Why can some ASCII characters not be expressed in the form '\uXXXX' in Java source code?

Question

I stumbled over this (again) today:

class Test {     char ok = '\n';     char okAsWell = '\u000B';     char error = '\u000A'; } 

It does not compile:

Invalid character constant in line 4.

The compiler seems to insist that I write '\n' instead. I see no reason for this, yet it's very annoying.

Is there a logical explanation why characters that have a special notation (like \t, \n, \r) must be expressed in that form in Java source?

Answer 1

Unicode characters are replaced by their value, so your line is replaced by the compiler with:

char error = ' '; 

which is not a valid Java statement.

This is dictated by the Language Specification:

A compiler for the Java programming language ("Java compiler") first recognizes Unicode escapes in its input, translating the ASCII characters \u followed by four hexadecimal digits to the UTF-16 code unit (§3.1) of the indicated hexadecimal value, and passing all other characters unchanged. Representing supplementary characters requires two consecutive Unicode escapes. This translation step results in a sequence of Unicode input characters.

This can lead to surprising stuff, for example, this is a valid Java program (it contains hidden unicode characters) - courtesy of Peter Lawrey:

public static void main(String[] args) {     for (char c‮h = 0; c‮h < Character.MAX_VALUE; c‮h++) {         if (Character.isJavaIdentifierPart(c‮h) && !Character.isJavaIdentifierStart(c‮h)) {             System.out.printf("%04x <%s>%n", (int) c‮h, "" + c‮h);         }     } }