Java中的4字节unicode字符 [英] 4 byte unicode character in Java

问题描述

我正在为我的自定义StringDatatype编写单元测试，并且需要写下4字节的unicode字符."\ U"-不起作用(非法转义字符错误)例如:U + 1F701(0xf0 0x9f 0x9c 0x81).如何将其写成字符串?

解决方案

Unicode代码点不是4个字节；它不是4个字节.它是一个整数(目前从U + 0000到U + 10FFFF).

您的4个字节是(很自然地)其UTF-8编码版本(；如果您的计算环境支持，则直接作为符号.

另请参见 CharsetDecoder 和 CharsetEncoder 类.

另请参见 String.codePointCount()，以及从Java 8开始的 String.codePoints()(继承自 CharSequence )./p>

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?

解决方案

A Unicode code point is not 4 bytes; it is an integer (ranging, at the moment, from U+0000 to U+10FFFF).

Your 4 bytes are (wild guess) its UTF-8 encoding version (edit: I was right).

You need to do this:

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 char is 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.

Given that a char is, in fact, a UTF-16 code unit and that there are string 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.

See also the CharsetDecoder and CharsetEncoder classes.

See also String.codePointCount(), and, since Java 8, String.codePoints() (inherited from CharSequence).

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