What are the Java semantics of an escaped number i

Question

Please explain what, exactly, happens when the following sections of code are executed:

int a='\15';
System.out.println(a);

this prints out 13;

int a='\25';
System.out.println(a);

this prints out 21;

int a='\100';
System.out.println(a);

this prints out 64.

Answer 1

The notation \nnn denotes an octal character code in Java. so int a = '\15' assigns the auto-cast'ed value of octal character 15 to a which is decimal 13.

Answer 2

The fact that you put the digits in quotes makes me suspect it is interpreting the number as a character literal. The digits that follow must be in octal.

Answer 3

You have assigned a character literal, which is delimited by single quotes, eg 'a' (as distinct from a String literal, which is delimited by double quotes, eg "a") to an int variable. Java does an automatic widening cast from the 16-bit unsigned char to the 32-bit signed int.

However, when a character literal is a backslash followed by 1-3 digits, it is an octal (base/radix 8) representation of the character. Thus:

• \15 = 1×8 + 5 = 13 (a carriage return; same as '\r')
• \25 = 2×8 + 5 = 21 (a NAK char - negative acknowledgement)
• \100 = 1×64 + 0×8 + 0 = 64 (the @ symbol; same as '@')

For more info on character literals and escape sequences, see JLS sections:

• 3.10.4: Character Literals
• 3.10.6: Escape Sequences for Character and String Literals

Quoting the BNF from 3.10.6:

OctalEscape:
    \ OctalDigit
    \ OctalDigit OctalDigit
    \ ZeroToThree OctalDigit OctalDigit

OctalDigit: one of
    0 1 2 3 4 5 6 7

ZeroToThree: one of
    0 1 2 3