StringTokenizer 的性能看来真的不用担心

一直以来,分析HTTP的Header使用的都是StringTokenizer,但是看过jdk中关于StringTokenizer的介绍:

StringTokenizer 是出于兼容性的原因而被保留的遗留类(虽然在新代码中并不鼓励使用它)。建议所有寻求此功能的人使用 String 的 split 方法或 java.util.regex 包。

开始以为 StringTokenizer 功能或性能不是很给力,但经过半天的测试,使用String.split()、StringUtils.split()、mySplit(我定制的)、StringTokenizer 进行对比,下面是结果:

 

测试结果表明: StringTokenizer 对一个字符串进行分组读取,速度是最快的。

 

通过查看jdk源码,StringTokenizer.java 和 String.java中的split()方法,可以看到:StringTokenizer在对数据分段读取的时候,通过当前索引和下一个索引,进行判断和读取:

class StringTokenizer implements Enumeration<Object> {
    private int currentPosition;
    private int newPosition;
    private int maxPosition;
    private String str;
    private String delimiters;
    private boolean retDelims;
    private boolean delimsChanged;

................

 

而 String.split(),这个支持正则表达式(这个很耗时),然后先进行分组,然后保存到ArrayList,然后再转换成数组:

 public String[] split(String regex, int limit) {
        /* fastpath if the regex is a
           (1)one-char String and this character is not one of the
              RegEx's meta characters ".$|()[{^?*+\\", or
           (2)two-char String and the first char is the backslash and
              the second is not the ascii digit or ascii letter.
        */
        char ch = 0;
        if (((regex.count == 1 &&
             ".$|()[{^?*+\\".indexOf(ch = regex.charAt(0)) == -1) ||
             (regex.length() == 2 &&
              regex.charAt(0) == '\\' &&
              (((ch = regex.charAt(1))-'0')|('9'-ch)) < 0 &&
              ((ch-'a')|('z'-ch)) < 0 &&
              ((ch-'A')|('Z'-ch)) < 0)) &&
            (ch < Character.MIN_HIGH_SURROGATE ||
             ch > Character.MAX_LOW_SURROGATE))
        {
            int off = 0;
            int next = 0;
            boolean limited = limit > 0;
            ArrayList<String> list = new ArrayList<>();
            while ((next = indexOf(ch, off)) != -1) {
                if (!limited || list.size() < limit - 1) {
                    list.add(substring(off, next));
                    off = next + 1;
                } else {    // last one
                    //assert (list.size() == limit - 1);
                    list.add(substring(off, count));
                    off = count;
                    break;
                }
            }
            // If no match was found, return this
            if (off == 0)
                return new String[] { this };

            // Add remaining segment
            if (!limited || list.size() < limit)
                list.add(substring(off, count));

            // Construct result
            int resultSize = list.size();
            if (limit == 0)
                while (resultSize > 0 && list.get(resultSize-1).length() == 0)
                    resultSize--;
            String[] result = new String[resultSize];
            return list.subList(0, resultSize).toArray(result);
        }
        return Pattern.compile(regex).split(this, limit);
    }

所以,String.split()快不到哪里去。

 

2012-02-29

 

posted @ 2015-06-17 14:24  personnel  阅读(929)  评论(0编辑  收藏  举报
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