一个MagiskHide

勾选之后会重启，重启后，点击盾牌的那个分栏

点击第一个

然后勾选想要隐藏检测root的应用

再次打开app就不会闪退了，成功!

2.抓包

这个就不用说了，尿壶直接就能抓到，没有ssl。

有一个sign，但是很简单，就不说了。

返回值是一堆数字+字母，看来回来还做了一些特殊处理，算法部分主要说这个吧。

3.脱壳

众所周知，开启了MagiskHide无法运行frida脱壳脚本，那就先把面具隐藏关掉，再运行脱壳脚本，虽然会闪退，但是还是能脱下来，只不过代码有的地方可能会有问题(算法部分会提及，看java代码+尝试用了一上午才修改成正确的java代码，焯！)

这里分享两个好用的脱壳脚本

一个是yang神的 github.com/lasting-yan…

一个是葫芦娃大佬的 github.com/hluwa/FRIDA…

算法部分

重头戏，论如何被不知道是不是因为检测了root所以壳没脱好伤到了本体的java代码 折磨了一个上午。

首先搜索请求头中的参数，直接就找到了，然后跟调用

发现返回值处理成正确的数据在这个函数里

看了下f，是"AES" ，看来是个AES加密，hook一下入参和返回值，密钥也拿到了，结果也是正确的，那么重点就是他对返回的加密数据，使用a函数做了些处理，进去看看。

大致看一下，取了返回加密数据的长度，做了一个判断，然后进入循环，循环是本篇文章的重点，首先charAt()方法就是根据索引取字符串的对应值，然后看一下j方法

再跟

具体就是做了一些判断和加减法，python中可以使用ord()函数将字母或数字转为对应的 ASCII 数值。然后作为3里面的入参codePoint传进去，第二个参数是固定值16，因为返回的加密参数就是0-9和a-z的组合，所以ASCII不会超过128，把 >=128的if逻辑删掉就好了，然后用python还原。

这里贴一下，我还原后的整体代码

第二张图的a方法

def data_decrypt(self, decrData):
    length = len(decrData)
    if (length&1) != 0:
        decrData = "0" + decrData
        length = len(decrData)
    ss2 = []
    for i in range(length >> 1):
        ss2.append(0)
    i = 0
    i2 = 0
    while i<length-1:
        i3 = i +1
        i = i3 +1
        ss2[i2] = str((self.r_digit(ord(decrData[i]),16) << 4) | (self.r_digit(ord(decrData[i3]),16) & 255))
        i2+=1
    print(ss2)
复制代码

后面的123

def digit(self, codePoint, radix):
    result = -1
    if radix<2 | radix > 36:
        return -1
    if 48<=codePoint & codePoint<=57:
        result = codePoint-48
    if 97<=codePoint & codePoint<=122:
        result = codePoint - 87
    if 65<=codePoint & codePoint<=90:
        result = codePoint - 55
    if result < radix:
        return result
    return -1

def r_digit(self,codePoint,radix):
    dd = self.digit(codePoint, radix)
    if dd<0:
        return 0
    return dd
复制代码

好，还原之后运行，然后就报错了(我的心情:*(^*&()@&#)@(&#@*&^$)

Traceback (most recent call last):
  File "D:/work_soft/pycharm/__init__.py", line 91, in <module>
    print(eg.data_decrypt(data))
  File "D:/work_soft/pycharm/__init__.py", line 79, in data_decrypt
    ss2[i2] = str((self.r_digit(ord(decrData[i]),16) << 4) | (self.r_digit(ord(decrData[i3]),16) & 255))
IndexError: string index out of range
复制代码

很明显，报错提示的是decrData[i] 这个有问题，也就是最后一次循环，i的值是字符串的长度，我们都知道索引是从0开始，字符串的最后一位的索引是字符串的长度减1，所以这里就报错了。

做到这里我一度以为是我代码还原错了，于是把java代码放到idea里同样运行了一次，也报了同样的错误！当时人都迷茫了，这样的情况还是第一次遇到。

思路与做的尝试

1.控制变量

一开始我还不觉得是代码有问题(其实当然是代码有问题了！)，我以为是我hook的入参没有复制对，所以我又去hook了入参，当然是不成功了。

2.多次脱壳

这次我终于信了是代码有问题，所以又用了另一个脚本去脱壳，发现脱出来的代码还是完全一样，怎么能半路而退，这都没有涉及到so! 干！

3.修改java代码

我hook了第二张图的a方法，发现入参的长度是返回值的长度的两倍。

那是当然的，每次i3和i都是1，2、3，4...所以到最后就是字符串长度-1，字符串长度然后报错，那么我都-1不就完了，因为没道理索引从1开始，那不就跳过了字符串的第一位的字符了吗，这次运行出结果了，但是使用控制变量发现和正常的结果还是不一样。

又观察了一遍java代码，发现出问题的代码只可能在while循环，而j方法里的Character.digit(c2, 16)以及后面跟到的方法对比了一下java源码应该是没有魔改的，而while里面最大的运算就是这一行

bArr[i2] = (byte) (((j(charSequence.charAt(i), i) << 4) | j(charSequence.charAt(i3), i3)) & 255);
复制代码

所以我把i和i3的位置对调了一下，结果就一样了，我擦，看到结果和hook到的值是一样的时候简直就像小时候问老爸能不能玩会电脑，然后老爸的回复是玩会吧一样高兴啊

(然后就想到因为疫情快一年没回家了，哭= =、)

完整代码

# coding: utf-8

import base64
from Crypto.Cipher import AES


class EncryptDate:
    def __init__(self, key):
        self.key = key.encode("utf-8")  # 初始化密钥
        self.length = AES.block_size  # 初始化数据块大小
        self.aes = AES.new(self.key, AES.MODE_ECB)  # 初始化AES,ECB模式的实例
        # 截断函数，去除填充的字符
        self.unpad = lambda date: date[0:-ord(date[-1])]

    def pad(self, text):
        """
        #填充函数，使被加密数据的字节码长度是block_size的整数倍
        """
        count = len(text.encode('utf-8'))
        add = self.length - (count % self.length)
        entext = text + (chr(add) * add)
        return entext

    def digit(self, codePoint, radix):
        result = -1
        if radix<2 | radix > 36:
            return -1
        if 48<=codePoint & codePoint<=57:
            result = codePoint-48
        if 97<=codePoint & codePoint<=122:
            result = codePoint - 87
        if 65<=codePoint & codePoint<=90:
            result = codePoint - 55
        if result < radix:
            return result
        return -1

    def r_digit(self,codePoint,radix):
        dd = self.digit(codePoint, radix)
        if dd<0:
            return 0
        return dd


    def data_decrypt(self, decrData):
        length = len(decrData)
        if (length&1) != 0:
            decrData = "0" + decrData
            length = len(decrData)
        ss2 = []
        for i in range(length >> 1):
            ss2.append(0)
        i = 0
        i2 = 0
        while i<length-1:
            i3 = i +1
            i = i3 +1
            ss2[i2] = str((self.r_digit(ord(decrData[i3-1]),16) << 4) | (self.r_digit(ord(decrData[i-1]),16) & 255))
            i2+=1
        print(ss2)

        # b = ''.join(ss2).encode(encoding="utf8")

        # msg = self.aes.decrypt(b).decode("utf8")
        # return self.unpad(msg)

eg = EncryptDate("rFQJTvMFzm8MtBRy")  # 这里密钥的长度必须是16的倍数
data = '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print(eg.data_decrypt(data))
复制代码

然后现在就是到了最后一步

现在已经得到了a(str)，就是我代码中打印的ss2，那么在python中怎么使用一个列表(java是一个字节数组)进行aes解密呢，我再研究研究，研究出来会修改文章，大佬看到可以指点一下吗(图穷匕见)

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看了一下午终于解决了，原来把列表里的之前的字符串改成int，然后bytes()转一下，就可以了。算法学的不好啊还是，= =。

def data_decrypt(self, decrData):
    length = len(decrData)
    if (length&1) != 0:
        decrData = "0" + decrData
        length = len(decrData)
    ss2 = []
    for i in range(length >> 1):
        ss2.append(0)
    i = 0
    i2 = 0
    while i<length-1:
        i3 = i +1
        i = i3 +1
        ss2[i2] = int((self.r_digit(ord(decrData[i3-1]),16) << 4) | (self.r_digit(ord(decrData[i-1]),16) & 255))
        i2+=1
    print(ss2)

    b = bytes(ss2)
    msg = self.aes.decrypt(b).decode("utf8")
    return self.unpad(msg)
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