java Java中的4字节Unicode字符
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4 byte unicode character in Java
提问by Constantine
I am writing unit tests for my custom StringDatatype, and I need to write down 4 byte unicode character. "\U" - not working (illegal escape character error) for example: U+1F701 (0xf0 0x9f 0x9c 0x81). How it can be written in a string?
我正在为我的自定义 StringDatatype 编写单元测试，我需要写下 4 字节的 unicode 字符。"\U" - 不工作（非法转义字符错误），例如：U+1F701 (0xf0 0x9f 0x9c 0x81)。怎么可以写成字符串？
回答by fge
A Unicode code point is not 4 bytes; it is an integer (ranging, at the moment, from U+0000 to U+10FFFF).
Unicode 代码点不是 4 个字节；它是一个整数（目前范围从 U+0000 到 U+10FFFF）。
Your 4 bytes are (wild guess) its UTF-8 encoding version (edit: I was right).
你的 4 个字节是（猜测）它的 UTF-8 编码版本（编辑：我是对的）。
You need to do this:
你需要这样做：
final char[] chars = Character.toChars(0x1F701);
final String s = new String(chars);
final byte[] asBytes = s.getBytes(StandardCharsets.UTF_8);
When Java was created, Unicode did not define code points outside the BMP (ie, U+0000 to U+FFFF), which is the reason why a charis only 16 bits long (well, OK, this is only a guess, but I think I'm not far off the mark here); since then, well, it had to adapt... And code points outside the BMP need two chars (a leading surrogate and a trailing surrogate -- Java calls these a high and low surrogate respectively). There is no character literal in Java allowing to enter code points outside the BMP directly.
Java 创建的时候，Unicode 没有定义 BMP 之外的码位（即 U+0000 到 U+FFFF），这就是 achar只有 16 位长的原因（好吧，好吧，这只是一个猜测，但我认为我在这里不远）；从那时起，它必须适应……并且 BMP 之外的代码点需要两个字符（前导代理和尾随代理——Java 分别将它们称为高代理和低代理）。Java 中没有字符文字允许直接在 BMP 之外输入代码点。
Given that a charis, in fact, a UTF-16 code unit and that there arestring literals for these, you can input this "character" in a String as "\uD83D\uDF01"-- or directly as the symbol if your computing environment has support for it.
鉴于 achar实际上是一个 UTF-16 代码单元，并且有这些字符串文字，您可以在字符串中输入这个“字符”作为"\uD83D\uDF01"-- 或者如果您的计算环境支持它，则直接作为符号输入。
See also the CharsetDecoderand CharsetEncoderclasses.
另请参见CharsetDecoder和CharsetEncoder类。
See also String.codePointCount(), and, since Java 8, String.codePoints()(inherited from CharSequence).
另请参见String.codePointCount()，并且，自 Java 8 起，String.codePoints()（继承自CharSequence）。
回答by Andrew
String s = "";
String s = "";
Technically this is one character. But be careful s.length()will returns 2. Also java won't compile String s = ''. Java don't promise you that String.length()shall returns exact number of characters, it returns just number of java-chars required for store this string.
从技术上讲，这是一个字符。但要小心s.length()会返回 2. 而且 java 不会编译String s = ''。Java 不向您保证String.length()将返回确切数量的字符，它仅返回存储此字符串所需的 java 字符数。
Real number of characters can be obtained from s.codePointCount(0, s.length()).
可从 中获得实际字符数s.codePointCount(0, s.length())。
