A tricky problem

[rituk Kapoor]

If we are doing a operation like this(as shown below), we need to cast to char type. I'm getting the same output in both the case. Can anyone pls explain this to me.
1. what is actually happening at stage 1(commented)?
2. Why is it display "?" in both the cases?

byte b =-20;
short s= -34;
char c1 = (char)b; //1
char c2 = (char)s;
System.out.println(c1);
System.out.println(c2);

Output is :
?
?

Regards;
Ritu

[Barry Gaunt]

The eight bit representation of the byte containing -20 is "1110 1100" (the twos-complement representation of -20). At line //1 it is converted to a 16 bit wide char with sign extension: "1111 1111 1110 1100".

The ? is just an indicator that the character represented by the bit pattern cannot be printed on the output stream.

Try the following:

public class Explain {
  public static void main(String[] args) {
    byte b =-20;
    short s= -34;
    char c1 = (char)b; //1
    char c2 = (char)s;
    System.out.println(b + " is really " + Integer.toBinaryString(b & 0xFF));
    System.out.println(c1 + " is really " + Integer.toBinaryString(c1 & 0xFFFF));
    System.out.println(s + " is really " + Integer.toBinaryString(s & 0xFFFF));
    System.out.println(c2 + " is really " + Integer.toBinaryString(c2 & 0xFFFF));
  }
}  

[ September 02, 2005: Message edited by: Barry Gaunt ]
[ September 02, 2005: Message edited by: Barry Gaunt ]

[Karthik Hariharan]

Barry is correct. There is a finite list of ASCII characters mapped to particular integer values. If you try to cast any integer outside this valid list of ASCII values, you will get a ? displayed.

[rituk Kapoor]

Thanks to both of you. Specially Barry.