Sign extension doubt?

[Deepak Borania]

public class test {

    public static void main(String[] args) {
	byte b=-1;
	char c= (char)(b);		//(1)
	System.out.println((int)c);	//output: 65535
        c=(char)(b&0xff);		(2)
	System.out.println((int)c);	//output: 255


    }

}

I understand that the output 655535 is caused due to sign extension in statement (1), but what is happening in statement (2) that prevents this. Width of byte is 8 bits, so 'anding' it with 0xff should just result in byte value unchanged, shouldn't it?

BTW how can I print 'byte' as binary or hex strings?

[Henry Wong]

Width of byte is 8 bits, so 'anding' it with 0xff should just result in byte value unchanged, shouldn't it?

Not really. 0xff is a int literal. So the result is indeed the original bit value from the byte, but now stored in an int.

Henry

[Henry Wong]

BTW how can I print 'byte' as binary or hex strings?

One option is to first convert it to a binary or hex string using the java.lang.Integer class. Another option is to use the System.out.printf() method call.

Henry

[Deepak Borania]

Thanks Henry. I was kind of sleepless night because I couldn't understand it.

[santhosh.R gowda]

char c= (char)(b); //(1)
System.out.println((int)c); //output: 65535

Can you please tell me how we get the out put 65535 i'm confused

[Campbell Ritchie]

Does this Java™ Language Specification section help at all? Note, if you wind back about three lines, you see that

char . . . values are 16-bit unsigned integers

If you need more explanation, please ask again.

[Campbell Ritchie]

santhosh.R gowda wrote:char c= (char)(b); //(1)

You mean -1 surely? There is a lot of difference between 1 and -1. Thirty-one "1" bits' worth, in fact.

[Rob Spoor]

The (1) is not the value but a marker, indicating the line is special.

[Campbell Ritchie]

Rob Prime wrote:The (1) is not the value but a marker, indicating the line is special.

Thank you, Rob. I should know better than to try reading that sort of code after beer

[Henry Wong]

santhosh.R gowda wrote:

char c= (char)(b); //(1)
System.out.println((int)c); //output: 65535

Can you please tell me how we get the out put 65535 i'm confused

Easy part, a byte is signed 8 bit value, and a char is a unsigned 16 bit value. Hard part, you need to understand how Java stores negative numbers... see twos complement...

http://en.wikipedia.org/wiki/Two's_complement

Once you understand twos complement, and how a byte is sign extended when casting to a char, it should be straightforward. However, if you are still confused (after you get up to speed on twos complement), come back here, and elaborate what you don't understand.

Henry

[santhosh.R gowda]

come back here, and elaborate what you don't understand.

Dear sir, as you stated above that

   
byte b=-1;
char c= (char)b;
System.out.println((int)c); //output: 65535 
 

please tell me what is the integer value stored in c before type casting into int.
please rectify the below code

byte b=-1;
// -1 means 1111 1111 stored in byte which is two's complement form
//this has to be extended to char which is 16 bit
char c =(char)b;
//1111 1111 1111 1111 1111
// so the value is 65536; but the maximum value is 65535 

    

please give me debugging details

[Rob Spoor]

Just as 1111 1111 is 255, 1111 1111 1111 1111 is 65535. After all, its 32768 + 16384 + 8192 + 4096 + 2048 + 1024 + 512 + 256 + 128 + 64 + 32 + 16 + 8 + 4 + 2 + 1.

[santhosh.R gowda]

Dear all

byte b=-1;
int i=(int)b;
System.out.println(i);//output will be -1

how come this even though the byte has to sign extension like char in above

[Henry Wong]

santhosh.R gowda wrote:Dear all

byte b=-1;
int i=(int)b;
System.out.println(i);//output will be -1

how come this even though the byte has to sign extension like char in above

b = 1111 1111
i (sign extend b to length of int) = 1111 1111 1111 1111 1111 1111 1111 1111

Remember an int is signed, and a char isn't (when applying twos complement)....

Henry

[Deepak Borania]

Take look in "Java Puzzlers: Traps, Pitfalls, and Corner Cases"- Puzzle 6 for more info on the topic.
It was the puzzle from which this question orignated anyway.