help me guys i got one simple doubt

[funkygarzon]

class HexByte {
static public void main(String args[]) {
char hex[] = {
'0', '1', '2', '3', '4', '5', '6', '7',
'8', '9', 'a', 'b', 'c', 'd', 'e', 'f'
};
byte b = (byte) 0xf1;
System.out.println("b = 0x" + hex[(b >> 4) & 0x0f] + hex[b & 0x0f]);
}
}

output :

b = 0xf1

please pleas tell me the logic for the above program please ?

[funkygarzon]

please guys just tell me the logic "how the each line is workinf " ..sorry for my noob question

[JosAH]

please guys just tell me the logic "how the each line is workinf " ..sorry for my noob question

What does your text book say about the '>>' and '&' operators?

kind regards,

Jos

[funkygarzon]

What does your text book say about the '>>' and '&' operators?

kind regards,

Jos

thanks for your reply buddy . Actually buddy , i know it is ">> is right shift " and "& is AND operator "

And i also understand the shift right ( b >> 4 ) shifts the bit s towards right for four position .

But my doubt is ; i could not trace out the logic going behind

"byte b = (byte) 0xf1;"

the above line :confused:

[JosAH]

But my doubt is ; i could not trace out the logic going behind

"byte b = (byte) 0xf1;"

the above line :confused:

The value 0xf1 is an int value (241 decimal) and the value is too large to fit in a single byte; a byte can only store values -128 ... 127 so you have to explicitly cast the value to a byte type to store it in a byte.

kind regards,

Jos

[funkygarzon]

The value 0xf1 is an int value (241 decimal) and the value is too large to fit in a single byte; a byte can only store values -128 ... 127 so you have to explicitly cast the value to a byte type to store it in a byte.

kind regards,

Jos

thank you very much for your reply buddy . Actually i know java automatically converts values to int , so that 's why we are casting it to byte .

1. my doubt is "0xf1 = 11110001" right ? it is having only 8 bits and how are saying this as "251" :confused::confused:

then after casting this "0xf1" to byte , how it will be look in binary form ? will it won't be same ?

[JosAH]

thank you very much for your reply buddy . Actually i know java automatically converts values to int , so that 's why we are casting it to byte .

1. my doubt is "0xf1 = 11110001" right ? it is having only 8 bits and how are saying this as "251" :confused::confused:

then after casting this "0xf1" to byte , how it will be look in binary form ? will it won't be same ?

241, not 251; and yes, 0xf1 in hexadecimal notation equals 11110001 in binary notation. The value 241 is too large for a byte; a byte can only store numbers in the range -128 ... 127 so the compiler will protest if you do:

[code]
byte b = 0xf1;
/code]

So we have to explicitly cast it to a byte type, but the eight bits 11110001 fit in the eight bits of a byte so the cast actually doesn't do anything except for keeping the compiler's mouth shut, i.e. those eight bits 11110001 are stored in the byte type variable.

kind regards,

Jos

[funkygarzon]

241, not 251; and yes, 0xf1 in hexadecimal notation equals 11110001 in binary notation. The value 241 is too large for a byte; a byte can only store numbers in the range -128 ... 127 so the compiler will protest if you do:

[code]
byte b = 0xf1;
/code]

So we have to explicitly cast it to a byte type, but the eight bits 11110001 fit in the eight bits of a byte so the cast actually doesn't do anything except for keeping the compiler's mouth shut, i.e. those eight bits 11110001 are stored in the byte type variable.

kind regards,

Jos

oooh great buddy ... thanks a lot for your wonderful info .

I have predicted some logic behind this program , please tell me whether my prediction is correct or not buddy ? :confused::confused:;

so after seeing " 0x0f" jvm thinks that 241 will not suit into byte and then it will converts it to int like this

00000000 00000000 00000000 11110001

so after casting the above value changes like this

11110001

so when the jvm see 's " b>>4 " it will start doing right shift like this

 11111000 -> 1st shift 
          11111100 -> 2nd shift 
          11111110 -> 3rd shift 
          11111111 -> 4th shift

now my doubt arises after this :confused:

so after seeing "hex[(b >> 4) & 0x0f]" , what will jvm will do buddy ?:confused::confused::confused: ..please make me clear ..thanks in advance

[funkygarzon]

please guys help me .....i could not understand the final step " hex[(b >> 4) & 0x0f] " ? how does the -1 turns to 1 when we do AND operation ? ..thanks in advance

[JosAH]

please guys help me .....i could not understand the final step " hex[(b >> 4) & 0x0f] " ? how does the -1 turns to 1 when we do AND operation ? ..thanks in advance

It'd be much faster if you'd use the "poor man's" debugger System.out.println( ... ). Print every intermediate step and see for yourself. Also, textbooks are valuable resources here.

kind regards,

Jos

[funkygarzon]

It'd be much faster if you'd use the "poor man's" debugger System.out.println( ... ). Print every intermediate step and see for yourself. Also, textbooks are valuable resources here.

kind regards,

Jos

ok buddy , thanks for your advice and info :) . ok let me close this thread :(