4 byte unicode character in Java

Question

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?

Accepted Answer

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`).

4 byte unicode character in Java

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