Unicode escape syntax in Java

Solution 1:

Interesting question. Section 3.3 of the JLS says:

UnicodeEscape:
    \ UnicodeMarker HexDigit HexDigit HexDigit HexDigit

UnicodeMarker:
    u
    UnicodeMarker u

which translates to \\u+\p{XDigit}{4}

and

If an eligible \ is followed by u, or more than one u, and the last u is not followed by four hexadecimal digits, then a compile-time error occurs.

So you're right, there can be one or more u after the backslash. The reason is given further down:

The Java programming language specifies a standard way of transforming a program written in Unicode into ASCII that changes a program into a form that can be processed by ASCII-based tools. The transformation involves converting any Unicode escapes in the source text of the program to ASCII by adding an extra u - for example, \uxxxx becomes \uuxxxx - while simultaneously converting non-ASCII characters in the source text to Unicode escapes containing a single u each.

This transformed version is equally acceptable to a Java compiler and represents the exact same program. The exact Unicode source can later be restored from this ASCII form by converting each escape sequence where multiple u's are present to a sequence of Unicode characters with one fewer u, while simultaneously converting each escape sequence with a single u to the corresponding single Unicode character.

So this input

 \u0020ä

becomes

 \uu0020\u00e4

The first uu means here "this was a unicode escape sequence to begin with" while the second u says "An automatic tool converted a non-ASCII character to a unicode escape."

This information is useful when you want to convert back from ASCII to unicode: You can restore as much of the original code as possible.

Solution 2:

It means you can add as many u as you want - for example these lines are equivalent:

char ch = '\u00A5';
char ch = '\uuuuu00A5';
char ch = '\uuuuuuuuuuuuuuuuuu00A5';

(and all compile)