前言
第二次遇到鸿蒙逆向的题，感觉挺新奇，琢磨了一中午，虽然中间有些出题人疏忽的小插曲，不过修复的速度也挺快，也是顺利拿了个一血哈哈
分析
题目给了一个shctf.app文件，对于文件结构可以查看官方文档的描述 应用程序包基础知识
实际上同安卓类似，使用解压缩软件即可解压，得到一个.hap文件和一个pack.info
HAP（Harmony Ability Package）是应用安装和运行的基本单元。HAP包是由代码、资源、第三方库、配置文件等打包生成的模块包，其主要分为两种类型：entry和feature。
entry：应用的主模块，作为应用的入口，提供了应用的基础功能。
feature：应用的动态特性模块，作为应用能力的扩展，可以根据用户的需求和设备类型进行选择性安装。
应用程序包可以只包含一个基础的entry包，也可以包含一个基础的entry包和多个功能性的feature包。
然后继续解压hap文件
在etc目录下，可以看到一个module.abc文件
abc文件
方舟字节码（ArkCompiler Bytecode）文件，是ArkCompiler的编译工具链以源代码作为输入编译生成的产物，其文件后缀名为.abc。在发布时，abc文件会被打包到HAP中。
接下来就是想办法如何去反编译这个abc文件，最后在github上发现一个基于jadx的开源abc反编译工具 abc-decompiler，虽然反编译效果还不是很完美，期待大佬们的完善
分析page/Index，在构造函数中初始化了secretKey
复制代码 隐藏代码
public Object Index(Object functionObject, Object newTarget, Index this, Object arg0, Object arg1, Object arg2, Object arg3, Object arg4, Object arg5) {
if ((0 == arg3 ? 1 : 0) != 0) {
arg3 = -1;
}
if ((0 == arg4 ? 1 : 0) != 0) {
arg4 = null;
}
Object obj = super(arg0, arg2, arg3, arg5);
if (("function" == typeof(arg4) ? 1 : 0) != 0) {
obj.paramsGenerator_ = arg4;
}
obj.__userInput = ObservedPropertySimplePU("", obj, "userInput");
obj.secretKey = "[f#fLw)??Pz?#9w)Du[ks[q[#w4D?4P4UJf,kU[f.rDkfwrDtq...)?J.#rP4[qrPDJkkJ|.9J|qffU?k|D4P4P[wkk.)k?JUJ[k#9kww[r??wUfw|PkrPUf.P#f.P#.PwJ4f4q.PU4UPDr9.[9fJ#PqP)cDDffJPDrJ.J4qPP[r[.JfJ4f|?U9#";
obj.setInitiallyProvidedValue(arg1);
obj.finalizeConstruction();
return obj;
}
获取用户输入，并加密，和secretKey对比，如果校验通过则创建一个迷宫，然后写入What文件中
复制代码 隐藏代码
public Object validateAndCreateMaze(Object functionObject, Object newTarget, Index this, Object arg0) {
newlexenv(2);
_lexenv_0_0_ = newTarget;
_lexenv_0_1_ = this;
Object newobjrange = import { encode } from "@bundle:com.welcome.shctf/entry/ets/model/encode"();
if ((_lexenv_0_1_.secretKey == newobjrange.encode(arg0) ? 1 : 0) == 0) {
Object promptAction = import { default as promptAction } from "@ohos:promptAction";
promptAction.showToast(createobjectwithbuffer(["message", "口令错误！"]));
return null;
}
Object newobjrange2 = import { maze } from "@bundle:com.welcome.shctf/entry/ets/model/maze"();
r12 = getContext(_lexenv_0_1_).filesDir + "/What";
try {
Object fileIo = import { default as fileIo } from "@ohos:file.fs";
Object openSync = fileIo.openSync(r12, import { default as fileIo } from "@ohos:file.fs".OpenMode.READ_WRITE | import { default as fileIo } from "@ohos:file.fs".OpenMode.CREATE);
Object CreateMaze = newobjrange2.CreateMaze(arg0);
Object map = CreateMaze.map(#11009740135539449431#);
Object join = map.join("\n");
Object fileIo2 = import { default as fileIo } from "@ohos:file.fs";
fileIo2.writeSync(openSync.fd, join);
Object promptAction2 = import { default as promptAction } from "@ohos:promptAction";
Object obj = promptAction2.showToast;
Object obj2 = createobjectwithbuffer(["message", 0, "duration", 5000]);
obj2.message = "口令正确！\n等下！好像创建了什么东西？\n" + r12 + "";
obj(obj2);
return null;
} catch (ExceptionI0 unused) {
Object promptAction3 = import { default as promptAction } from "@ohos:promptAction";
Object obj3 = promptAction3.showToast;
Object obj4 = createobjectwithbuffer(["message", 0]);
obj4.message = "文件操作失败: " + r12.message + "";
obj3(obj4);
return null;
}
}
先对model/encode方法进行分析，一些常规的运算，单字节加密
复制代码 隐藏代码
public Object #9151489524504893103#encode(Object functionObject, Object newTarget, encode this, Object arg0) {
Object obj = [Object];
Object obj2 = obj;
throw(obj);
Object obj3 = getiterator(arg0);
Object obj4 = obj3.next;
while (true) {
Object callthisN = obj4();
throw.ifnotobject(callthisN);
if (istrue(callthisN.done) != null) {
break;
}
Object obj5 = callthisN.value;
try {
obj2.push(String.fromCharCode((((114514 * (obj5.charCodeAt(0) - 32)) + 1919810) % 95) + 32));
} catch (ExceptionI0 unused) {
if (istrue(0) == null) {
try {
obj2 = obj3.return;
} catch (ExceptionI0 unused2) {
}
if ((0 == obj2 ? 1 : 0) == 0) {
obj2();
throw(obj5);
}
}
throw(obj5);
}
}
return obj2.join("");
}
可以直接写个爆破脚本来解决
复制代码 隐藏代码
s = "[f#fLw)??Pz?#9w)Du[ks[q[#w4D?4P4UJf,kU[f.rDkfwrDtq...)?J.#rP4[qrPDJkkJ|.9J|qffU?k|D4P4P[wkk.)k?JUJ[k#9kww[r??wUfw|PkrPUf.P#f.P#.PwJ4f4q.PU4UPDr9.[9fJ#PqP)cDDffJPDrJ.J4qPP[r[.JfJ4f|?U9#"
f = ""
for i in range(len(s)):
for j in range(32, 128):
tmp = ((114514*(j - 32)+1919810)%95)+32
if s[i] == chr(tmp):
f += chr(j)
break
print(f)
由于手上没有鸿蒙的机子，也没有调试hap的方法，因此只能继续静态分析他是如何创建迷宫的
分析model/maze，在构造函数中初始化了迷宫数据
复制代码 隐藏代码
public Object maze(Object functionObject, Object newTarget, maze this) {
Object[] objArr = [Object];
objArr[0] = createarraywithbuffer(["#", "#", "#", "#", "#", "#", "#", "#", "#", "#", "#", "#", "#", "#", "#", "#", "#", "#", "#", "#", "#", "#", "#", "#", "#", "#", "#", "#", "#", "#", "#", "#", "#", "#", "#", "#", "#", "#", "#", "#", "#", "#", "#", "#", "#", "#", "#", "#", "#", "#", "#", "#", "#", "#", "#", "#", "#", "#", "#", "#", "#", "#", "#", "#", "#", "#", "#", "#", "#", "#", "#", "#", "#", "#", "#", "#", "#", "#", "#", "#", "#", "#", "#", "#", "#"]);
objArr[1] = createarraywithbuffer(["#", " ", "#", " ", " ", " ", " ", " ", "#", " ", " ", " ", " ", " ", " ", " ", " ", " ", "#", " ", " ", " ", " ", " ", " ", " ", "#", " ", " ", " ", " ", " ", "#", " ", " ", " ", " ", " ", " ", " ", " ", " ", " ", " ", " ", " ", " ", " ", " ", " ", "#", " ", " ", " ", " ", " ", " ", " ", "#", " ", " ", " ", "#", " ", " ", " ", "#", " ", " ", " ", "#", " ", " ", " ", "#", " ", " ", " ", " ", " ", " ", " ", " ", " ", "#"]);
......
this.maze = objArr;
return this;
}
CreateMaze方法中设置了迷宫的起点(S)和终点(E)，并将我们输入的内容填充在迷宫中
复制代码 隐藏代码
public Object CreateMaze(Object functionObject, Object newTarget, maze this, Object arg0) {
this.maze[77][1] = "E";
this.maze[1][83] = "S";
return this.FillF1ag(this.maze, arg0);
}
FillF1ag方法遍历迷宫所有格子，判断是否为空格，然后调用CheckGround再判断，由于反编译工具还不完善，反编译结果看着还比较难懂，手动优化一下可以发现他进行了边界检查，以及返回四个方向空格的个数，当个数超过2时，FillFlag就会将输入内容填充进去
sishen 
