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.
Char Literal: Java literal is specified as an integer literal representing the Unicode value of a char. This integer can be specified in octal, decimal, and hexadecimal, ranging from 0 to 65535. For example, char ch = 062; Escape Sequence: Every escape char can be specified as char literal.
Character literals are enclosed in single quotation marks. Any printable character, other than a backslash (\), can be specified as the single character itself enclosed in single quotes. Some examples of these literals are 'a', 'A', '9', '+', '_', and '~'.
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:
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
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