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

Question

我最近利用了前瞻性正则表达式的功能来分割字符串:

"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,如下所示:

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Strings(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a结束时发生空匹配,否则这种逻辑很有效String.如果空匹配确实发生在a的结尾String,它将最终跳过该字符.这个部分应该是什么样的(注意>=- > >):

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

Splitter当模式匹配空字符串时,Guava 似乎有一个bug.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());
}

你得到输出3,3,使它看起来像匹配8.因此它只是在那里分裂abc.

您可以使用eg Pattern#split(String)似乎给出正确的输出:

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