Java Character Values

Question

Hi ,
Please look into my following java programs:
public class Test{
public static void main(String args[]){
StringBuffer sb = new StringBuffer();
for(int i = 32;i<256;i++){
   sb.append((char)i);
   doThis(sb.toString());
public static void doThis(String str){
  byte arr[] = str.getBytes();
  int len = arr.length;
     for(int i =0;i<len;i++){
      char ch = (char)arr;
int ii = (int)ch;
System.out.println("character "+ch+ " value "+ ii);
when I run this program I get the output
as
character A value 65
once the character value reaches 127, then it prints as follow for next 32 enteries that is untill character actual value is 159..
character ? value 63
after characters actual value becomes more than 159 that it gives me huges vaules like as below:
character ? value 65440
character ? value 65535
any idea why?
I want to print the exact value of the characters?
thanks & Regards,
Akhil

Answer 1

char values are not the same thing as byte values in Java. Characters take 16 bit UNICODE values which represent not just the regular latin alphabet but all kind of exotic characters including Greek, Arabic, Hebrew, Chinese and so on.
When reading and writing text files these UNICODE characters have to be translated from or to byte values. The translation can use many different schemes. There are character encoding schemes designed for various national requirements, for example one where the bytes in the range 128-255 represent Greek characters. There are also international schemes like UTF-8. If you do an operation that changes bytes to characters or characters to bytes (like getBytes()) you generally have the option of specifying a scheme. If not the system uses a default based on your machine's location information.
In your case I guess the default is ASCII in which only characters up to 127 have valid mappings.
Had you put
byte arr[] = str.getBytes("ISO-LATIN-1");I think you'd get something more like what you expected.