Why is '\117' a valid character literal in Java?

Question

I've seen this char defined as char ch = '\117'

What kind of representation is '\117' in?

I know escaped-sequence is '\n', for instance, or unicode is `\udddd', where d is a single hex digit, but I've never seen such thing as '\117' in my entire life! Surprisingly, it does compile! (And the output is O)

See [Escape Sequences for Character and String Literals](http://docs.oracle.com/javase/specs/jls/se7/html/jls-3.html#jls-3.10.6). — user207421, Feb 19 '13 at 05:42
This is octal notation. See [Why do Java octal escapes only go up to 255?](http://stackoverflow.com/q/9543026/111424), especially [rob mayoff's answer](http://stackoverflow.com/a/9543611/111424). It's probably for compatibility with older languages and programmers. — Iain Samuel McLean Elder, Feb 19 '13 at 12:00

score 11 · Accepted Answer · answered Feb 19 '13 at 05:43

11

This is the octal representation for ascii. You can see lots more values of it here: http://donsnotes.com/tech/charsets/ascii.html

answered Feb 19 '13 at 05:43

Daniel Kaplan

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score 8 · Answer 2 · answered Feb 19 '13 at 05:42

8

It's in octal, a holdover from C/C++.

answered Feb 19 '13 at 05:42

ldav1s

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score 3 · Answer 3 · answered Feb 19 '13 at 05:42

3

That is because its the Octal representation of captial O character.

If you try to print your char ch='\117'; , you will see that it prints O.

answered Feb 19 '13 at 05:42

Rahul

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score 2 · Answer 4 · answered Feb 19 '13 at 05:42

2

It's a Octal value for character "O", when I did system.out.println(..) I got this output:

char ch = '\117';
System.out.println("Char is: " + ch);

Output:

Char is : O

answered Feb 19 '13 at 05:42

Pradeep Simha

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It's more than that, it's an ASCII value in octal. – user207421 Feb 19 '13 at 05:44
Sorry, updated same :) @EJP – Pradeep Simha Feb 19 '13 at 05:44

Why is '\117' a valid character literal in Java?

4 Answers4

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