Question

我最近利用了预见正则表达式的强大功能来分割字符串：

"abc8".split("(?=\\d)|\\W")

如果打印到控制台，则此表达式返回：

[abc, 8]

对此结果非常满意，我想将其转移到Guava进行进一步开发，看起来像这样：

Splitter.onPattern("(?=\\d)|\\W").split("abc8")

令我惊讶的是输出变为：

[abc]

为什么？

Answer 1

你发现了一个错误！

System.out.println(s.split("abc82")); // [abc, 8]
System.out.println(s.split("abc8"));  // [abc]

这是Splitter用于实际拆分String s（Splitter.SplittingIterator::computeNext）的方法：

@Override
protected String computeNext() {
  /*
   * The returned string will be from the end of the last match to the
   * beginning of the next one. nextStart is the start position of the
   * returned substring, while offset is the place to start looking for a
   * separator.
   */
  int nextStart = offset;
  while (offset != -1) {
    int start = nextStart;
    int end;

    int separatorPosition = separatorStart(offset);

    if (separatorPosition == -1) {
      end = toSplit.length();
      offset = -1;
    } else {
      end = separatorPosition;
      offset = separatorEnd(separatorPosition);
    }

    if (offset == nextStart) {
      /*
       * This occurs when some pattern has an empty match, even if it
       * doesn't match the empty string -- for example, if it requires
       * lookahead or the like. The offset must be increased to look for
       * separators beyond this point, without changing the start position
       * of the next returned substring -- so nextStart stays the same.
       */
      offset++;
      if (offset >= toSplit.length()) {
        offset = -1;
      }
      continue;
    }

    while (start < end && trimmer.matches(toSplit.charAt(start))) {
      start++;
    }
    while (end > start && trimmer.matches(toSplit.charAt(end - 1))) {
      end--;
    }

    if (omitEmptyStrings && start == end) {
      // Don't include the (unused) separator in next split string.
      nextStart = offset;
      continue;
    }

    if (limit == 1) {
      // The limit has been reached, return the rest of the string as the
      // final item.  This is tested after empty string removal so that
      // empty strings do not count towards the limit.
      end = toSplit.length();
      offset = -1;
      // Since we may have changed the end, we need to trim it again.
      while (end > start && trimmer.matches(toSplit.charAt(end - 1))) {
        end--;
      }
    } else {
      limit--;
    }

    return toSplit.subSequence(start, end).toString();
  }
  return endOfData();
}

感兴趣的领域是：

if (offset == nextStart) {
  /*
   * This occurs when some pattern has an empty match, even if it
   * doesn't match the empty string -- for example, if it requires
   * lookahead or the like. The offset must be increased to look for
   * separators beyond this point, without changing the start position
   * of the next returned substring -- so nextStart stays the same.
   */
  offset++;
  if (offset >= toSplit.length()) {
    offset = -1;
  }
  continue;
}

除非在String结束时发生空匹配，否则此逻辑很有效。如果空匹配发生在String的末尾，则最终会跳过该字符。这部分应该是什么样的（注意>= - ＆gt; >）：

if (offset == nextStart) {
  /*
   * This occurs when some pattern has an empty match, even if it
   * doesn't match the empty string -- for example, if it requires
   * lookahead or the like. The offset must be increased to look for
   * separators beyond this point, without changing the start position
   * of the next returned substring -- so nextStart stays the same.
   */
  offset++;
  if (offset > toSplit.length()) {
    offset = -1;
  }
  continue;
}

Answer 2

当模式匹配空字符串时，Guava Splitter似乎有一个错误。如果您尝试创建Matcher并打印出匹配的内容：

Pattern pattern = Pattern.compile("(?=\\d)|\\W");
Matcher matcher = pattern.matcher("abc8");
while (matcher.find()) {
    System.out.println(matcher.start() + "," + matcher.end());
}

您获得输出3,3，使其看起来与8匹配。因此，它只是在那里分割，结果只有abc。

您可以使用例如Pattern#split(String)似乎给出了正确的输出：

Pattern.compile("(?=\\d)|\\W").split("abc8")

Guava Splitter.onPattern（..）。split（）与String.split（..）有什么不同？

2 个答案: