Dan's question:Topic. Literals

[Lakshmi Saradha]

The question is
class MCZ24 {
public static void main (String[] args) {
char a = 061; // 1
char b = '\61'; // 2
char c = '\061'; // 3
char d = 0x0031; // 4
char e = '\u0031'; // 5
System.out.print(""+a+b+c+d+e);
}}
A compile-time error is generated at which line?
a. 1
b. 2
c. 3
d. 4
e. 5
f. None of the above
The answer is f). I do agree that the answer is f). The explaination for the answer is..
"All of the declarations are legal. The first three ( 061, '\61', '\061' ) are declared in octal format. The fourth (0x0031) is declared as a hexadecimal literal. The fifth ('\u0031') is a Unicode escape sequence."
My question is "Is '\61' an octal format? Is it not a decimal format?".

[Dan Chisholm]

Lakshmi,
Java has Unicode escapes and octal escapes. An octal escape can begin with an optional leading zero. An octal escape does not require a leading zero.
There are no decimal escapes.
An octal integer literal does require a leading zero.
The following program prints the decimal value 8.

class A {
  public static void main (String[] s) {
    char c1 = '\10';
    System.out.println((int)c1);
  }
}

[Lakshmi Saradha]

Thank you Dan..