《JavaScript AST其实很简单》五、Step4-平坦化控制流

漁滒发表于 2020-12-31 22:09

## 平坦化控制流
上一篇我们说到去除了if判断下的废代码，那么本篇文章为本阶段的最后一篇，可能平坦化控制流不知道是什么，那么直接看看下面截图就知道了
![在这里插入图片描述](https://img-blog.csdnimg.cn/20201231210658414.jpg?x-oss-process=image/watermark,type_ZmFuZ3poZW5naGVpdGk,shadow_10,text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3pqcTU5Mjc2NzgwOQ==,size_16,color_FFFFFF,t_70#pic_center)
可以看到，这里代码的都是有用的代码，不同于上一篇存在废代码，但是代码的执行顺序并不是按照一般的从上到下执行，这对代码阅读并不友好。

```javascript
var _0x51adbe = '3|4|0|5|1|2'['split']('|'), _0x50da1a = 0;
while (!![]) {
switch (_0x51adbe) {
case '0':
var _0x89f607 = _0x5c99f5 || _0x527778;
continue;
case '1':
_0x527778['toString'] = _0x89f607['toString']['bind'](_0x89f607);
continue;
case '2':
_0x5c99f5 = _0x527778;
continue;
case '3':
var _0x527778 = _0x5a83cd['constructor']['prototype']['bind'](_0x5a83cd);
continue;
case '4':
var _0x55b3ce = _0x5b4bc9;
continue;
case '5':
_0x527778['__proto__'] = _0x5a83cd['bind'](_0x5a83cd);
continue;
}
break;
}
```

先看一下上面代码的规律，可以看到先定义了一个字符串，并且使用split切割，作为代码执行的顺序，再定义一个临时变量。跟着就是一个while循环，下面就是我们需要的代码了。
看起来好像是最困难的，但是经过分析后发现，其实只是比上一篇难一点点，属于第二简单的。将上面代码做一下语法的分析
![在这里插入图片描述](https://img-blog.csdnimg.cn/20201231213036251.jpg?x-oss-process=image/watermark,type_ZmFuZ3poZW5naGVpdGk,shadow_10,text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3pqcTU5Mjc2NzgwOQ==,size_16,color_FFFFFF,t_70#pic_center)
可以看到，这部分的代码块包含两部分。一部分是VariableDeclaration节点，一部分是WhileStatement节点。前面VariableDeclaration节点就是定义字符串的部分，所以我们要首先遍历所有BlockStatement（代码块）的节点，找到第一个节点的类型是VariableDeclaration，第二个节点是WhileStatement，并且VariableDeclaration节点一次定义了两个变量，且第一个变量的的属性赋值是一个CallExpression
![在这里插入图片描述](https://img-blog.csdnimg.cn/20201231215801236.jpg?x-oss-process=image/watermark,type_ZmFuZ3poZW5naGVpdGk,shadow_10,text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3pqcTU5Mjc2NzgwOQ==,size_16,color_FFFFFF,t_70#pic_center)
那么我们就可以递归遍历找到所有这类型的代码块，然后我们最终也是要返回一个代码块的，那么我们就创建一个空列表，用来保存每一个case下的代码。获取第一个变量的字符串，可以直接在python里面的split函数切分，得到一个顺序的列表
![在这里插入图片描述](https://img-blog.csdnimg.cn/20201231220146527.png?x-oss-process=image/watermark,type_ZmFuZ3poZW5naGVpdGk,shadow_10,text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3pqcTU5Mjc2NzgwOQ==,size_16,color_FFFFFF,t_70#pic_center)
这里可以看到，case部分的代码在cases节点下是按顺序的，那么就可以按照前面切分得到列表的顺序来遍历case，将里面的代码部分添加到前面我们新建的空列表里面。当所有case遍历完以后，也得到我们最终的代码块数组，将原本整个节点部分的的代码块替换为后面添加了代码的列表。这时就完成了平坦化控制流。

运行替换后格式化一下代码
![在这里插入图片描述](https://img-blog.csdnimg.cn/20201231220704440.jpg?x-oss-process=image/watermark,type_ZmFuZ3poZW5naGVpdGk,shadow_10,text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3pqcTU5Mjc2NzgwOQ==,size_16,color_FFFFFF,t_70#pic_center)

可以看到代码已经完成反混淆的工作，此时阅读代码已经不再困难了。
**备注：源文件见最下方附件内的ob_decrypt.txt**

不苦小和尚 发表于 2020-12-31 22:50

天空宫阙发表于 2020-12-31 22:43
抖音web端现在不返回重要参数tac所以搞定了也拿不到数据，并且那个代码属于精心设计的大型控制流平坦化代 ...

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               else if (_ > 1) {
                     if ((_ = x) > 13)
                        z[++S] = !1;
                     else if (_ > 6)
                        C = z,
                        z = zinstanceof C;
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                        C = z,
                        z = z % C;
                     else if (_ > 2)
                        z ? $ += 4 : $ += 2 * (j = G[$]) - 2;
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                        for (j = G[$],
                        C = "",
                        D = n.q; D < n.q; D++)
                           C += String.fromCharCode(l ^ n.p);
                        z[++S] = C,
                        $ += 4
                     }
               } else if (_ > 0) {
                     (_ = x) > 12 ? (C = z,
                     q = z,
                     z[++S] = C,
                     z[++S] = q) : _ > 3 ? (C = z,
                     z = z == C) : _ > 1 ? (C = z,
                     z = z + C) : _ > -1 && (z[++S] = h)
               } else {
                     (_ = x) > 13 ? (z[++S] = G[$],
                     $ += 4) : _ > 11 ? (C = z,
                     z = z >> C) : _ > 9 ? (j = G[$],
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                     C = z,
                     c = C) : _ > 7 ? (j = G[$],
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               _ = 3 & x;
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               _ > 2)
                     (_ = x) > 13 ? (z[++S] = G[$],
                     $ += 8) : _ > 11 ? (C = z,
                     z = z >>> C) : _ > 9 ? z[++S] = !0 : _ > 7 ? (j = G[$],
                     $ += 2,
                     z = z) : _ > 0 && (C = z,
                     z = z < C);
               else if (_ > 1) {
                     (_ = x) > 10 ? (j = G[$],
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                     z = z ^ C) : _ > 6 && (C = z)
               } else if (_ > 0) {
                     if ((_ = x) < 3) {
                        A = 0,
                        k = z.length,
                        E = z;
                        z[++S] = function() {
                           var b = A < k;
                           if (b) {
                                 var a = E;
                                 z[++S] = a
                           }
                           z[++S] = b
                        }
                     } else
                        _ < 5 ? (j = G[$],
                        $ += 2,
                        C = c,
                        z[++S] = C) : _ < 7 ? z = ++z : _ < 9 && (C = z,
                        z = zin C)
               } else {
                     if ((_ = x) < 2) {
                        for (j = G[$],
                        _ = "",
                        D = n.q; D < n.q; D++)
                           _ += String.fromCharCode(l ^ n.p);
                        _ = +_,
                        $ += 4,
                        z[++S] = _
                     } else
                        _ < 4 ? (C = z,
                        z = z - C) : _ < 6 ? (C = z,
                        z = z === C) : _ < 15 && (C = z,
                        z = z,
                        z = C)
               }
            }
         }
   return
}
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   var l, t;
   null == c && (c = this),
   d && !d.d && (d.d = 0,
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   d = {});
   var i = {}
      , o = i.d = d ? d.d + 1 : 0;
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         i = d;
   for (t = 0,
   o = i.length = f.length; t < o; t++)
         i = f;
   return n && !B && K(b, a, 2 * e),
   B ? L(b, a, e, 0, i, c, null, 1) : L(b, a, e, 0, i, c, null, 0)
}
}
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[, , "undefined" != typeof exports ? exports : void 0, "undefined" != typeof module ? module : void 0]);

不苦小和尚 发表于 2020-12-31 22:54

不苦小和尚发表于 2020-12-31 22:50
_$jsvmprt = function(b, a, e) {
function f() {
if ("und ...
https://creator.douyin.com/creator-micro/home，这个是网址，控制台输出window.byted_acrawler.sign('','asd')
window.byted_acrawler.sign这个方法被混淆了
打印出来的就是siguare
"_02B4Z6wo00d01EMC5fgAAIBA-8Sbp1GoJ-xDB-FAAE8yR2bt5DdM.TS4Jhe5e6EVoHjCBMNBIlAlHrsFCFtWmviV2P6iWFkcopMFt.ppZI8WrLHGFL.dpqKxrr1rrS6Fsi5I6RY0VByNT5sgc7"

不苦小和尚 发表于 2020-12-31 22:35

能讲下抖音那个siguare混淆怎么还原吗

天空宫阙 发表于 2020-12-31 22:43

不苦小和尚发表于 2020-12-31 22:35
能讲下抖音那个siguare混淆怎么还原吗

抖音web端现在不返回重要参数tac所以搞定了也拿不到数据，并且那个代码属于精心设计的大型控制流平坦化代码，还原的难度上天，一般只能补充其检测的浏览器属性

不苦小和尚 发表于 2020-12-31 22:51

天空宫阙发表于 2020-12-31 22:43
抖音web端现在不返回重要参数tac所以搞定了也拿不到数据，并且那个代码属于精心设计的大型控制流平坦化代 ...

是发布视频那个siguare，不是需要tac的

漁滒发表于 2021-1-1 00:36

本帖最后由漁滒于 2021-1-1 00:45 编辑

天空宫阙发表于 2020-12-31 22:43
抖音web端现在不返回重要参数tac所以搞定了也拿不到数据，并且那个代码属于精心设计的大型控制流平坦化代 ...
其实有一个骚操作，就是多线程一直请求web的接口，多请求几十次以后，有可能会有视频数据返回{:17_1068:}

好久丿好酒 发表于 2021-1-1 07:58

siguare混淆怎么还原还原处出来

天空宫阙 发表于 2021-1-1 09:22

漁滒发表于 2021-1-1 00:36
其实有一个骚操作，就是多线程一直请求web的接口，多请求几十次以后，有可能会有视频数据返回 ...

可以，还有这操作。抖音web端也算是风控拉满连自己也不放过

漁滒发表于 2021-1-1 11:20

好久丿好酒发表于 2021-1-1 07:58
siguare混淆怎么还原还原处出来

赋空值就可以，不校验的

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《JavaScript AST其实很简单》五、Step4-平坦化控制流