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.
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