Question

考虑以下两个例子：

    testFind("\\W.*", "@ this is a sentence");
    testFind(".*", "@ this is a sentence");

这是我的testFind方法

 private static void testFind(String regex, String input) {
    Pattern pattern = Pattern.compile(regex);
    Matcher matcher = pattern.matcher(input);
    int matches = 0;
    int nonZeroLengthMatches = 0;

    while (matcher.find()) {
        matches++;
        String matchedValue = matcher.group();
        if (matchedValue.length() > 0) {
            nonZeroLengthMatches++;
        }
        System.out.printf("Matched startIndex= %s, endIndex= %s, value: '%s'\n",
                matcher.start(), matcher.end(), matchedValue);

    }

    System.out.printf("Total non zero length matches = %s/%s \n", nonZeroLengthMatches, matches);
}

这是输出：

 ---------------------
   Regex: '\W.*', Input: '@ this is a sentence'
   Matched startIndex= 0, endIndex= 20, value: '@ this is a sentence'
   Total non zero length matches = 1/1 
   ---------------------
   Regex: '.*', Input: '@ this is a sentence'
   Matched startIndex= 0, endIndex= 20, value: '@ this is a sentence'
   Matched startIndex= 20, endIndex= 20, value: ''
   Total non zero length matches = 1/2

根据这个：https://docs.oracle.com/javase/7/docs/api/java/util/regex/Pattern.html

贪婪的量词 ..... X * X，零次或多次

我的问题是为什么在regex =“\ W. *”的情况下匹配器没有给出零长度匹配？

Answer 1

因为"\W.*"表示："\W" - 非单词字符加上".*" - 任意字符为零或更多次，所以只有'@...'等于此模式{{1 }}，但"\W.*"没有匹配。

当Matcher #find返回false时

1 个答案: