While converting byte array to hexstring

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hi All,

I got a piece of code from google where i want to convert a byte array(randomly generated) to hex string. i dint understand few code of lines..can any one help please.??

The three functions i used :

Java

public static String CellKeysGeneration() {

        byte[] btba = new byte[5];

        Random r = new Random();

        r.nextBytes(btba);

        for (int i = 0; i < btba.length; i++) {

            btba[i] = btba[i];

        }
        String str = tohexString(btba);
        return str;
    }

    public static String tohexString(byte[] bytes) {

        StringBuffer sb = new StringBuffer(bytes.length * 2);

        for (int i = 0; i < bytes.length; i++) {
            sb.append(tohex(bytes[i] >> 4));   
            sb.append(tohex(bytes[i]));

        }

        return sb.toString();

    }

    public static char tohex(int nibble) {

        final char[] hexdigit = { '0', '1', '2', '3', '4', '5', '6', '7', '8',
                '9', 'A', 'B', 'C', 'D', 'E' , 'F' };

        return hexdigit[nibble & 0xF];
    }

1) I din't understand why always "and" operation is done only with 0xF.
2) In tohexString(), why ">>" operation with 4 is done?

Can any one help me to understand tohexString() and toHex() methods :)

Posted 17-Jan-13 2:36am

IICTECH

Updated 17-Jan-13 2:53am

Richard MacCutchan

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Richard MacCutchan · Answer 1 · 2013-01-17T03:03:00

Because the value passed to the tohex() method is always a 4-bit value.
To extract the leftmost 4 bits of the byte and pass it to the tohex() method.

If you step through the code it goes something like:

Select the next byte in sequence
Right shift 4 bits and pass the resulting 4 bits (the left half) to the tohex() method
Append the resulting character to the string
Pass this byte to the tohex() method which masks off the right half
Append the resulting character to the string

So the hex value 0xE7 gets shifted right passed as 0x0E which is masked with 0x0F giving 0x0E. Next the value 0xE7 gets passed, which is masked with 0x0F giving 0x07.

CPallini · Answer 2 · 2013-01-17T03:22:00

Quote:
1) I din't understand why always "and" operation is done only with 0xF.

Such operation masks out (discards) the 4 most significant bits of the byte, because 0xF is 00001111 in binary representation. (e.g. 10101001 & 00001111 = 00001001).

Quote:
2) In tohexString(), why ">>" operation with 4 is done?

This 'moves the right nibble to the left', that is the 4 most significant bits of the byte replace the 4 least significant ones (the right shift operator >> moves bits to the right). Hence, for instance, 0xAB becomes 0x0A.

While converting byte array to hexstring

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