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      "body": "<div  id='write'  class = 'is-mac'><p>&nbsp;</p><h1>准备资料</h1><ul><li><a href='https://github.com/kevinejohn/blind-signatures/blob/master/rsablind.js'>盲签名源码</a></li><li><a href='https://baike.baidu.com/link?url=9P_8pS64n4GHB0Vk9w7WP0KeYh_-VS5zJ0k_GKr_ovXFjxwxvgP9ES2aJxuA1jcAlbypJE0NwvDXh_Hi3rVlvWv0HTlqmcZmpsE62jjj3J_Nnp84kL1_u3UCql8ZxFDi'>取模运算规则</a></li><li><a href='https://zh.wikipedia.org/wiki/%E6%AC%A7%E6%8B%89%E5%AE%9A%E7%90%86_(%E6%95%B0%E8%AE%BA)'>欧拉定理</a></li><li><a href='http://www.ruanyifeng.com/blog/2013/06/rsa_algorithm_part_one.html'>RSA算法原理</a></li></ul><p>&nbsp;</p><h2>资料提炼</h2><ol start='' ><li><p>通过阅读<a href='https://github.com/kevinejohn/blind-signatures/blob/master/rsablind.js'>盲签名源码</a>可以知道，当一个明文<code>m</code>经过盲化、签名，最后再去盲的操作后的最终结果为：</p><blockquote>((((m(r<sup>e</sup> % n)) % n)<sup>d</sup> %n)r<sup>-1</sup>) % n</h1></blockquote><p><em><code>m</code>为明文</em></p><p><em><code>r</code>为随机数</em></p><p><em><code>n</code>和<code>e</code>组成公钥，<code>n</code>和<code>d</code>组成私钥，根据RSA加密算法定义可知：<code>e</code>和<code>d</code>满足：ed % φ(n) = 1</em></p><p><em>r<sup>-1</sup>为r对于n的模反元素，满足：(r<sup>-1</sup>r) % n = 1。</em></p></li><li><p>从<a href='https://zh.wikipedia.org/wiki/%E6%AC%A7%E6%8B%89%E5%AE%9A%E7%90%86_(%E6%95%B0%E8%AE%BA)'>欧拉定理</a>得到公式（a与n互质）</p><blockquote><p>a<sup>𝜑(n)</sup> % n = 1</p></blockquote></li><li><p>从<a href='https://baike.baidu.com/link?url=9P_8pS64n4GHB0Vk9w7WP0KeYh_-VS5zJ0k_GKr_ovXFjxwxvgP9ES2aJxuA1jcAlbypJE0NwvDXh_Hi3rVlvWv0HTlqmcZmpsE62jjj3J_Nnp84kL1_u3UCql8ZxFDi'>取模运算规则</a>中列出以下对于本次证明有用的公式：</p><table><thead><tr><th style='text-align:left;' >说明</th><th style='text-align:left;' >公式</th><th style='text-align:center;' >编号</th></tr></thead><tbody><tr><td style='text-align:left;' >结合律</td><td style='text-align:left;' >((a*b) % n * c) % n = (a * (b*c) % n) % n</td><td style='text-align:center;' >1</td></tr><tr><td style='text-align:left;' >模结合律</td><td style='text-align:left;' >a<sup>b</sup> % n = ((a % n)<sup>b</sup>) % n</td><td style='text-align:center;' >2</td></tr><tr><td style='text-align:left;' >乘法分配律</td><td style='text-align:left;' >(a * b) % n = (a % n * b % n) % n</td><td style='text-align:center;' >3</td></tr></tbody></table></li></ol><h1>证明过程</h1><p>由盲签名的原理可知，我们需要证明：</p><blockquote><p>((((m(r<sup>e</sup> % n)) % n)<sup>d</sup> %n)r<sup>-1</sup>) % n 等价于 m<sup>d</sup> % n</p></blockquote><p>((((m(r<sup>e</sup> % n)) %n)<sup>d</sup> %n)r<sup>-1</sup>) % n&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;公式2</p><p>= (((m(r<sup>e</sup> % n))<sup>d</sup> % n)r<sup>-1</sup>) % n \t\t\t\t</p><p>= (((m<sup>d</sup> * (r<sup>e</sup> % n)<sup>d</sup>) %n)r<sup>-1</sup>) % n&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;公式3</p><p>= (((m<sup>d</sup>%n * (r<sup>e</sup> % n)<sup>d</sup> %n) %n)r<sup>-1</sup>) % n&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;公式2</p><p>= ((((m<sup>d</sup> %n) * (r<sup>ed</sup> %n)) %n)r<sup>-1</sup>) % n&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;公式3</p><p>= (((m<sup>d</sup>  * r<sup>ed</sup>) %n) * r<sup>-1</sup>) % n&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;公式1</p><p>= (m<sup>d</sup>  * (r<sup>ed</sup> r<sup>-1</sup>) %n) % n</p><p>&nbsp;</p><p>即需要证明 (r<sup>ed</sup> r<sup>-1</sup>) %n = 1，将已知条件 ed = k * φ(n) + 1代入有：</p><p>(r<sup>ed</sup> r<sup>-1</sup>) %n</p><p>= (r<sup>kφ(n)+1</sup> r<sup>-1</sup>) %n</p><p>=(r<sup>kφ(n)</sup> * rr<sup>-1</sup>) %n&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;公式3</p><p>= (r<sup>kφ(n)</sup>%n *(rr<sup>-1</sup>)%n) %n\t&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;模反元素的定义</p><p>= (r<sup>kφ(n)</sup>%n * 1) %n</p><p>= ((r<sup>φ(n)</sup> …… r<sup>φ(n)</sup>) %n ) %n\t\t\t\t\t</p><p>&nbsp;</p><p>设r<sup>φ(n)</sup>=p，p%n=q，则有：</p><p>((r<sup>φ(n)</sup> …… r<sup>φ(n)</sup>) %n ) %n </p><p>= ((p …… p) %n ) %n &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;公式3的变形</p><p>= ((q(q(q(……(q*q) %n……) %n) %n) %n) %n</p><p>&nbsp;</p><p>通过阅读<a href='https://github.com/kevinejohn/blind-signatures/blob/master/rsablind.js'>盲签名源码</a>第40行至第47行可知，<code>r</code>与<code>n</code>互质，满足欧拉定理。则有r<sup>φ(n)</sup>%n=1，即q=1，原式得证。</p><p>&nbsp;</p><h1>盲签原理</h1><p>为什么我们要证明</p><blockquote><p>((((m(r<sup>e</sup> % n)) % n)<sup>d</sup> %n)r<sup>-1</sup>) % n 等价于 m<sup>d</sup> % n</p></blockquote><p>因为根据RSA加解密过程，我们知道，加密过程为：</p><blockquote><p>c = m<sup>e</sup> % n</p></blockquote><p>解密过程为：</p><blockquote><p>m = c<sup>d</sup> % n</p></blockquote><p>&nbsp;</p><p>我们传给服务器的是盲化后的内容，即：</p><blockquote><p>(m(r<sup>e</sup> % n)) % n</p></blockquote><p>服务器用私钥将这个内容做加密操作，然后我们再将加密后的内容做去盲处理就拿到了服务器对原文的加密结果。其他人就可以用公钥来校验这个结果。</p><p>&nbsp;</p><h1>安全性</h1><p>我们来看看交给服务器的盲化后的字串：</p><blockquote><p>(m(r<sup>e</sup> % n)) % n</p></blockquote><p>式子里总共有四个未知量：m、r、e、n，服务器只知道e和n，没有r求不出m。</p><p>&nbsp;</p><p>再来考虑一下r被穷举的难度有多大呢？</p><p>通过阅读<a href='https://github.com/kevinejohn/blind-signatures/blob/master/rsablind.js'>盲签名源码</a>第40行至第47行可知（尤其是其中的第41行），r的选取需满足以下要求：</p><blockquote><p>1 &lt; r &amp;&amp; r &lt; min(n, 2<sup>80</sup>) &amp;&amp; r与n互质 &amp;&amp; r ∈ N</p></blockquote><p>现在的RSA算法中的n一般都是1024位，那么r就是小于2<sup>80</sup>且与n互质的正整数。</p><p>&nbsp;</p><p>那这样的数有多少个呢？</p><p>根据<a href='https://zh.wikipedia.org/wiki/%E6%AC%A7%E6%8B%89%E5%87%BD%E6%95%B0'>资料</a>我们知道小于n且与n互质的正整数的个数叫做<strong>欧拉函数</strong>，写作<strong>φ(n)</strong>。在RSA算法里，p、q是任意两个质数，φ(n)满足下面的条件：</p><blockquote><p>n = pq</p><p>φ(n) = (p-1)(q-1)</p></blockquote><p>当n足够大时，并假设p、q亦足够大（事实上p、q也确实足够大），那么φ(n)取其二次方项：</p><blockquote><p>φ(n) = (p-1)(q-1) = pq - p - q + 1≈ pq = 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      "body": "很多软件试用期结束后，通常都是会弹出一个对话框，然后告诉你试用期已经结束，底下再来两个按钮，一个是**退出**，另一个是**购买**。点退出就直接退出程序，点购买就浏览器跳转到他们官网的购买页面。由于该对话框强制获取了软件的焦点，所以不点任意一个按钮的话是没有办法做其他操作的。\n\n很早以前使用的试用版Reveal就遇到了这个问题，当时心血来潮，想重新温习一下破解过程，决定自己动手试试。下面便是**Reveal 1.6.1**版本的破解过程：\n\n\n\n## 绕过过期对话框逻辑\n<br>\n\n现在理清了问题，再来看有哪些方案可以绕过这个对话框：\n\n* 因为有试用期，那么启动软件后肯定有判断是否过期的逻辑：\n\n   > 从用户沙盒里取出第一次使用软件的时间点，然后和当前的时间点做个差值，最后拿这个差值和试用期时间做个比较，然后根据比较结果判断是否过期。\n\n   所以我们的方案就是：反汇编后，猜测一些关键字：`trial`、`expire`等。通\t过这些关键字搜索相应的可能函数，然后再在判断是否过期的`if-else`语句里将逻辑置反即可。\n   \n\n * 从弹出的过期对话框入手。之所以不能使用软件了，就是因为对话框吸引了焦点，所以我们只需要隐藏掉对话框即可。但是_用汇编写隐藏对话框_的难度太大，可以从展示对话框的逻辑那里置反即可。不过直接搜出展示逻辑不大可能，我们就先通过关键字找到该对话框的viewcontroller，然后在里面随便设个断点，然后启动调试，等跑到断点处时，查看调用堆栈，这样就能找到show这个过期view的逻辑在哪个函数里了，然后进那个函数将show的逻辑置反。虽然数据里依旧是过期的，但是却能和正常软件一样使用了。\n\n我这里选择的是第2种方法，在`IBATrialModeReminderWindowController windowDidLoad`里面随便找一处设个断点。然后找到并将显示的`if`逻辑置反(`Alt`+`A`编辑汇编命令)。注意修改汇编代码的时候要保持byte位数与改之前一致，空出来的位置用`nop`填充，但是不能多出byte了，多了很容易就会导致软件直接crash。\n\n\n\n## 添加codesign\n<br>\n改好过期逻辑后，`⌘+⇧+E`保存修改后的二进制文件。这时会弹出下面的提示：\n![20934E1E-2F17-4550-AAD1-B5A5375F561C.png](https://steemitimages.com/DQmWBxAi8AoTRQJFnWhKSH43QtjGqqAjHEZRt3y5o3U5XBH/20934E1E-2F17-4550-AAD1-B5A5375F561C.png)\n\n这里选择`否`，我们要自己添加一个`codesign`，先自己随便建一个根证书，`钥匙串访问` =\\> `证书助理` =\\> `创建证书`，然后按下图设置后一路`下一步`狂点。\n![4ABC60C4-3B3F-468D-A902-B880F2B9DDA8.png](https://steemitimages.com/DQmeYzstJpi7pn46XBFiekGtZvZYQgEnGPDPmE9AXi3QK8L/4ABC60C4-3B3F-468D-A902-B880F2B9DDA8.png)\n\n后续具体操作可参照[这里](http://forums.macnn.com/79/developer-center/355720/how-re-sign-apples-applications-once/)。\n\n证书弄好后，就要给我们新修改的文件加codesign了，打开终端，cd到新文件所在目录，然后执行\n\n```bash\n codesign -f -s gdb-cert Reveal \n```\n<br>\n其中的`gdb-cert`就是上图创建证书的时候取的名字，`Reveal`就是你新生成的二进制文件的名字。\n\n\n\n## 绕过自身codesign验证\n<br>\ncodesign做好后，备份下原来的二进制文件，然后替换原来的。启动Reveal会弹出错误提示，说是codesign不对，然后直接退出了：\n![8AE2396D-0769-4686-A3DF-C6BB82AE56E8.png](https://steemitimages.com/DQmPFWKjp9YgKuiRMRcC1vbp8PvvYEd3SBaHuLB4GSEihLu/8AE2396D-0769-4686-A3DF-C6BB82AE56E8.png)\n\n这是因为Reveal做了自身验证，就是防止代码被更改，不过这个验证也是自己业务逻辑的验证，所以仿照第一步的方法，根据相关关键字搜索相关代码。我用的关键字是`codesign`，果然就搜到了一个名叫`IBAAppDelegate verifyCodeSignature`的函数：\n![7D9E70E9-C1DB-4168-ABDF-0F1386F636F5.png](https://steemitimages.com/DQmWgC8v6djbb2gKyVoL8Juqw39C3WgW1FgP57pSyJTGEXg/7D9E70E9-C1DB-4168-ABDF-0F1386F636F5.png)\n\n点进去根据伪代码能很轻松的就找到`if`逻辑：\n![992C6075-1B39-4AC7-8307-B67358A2283B.png](https://steemitimages.com/DQmanuovuqsvEQxiyr8tGunb7uszBGoAhtsRjeSZSCEwnqP/992C6075-1B39-4AC7-8307-B67358A2283B.png)\n\n![9032C90C-0DDF-461C-AEEA-7CFC1A41C2CA.png](https://steemitimages.com/DQmPswVsMZeqQXFcfuCmKeE8BdBnE4F5cw7k7b1FTTptJV5/9032C90C-0DDF-461C-AEEA-7CFC1A41C2CA.png)\n\n可以置反`if`逻辑，不过我们这里更简单，直接`jmp`到最后：\n![7F0DCDBC-154F-4B71-BF7B-29BBEECE42FC.png](https://steemitimages.com/DQmU6xZPRJcaeKVTwiHzRgMF6FS1wQzEZ5pxiaf5N5A4B6Y/7F0DCDBC-154F-4B71-BF7B-29BBEECE42FC.png)\n\n刚好保持位数一致。\n\n这下再导出、codesign、替换，最后运行OK。\n\n其实绝大部分的试用软件都可以通过这种方法达到破解的目的，只要有足够的耐心、细心再加一点点的运气~",
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  "trx_id": "8ed97b5326bcd0eb8823235fc049aa8563c65218",
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      "permlink": "56g5wz",
      "title": "记一次清除恶意脚本的过程",
      "body": "最近几天发现在用chrome的时候遇到很诡异的问题，打开网页后，点击网页任意地方都会被重定向到其他页面，然后手动返回原网页又能正常使用。但是时间一长网页就会经常卡顿，是那种非正常的卡顿。chrome、safari、firefox都有这毛病，但是唯独canary没有。怀疑极有可能是有什么恶意脚本在作怪，因为canary属于小众浏览器就放过了。趁着快周末有时间，决心亲自动手找出毒瘤。\n\n随便打开一个网页，发现在其body最后有这段代码：\n![676F1182-1D9E-40C4-8313-C2619BFDC768.png](https://steemitimages.com/DQmdenMhA1tRsxtLByFxTrg5YsPFRLSQjoSarokvaJg3agy/676F1182-1D9E-40C4-8313-C2619BFDC768.png)\n\n在这里网页被嵌入了一个宽度100%、高度100%、不可见的带href的frame，难怪点击任意地方都会被重定向，为了满足好奇心，去看了下被转到哪了：\n![81FC26D7-02EE-4153-8B95-8BC4C5A18841.png](https://steemitimages.com/DQmPMWj1YpWfN9iHXhmAcVuJ9nYBz69memPcqD2KzrVX9xd/81FC26D7-02EE-4153-8B95-8BC4C5A18841.png)\n恩，是淘宝某店铺的页面…\n\n现在需要找出这段代码是怎么被插进网页的。顺藤摸瓜，找到一堆相关的js脚本，其中看到这个：\n![5245098E-D02A-45A5-9F11-60EE204CB217.png](https://steemitimages.com/DQmPCxn42WyYXFEKvq2jSPtMKeUEDWes52AP9rMC2drVx1D/5245098E-D02A-45A5-9F11-60EE204CB217.png)\n上图代码里写着的目标浏览器里并没有包含canary，或许这就解释了为啥其他浏览器都中招，唯独canary没事了吧。\n\n接着往下走，发现了一大堆的js，它们环环相扣，链式地下载彼此，下图里的绝大多数都是最后恶意下载的脚本：\n![9F78F7D7-BFA1-44F7-8DCB-74A4948F15B6.png](https://steemitimages.com/DQmTyUXp67SMvoHV2a4NwW3P9JK4v2gnn34jZG8X3yM65jv/9F78F7D7-BFA1-44F7-8DCB-74A4948F15B6.png)\n\n分析了半天，最终发现这堆脚本的鼻祖起源于这行代码：\n![66F0A374-F1E3-4213-ADD9-BD5F11410E6C.png](https://steemitimages.com/DQmceWgof6XwVEH4CNJ9BvSDy6g3XpSrTWFWm6weBAMxKjA/66F0A374-F1E3-4213-ADD9-BD5F11410E6C.png)\n\n这个`secure.surfbuyermac.com`也不知道是什么玩意，搞不懂是怎么插到页面里的，起初怀疑是浏览器的恶意插件导致的，于是停掉所有chrome、safari的插件，结果发现问题依旧存在。接着怀疑可能是电脑本身的恶意脚本导致的，于是调起万能的`Spotlight `：\n![0A0EEAC1-4463-412C-BFDB-1451D028FB90.png](https://steemitimages.com/DQmTWvqF1LbxfRiMV3aRFfAa4EkfbXyMH9BzPQWNpbmAU3Z/0A0EEAC1-4463-412C-BFDB-1451D028FB90.png)\n\n去`Applications`瞧一眼：\n![3087670C-9E9E-475E-B789-71F5024E4A22.png](https://steemitimages.com/DQmTS84FfEGSXoWcJczQxvYdoEy1pJjW98tr6qCWaqyTPop/3087670C-9E9E-475E-B789-71F5024E4A22.png)\n\n为了确认是不是它，再去终端看看有没有在自启动：\n![FE92A9BB-9CFF-4BDE-AE08-0058EE8D0220.png](https://steemitimages.com/DQmRVZpDMmFqLihmjJPnK2YdUb4mRBLmL8gDMZZEe9QepHk/FE92A9BB-9CFF-4BDE-AE08-0058EE8D0220.png)\n\n妈蛋，总算逮着你了！kill掉相关进程后，再打开网页，哈，整个世界都清净了~：\n![449B2CB0-11E7-4B4F-8AA4-20BE40861640.png](https://steemitimages.com/DQmUz88Yy1uH32byk5e2ZrBunAPPBrXpfWjZ5io7PQhS9yW/449B2CB0-11E7-4B4F-8AA4-20BE40861640.png)\n\n最后保管好那个`SurfBuyer`以后有时间拿来做样本好好研究研究。",
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