why "...." is valid?

[nowait shen]

would anyone like to tell me why the following 2 declaration statement are valid?
char b = '\61';
char c = '\061';
thanks!

[Marlene Miller]

The octal escape sequences \000 to \377 represent the same characters as the Unicode escape sequences \u0000 to \u00ff.
A character literal is expressed as a character or an escape sequence enclosed in single quotes: '1' '\u0031' '\061'
Try this,
char c1 = '\061';
char c2 = '\u0031'; //0x31 == 061
char c3 = '1';
char c4 = (char)49; //49 == 0x31 == 061
System.out.println(c1);
System.out.println(c2);
System.out.println(c3);
System.out.println(c4);
System.out.println("A very\061long\u0031string1of text");
int i1 = c1, i2 = c2, i3 = c3, i4 = c4;
System.out.println(i1);
System.out.println(i2);
System.out.println(i3);
System.out.println(i4);
[ July 27, 2003: Message edited by: Marlene Miller ]

[cyril vidal]

Just a little precision:

char c4 = (char)49; //49 == 0x31 == 061

In this case,

char c4 = 49  

is also legal.
Reason:

Java relaxes its assignement conversion rule when a literal int value is assigned to a narrower primitive type (byte, short, or char), provided the literal value falls within the legal range of primitive type

Here, 0 <= 49 <= 2^16 - 1, so narrower assignement is legal.

[Marlene Miller]

Thank you Cyril. I could not decide whether to include the cast or not. I decided to include it to subliminally suggest '\061' '\u0031' '1' are of type char whereas 49 is of type int. The first three are character literals and 49 is an integer literal. I am glad you added your remarks.