以 UTF-8 格式计算 Java 字符串的长度，而无需实际对其进行编码

java utf-8

2022-08-31 22:30:36

有谁知道标准Java库（任何版本）是否提供了一种计算字符串二进制编码长度的方法（在本例中是UTF-8），而无需实际生成编码输出？换句话说，我正在寻找一个有效的等价物：

"some really long string".getBytes("UTF-8").length

我需要为可能很长的序列化消息计算长度前缀。

答案 1

下面是一个基于 UTF-8 规范的实现：

public class Utf8LenCounter {
  public static int length(CharSequence sequence) {
    int count = 0;
    for (int i = 0, len = sequence.length(); i < len; i++) {
      char ch = sequence.charAt(i);
      if (ch <= 0x7F) {
        count++;
      } else if (ch <= 0x7FF) {
        count += 2;
      } else if (Character.isHighSurrogate(ch)) {
        count += 4;
        ++i;
      } else {
        count += 3;
      }
    }
    return count;
  }
}

此实现不能容忍格式错误的字符串。

以下是用于验证的 JUnit 4 测试：

public class LenCounterTest {
  @Test public void testUtf8Len() {
    Charset utf8 = Charset.forName("UTF-8");
    AllCodepointsIterator iterator = new AllCodepointsIterator();
    while (iterator.hasNext()) {
      String test = new String(Character.toChars(iterator.next()));
      Assert.assertEquals(test.getBytes(utf8).length,
                          Utf8LenCounter.length(test));
    }
  }

  private static class AllCodepointsIterator {
    private static final int MAX = 0x10FFFF; //see http://unicode.org/glossary/
    private static final int SURROGATE_FIRST = 0xD800;
    private static final int SURROGATE_LAST = 0xDFFF;
    private int codepoint = 0;
    public boolean hasNext() { return codepoint < MAX; }
    public int next() {
      int ret = codepoint;
      codepoint = next(codepoint);
      return ret;
    }
    private int next(int codepoint) {
      while (codepoint++ < MAX) {
        if (codepoint == SURROGATE_FIRST) { codepoint = SURROGATE_LAST + 1; }
        if (!Character.isDefined(codepoint)) { continue; }
        return codepoint;
      }
      return MAX;
    }
  }
}

请原谅紧凑的格式。

答案 2

使用番石榴的Utf8：

Utf8.encodedLength("some really long string")