小代码大智慧: FilenameUtils.getName 函数分析

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一、背景

最近用到了 org.apache.commons.io.FilenameUtils#getName 这个方法,该方法可以传入文件路径,获取文件名。
简单看了下源码,虽然并不复杂,但和自己设想略有区别,值得学习,本文简单分析下。

二、源码分析

org.apache.commons.io.FilenameUtils#getName

 /**
     * Gets the name minus the path from a full fileName.
     * <p>
     * This method will handle a file in either Unix or Windows format.
     * The text after the last forward or backslash is returned.
     * 
<pre>
     * a/b/c.txt --&gt; c.txt
     * a.txt     --&gt; a.txt
     * a/b/c     --&gt; c
     * a/b/c/    --&gt; ""
     * </pre>
* <p>
     * The output will be the same irrespective of the machine that the code is running on.
     *
     * @param fileName  the fileName to query, null returns null
     * @return the name of the file without the path, or an empty string if none exists.
     * Null bytes inside string will be removed
     */
    public static String getName(final String fileName) 
     // 传入 null 直接返回 null 
        if (fileName == null) 
            return null;
        
        
        // NonNul 检查
        requireNonNullChars(fileName);
       
       //  查找最后一个分隔符
        final int index = indexOfLastSeparator(fileName);

     // 从最后一个分隔符窃到最后
        return fileName.substring(index + 1);
    

2.1 问题1:为什么需要 NonNul 检查 ?

2.1.1 怎么检查的?

org.apache.commons.io.FilenameUtils#requireNonNullChars

   /**
     * Checks the input for null bytes, a sign of unsanitized data being passed to to file level functions.
     *
     * This may be used for poison byte attacks.
     *
     * @param path the path to check
     */
    private static void requireNonNullChars(final String path) 
        if (path.indexOf(0) >= 0) 
            throw new IllegalArgumentException("Null byte present in file/path name. There are no "
                + "known legitimate use cases for such data, but several injection attacks may use it");
        
    

java.lang.String#indexOf(int) 源码:

 /**
     * Returns the index within this string of the first occurrence of
     * the specified character. If a character with value
     * @code ch occurs in the character sequence represented by
     * this @code String object, then the index (in Unicode
     * code units) of the first such occurrence is returned. For
     * values of @code ch in the range from 0 to 0xFFFF
     * (inclusive), this is the smallest value <i>k</i> such that:
     * <blockquote><pre>
     * this.charAt(<i>k</i>) == ch
     * </pre></blockquote>
     * is true. For other values of @code ch, it is the
     * smallest value <i>k</i> such that:
     * <blockquote><pre>
     * this.codePointAt(<i>k</i>) == ch
     * </pre></blockquote>
     * is true. In either case, if no such character occurs in this
     * string, then @code -1 is returned.
     *
     * @param   ch   a character (Unicode code point).
     * @return  the index of the first occurrence of the character in the
     *          character sequence represented by this object, or
     *          @code -1 if the character does not occur.
     */
    public int indexOf(int ch) 
        return indexOf(ch, 0);
    

可知,indexOf(0) 目的是查找 ASCII 码为 0 的字符的位置,如果找到则抛出 IllegalArgumentException异常。
搜索 ASCII 对照表,得知 ASCII 值为 0 代表控制字符 NUT,并不是常规的文件名所应该包含的字符。

2.1.2 为什么要做这个检查呢?

null 字节是一个值为 0 的字节,如十六进制中的 0x00。
存在与 null 字节有关的安全漏洞。
因为 C 语言中使用 null 字节作为字符串终结符,而其他语言(Java,php等)没有这个字符串终结符;
例如,Java Web 项目只允许用户上传 .jpg 格式的图片,但利用这个漏洞就可以上传 .jsp 文件。
如用户上传 hack.jsp<NUL>.jpg 文件, Java 会认为符合 .jpg 格式,实际调用 C 语言系统函数写入磁盘时讲 当做字符串分隔符,结果将文件保存为 hack.jsp
有些编程语言不允许在文件名中使用 ··
<NUL>,如果你使用的编程语言没有对此处理,就需要自己去处理。
因此,这个检查很有必要。

代码示例:

package org.example;

import org.apache.commons.io.FilenameUtils;

public class FilenameDemo 
    public static void main(String[] args) 
        String filename= "hack.jsp\\0.jpg";
        System.out.println( FilenameUtils.getName(filename));
    

报错信息:

Exception in thread "main" java.lang.IllegalArgumentException: Null byte present in file/path name. There are no known legitimate use cases for such data, but several injection attacks may use it
	at org.apache.commons.io.FilenameUtils.requireNonNullChars(FilenameUtils.java:998)
	at org.apache.commons.io.FilenameUtils.getName(FilenameUtils.java:984)
	at org.example.FilenameDemo.main(FilenameDemo.java:8)

如果去掉校验:

package org.example;

import org.apache.commons.io.FilenameUtils;

public class FilenameDemo 
    public static void main(String[] args) 
        String filename= "hack.jsp\\0.jpg";
        
        // 不添加校验
        String name = getName(filename);
        
        // 获取拓展名
        String extension = FilenameUtils.getExtension(name);
        System.out.println(extension);
    

    public static String getName(final String fileName) 
        if (fileName == null) 
            return null;
        
        final int index = FilenameUtils.indexOfLastSeparator(fileName);
        return fileName.substring(index + 1);
    


Java 的确会将拓展名识别为 jpg

jpg

JDK 8 及其以上版本试图创建 hack.jsp\\0.jpg 的文件时,底层也会做类似的校验,无法创建成功。

大家感兴趣可以试试使用 C 语言写入名为 hack.jsp\\0.jpg 的文件,最终很可能文件名为 hack.jsp

2.2 问题2: 为什么不根据当前系统类型来获取分隔符?

查找最后一个分隔符
org.apache.commons.io.FilenameUtils#indexOfLastSeparator

 /**
     * Returns the index of the last directory separator character.
     * <p>
     * This method will handle a file in either Unix or Windows format.
     * The position of the last forward or backslash is returned.
     * <p>
     * The output will be the same irrespective of the machine that the code is running on.
     *
     * @param fileName  the fileName to find the last path separator in, null returns -1
     * @return the index of the last separator character, or -1 if there
     * is no such character
     */
    public static int indexOfLastSeparator(final String fileName) 
        if (fileName == null) 
            return NOT_FOUND;
        
        final int lastUnixPos = fileName.lastIndexOf(UNIX_SEPARATOR);
        final int lastWindowsPos = fileName.lastIndexOf(WINDOWS_SEPARATOR);
        return Math.max(lastUnixPos, lastWindowsPos);
    

该方法的语义是获取文件名,那么从函数的语义层面上来说,不管是啥系统的文件分隔符都必须要保证得到正确的文件名。
试想一下,在 Windows 系统上调用该函数,传入一个 Unix 文件路径,得不到正确的文件名合理吗?
函数设计本身就应该考虑兼容性。
因此不能获取当前系统的分隔符来截取文件名。
源码中分别获取 Window 和 Unix 分隔符,有哪个用哪个,显然更加合理。

三、Zoom Out

3.1 代码健壮性

我们日常编码时,要做防御性编程,对于错误的、非法的输入都要做好预防。

3.2 代码严谨性

我们写代码一定不要想当然。
我们先想清楚这个函数究竟要实现怎样的功能,而且不是做一个 “CV 工程师”,无脑“拷贝”代码。
同时,我们也应该写好单测,充分考虑各种异常 Case ,保证正常和异常的 Case 都覆盖到。

3.3 如何写注释

org.apache.commons.io.FilenameUtils#requireNonNullChars 函数注释部分就给出了这么设计的原因:This may be used for poison byte attacks.

注释不应该“喃喃自语”讲一些显而易见的废话。
对于容易让人困惑的设计,一定要通过注释讲清楚设计原因。

此外,结合工作经验,推荐一些其他注释技巧:
(1)对于稍微复杂或者重要的设计,可以通过注释给出核心的设计思路;
如: java.util.concurrent.ThreadPoolExecutor#execute

    /**
     * Executes the given task sometime in the future.  The task
     * may execute in a new thread or in an existing pooled thread.
     *
     * If the task cannot be submitted for execution, either because this
     * executor has been shutdown or because its capacity has been reached,
     * the task is handled by the current @link RejectedExecutionHandler.
     *
     * @param command the task to execute
     * @throws RejectedExecutionException at discretion of
     *         @code RejectedExecutionHandler, if the task
     *         cannot be accepted for execution
     * @throws NullPointerException if @code command is null
     */
    public void execute(Runnable command) 
        if (command == null)
            throw new NullPointerException();
        /*
         * Proceed in 3 steps:
         *
         * 1. If fewer than corePoolSize threads are running, try to
         * start a new thread with the given command as its first
         * task.  The call to addWorker atomically checks runState and
         * workerCount, and so prevents false alarms that would add
         * threads when it shouldn't, by returning false.
         *
         * 2. If a task can be successfully queued, then we still need
         * to double-check whether we should have added a thread
         * (because existing ones died since last checking) or that
         * the pool shut down since entry into this method. So we
         * recheck state and if necessary roll back the enqueuing if
         * stopped, or start a new thread if there are none.
         *
         * 3. If we cannot queue task, then we try to add a new
         * thread.  If it fails, we know we are shut down or saturated
         * and so reject the task.
         */
        int c = ctl.get();
        if (workerCountOf(c) < corePoolSize) 
            if (addWorker(command, true))
                return;
            c = ctl.get();
        
        if (isRunning(c) && workQueue.offer(command)) 
            int recheck = ctl.get();
            if (! isRunning(recheck) && remove(command))
                reject(command);
            else if (workerCountOf(recheck) == 0)
                addWorker(null, false);
        
        else if (!addWorker(command, false))
            reject(command);
    

(2)对于关联的代码,可以使用 @see 或者 @link 的方式,在代码中提供关联代码的快捷跳转方式。

    /**
     * Sets the core number of threads.  This overrides any value set
     * in the constructor.  If the new value is smaller than the
     * current value, excess existing threads will be terminated when
     * they next become idle.  If larger, new threads will, if needed,
     * be started to execute any queued tasks.
     *
     * @param corePoolSize the new core size
     * @throws IllegalArgumentException if @code corePoolSize < 0
     *         or @code corePoolSize is greater than the @linkplain
     *         #getMaximumPoolSize() maximum pool size
     * @see #getCorePoolSize
     */
    public void setCorePoolSize(int corePoolSize) 
        if (corePoolSize < 0 || maximumPoolSize < corePoolSize)
            throw new IllegalArgumentException();
        int delta = corePoolSize - this.corePoolSize;
        this.corePoolSize = corePoolSize;
        if (workerCountOf(ctl.get()) > corePoolSize)
            interruptIdleWorkers();
        else if (delta > 0) 
            // We don't really know how many new threads are "needed".
            // As a heuristic, prestart enough new workers (up to new
            // core size) to handle the current number of tasks in
            // queue, but stop if queue becomes empty while doing so.
            int k = Math.min(delta, workQueue.size());
            while (k-- > 0 && addWorker(null, true)) 
                if (workQueue.isEmpty())
                    break;
            
        
    

(2)在日常业务开发中,非常推荐讲相关的文档、配置页面链接也放到注释中,极大方便后期维护。
如:

    /**
     * 某某功能
     *
     * 相关文档:
     * <a href="https://blog.csdn.net/w605283073">设计文档</a>
     * <a href="https://blog.csdn.net/w605283073">三方API地址</a>
     */
    public void demo()
        // 省略

    

(4)对于工具类可以考虑讲给出常见的输入对应的输出。
org.apache.commons.lang3.StringUtils#center(java.lang.String, int, char)

 /**
     * <p>Centers a String in a larger String of size @code size.
     * Uses a supplied character as the value to pad the String with.</p>
     *
     * <p>If the size is less than the String length, the String is returned.
     * A @code null String returns @code null.
     * A negative size is treated as zero.</p>
     *
     * <pre>
     * StringUtils.center(null, *, *)     = null
     * StringUtils.center("", 4, ' ')     = "    "
     * StringUtils.center("ab", -1, ' ')  = "ab"
     * StringUtils.center("ab", 4, ' ')   = " ab "
     * StringUtils.center("abcd", 2, ' ') = "abcd"
     * StringUtils.center("a", 4, ' ')    = " a  "
     * StringUtils.center("a", 4, 'y')    = "yayy"
     * </pre>
     *
     * @param str  the String to center, may be null
     * @param size  the int size of new String, negative treated as zero
     * @param padChar  the character to pad the new String with
     * @return centered String, @code null if null String input
     * @since 2.0
     */
    public static String center(String str, final int size, final char padChar) 
        if (str == null || size <= 0) 
            return str;
        
        final int strLen = str.length();
        final int pads = size - strLen;
        if (pads <= 0) 
            return str;
        
        str = leftPad(str, strLen + pads / 2, padChar);
        str = rightPad(str, size, padChar);
        return str;
    

(5) 对于废弃的方法,一定要注明废弃的原因,给出替代方案。
如:java.security.Signature#setParameter(java.lang.String, java.lang.Object)

    /**
     * 省略部分
     * 
     * @see #getParameter
     *
     * @deprecated Use
     * @link #setParameter(java.security.spec.AlgorithmParameterSpec)
     * setParameter.
     */
    @Deprecated
    public final void setParameter(String param, Object value)
            throws InvalidParameterException 
        engineSetParameter(param, value);
    

四、总结

很多优秀的开源项目的代码设计都非常严谨,往往简单的代码中也蕴藏着缜密的思考。
我们有时间可以看看一些优秀的开源项目,可以从简单的入手,可以先想想如果自己写大概该如何实现,然后和作者的实现思路对比,会有更大收获。
平时看源码时,不仅要知道源码长这样,更要了解为什么这么设计。


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参考文章

《What does it mean to have a “file name with NULL bytes in serialized instances”?》

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