Which declarations are valid?

Question

Select the three correct answers (valid declarations).

(a) char a = '\u0061';

(b) char 'a' = 'a';

(c) char \u0061 = 'a';

(d) ch\u0061r a = 'a';

(e) ch'a'r a = 'a';

Answer: (a), (c) and (d)

Book:

A Programmer's Guide to Java SCJP Certification (Third Edition)

Can someone please explain the reason for the option (c) and (d) as the IDE (IntelliJ IDEA) is showing it in red saying:

Cannot resolve symbol 'u0063'

Portwin

Java allows unicode characters in source code: #4448680 – Chinquapin

Why are (c) and (d) not working for you? What error do you get? – Periwig

Java converts the \u0061 to the character a, that makes (c) valid. Since it is allowed and possible as a variable name, it is also allowed and possible in a type name, so ch\u0061r is just char, which makes (d) valid as well. – Rocketeer

"not working for me" is not very helpful (e.g. do you convert Unicode sequences?) If using a (standard) Java compiler this must work, is one of the first operations a compiler has to do. See: Java Language Specification 3.3. Unicode Escapes – Malek

I was wrong in saying "not working for me" as I did not compile as the IDE was showing is a 'Cannot resolve symbol'. But it works once compiled. The issue was just that it is not yet supported in IDE. Thanks – Portwin

not true, IDE is not showing that... at least not mine, also works fine with javac compiler... maybe some settings or so, but no idea WHICH IDE your are using. – Malek

@Carlos - I have updated the IDE name and also attached a screenshot of the code in the question. Thanks for asking. – Portwin

Perfect! I don't use IntelliJ, but it must have an option to set encoding, but I do not think that this would help. Maybe some InteliJ users can help, despite it really seems like an bug, eventually search/raise an issue at JetBrains – Malek

It is a known error (15 years!): IDEABKL-89 and IDEA-65898 – Malek

I wonder this is still not corrected... there are plenty duplicates for that ticket – Malek

Answer 1

The compiler can recognise Unicode escapes and translate them to UTF-16. ch\u0061r will become char which is a valid primitive type. It makes option D correct.

3.3. Unicode Escapes

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) for the indicated hexadecimal value, and passing all other characters unchanged.

\u0061 will be translated to a which is a valid Java letter that can be used to form an identifier. It makes option C correct.

3.8. Identifiers

An identifier is an unlimited-length sequence of Java letters and Java digits, the first of which must be a Java letter.

Identifier:
    IdentifierChars but not a Keyword or BooleanLiteral or NullLiteral
IdentifierChars:
    JavaLetter {JavaLetterOrDigit}
JavaLetter:
    any Unicode character that is a "Java letter"
JavaLetterOrDigit:
    any Unicode character that is a "Java letter-or-digit"

A "Java letter" is a character for which the method Character.isJavaIdentifierStart(int) returns true.

A "Java letter-or-digit" is a character for which the method Character.isJavaIdentifierPart(int) returns true.

The "Java letters" include uppercase and lowercase ASCII Latin letters A-Z (\u0041-\u005a), and a-z (\u0061-\u007a), and, for historical reasons, the ASCII dollar sign ($, or \u0024) and underscore (_, or \u005f). The dollar sign should be used only in mechanically generated source code or, rarely, to access pre-existing names on legacy systems. The underscore may be used in identifiers formed of two or more characters, but it cannot be used as a one-character identifier due to being a keyword.

Landpoor

Answer 2

\u0061 means a. You can use \u0061 instead of a, therefore:

char \u0061 = 'a';

is the same as

char a = 'a';

and

ch\u0061r a = 'a';

is the same as

char a = 'a';

Grieg