converting byte array of decimal value to hexadecimal string in java [duplicate]

Question

i m new to java. i want to convert a byte array of decimal value to hexadecimal string. my input byte array is [0, 0, 0, 0, 0, 0, 1, -28]. i m getting 00000000000001e4 instead of 0000001e4. plz help me to solve this problem

 public static String ConvetToHex(byte[] decValue) 
{

    String value = "";
    for(int i = 0;i<decValue.length;i++)
    {
         value = value+ Integer.toString((decValue[i] & 0xff) + 0x100, 16).substring(1);
    }
    return value;
}

Peter Lawrey · Accepted Answer · 2012-12-13 11:39:07Z

2

It looks correct to me. Eight bytes should turn into 16 hex characters. You can use

return new BigInteger(1, decValue).toString(16);

but it will produce the same output.

answered Dec 13, 2012 at 11:39

Peter Lawrey

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6 Comments

Sachin K S Over a year ago

but is there anyway to get in the specified format..?

Peter Lawrey Over a year ago

Sure you can but you will produce a number which is meaningless as there is no way to read the value correctly.

Peter Lawrey Over a year ago

For example in the format you suggest there is no way to tell if 1001 is 10, 01 (as it should be) or 10, 0, 1 or 1,0,0,1.

Sachin K S Over a year ago

i wrote the same code in .NET like value = value + Convert.ToInt32(decvalue[i]).toString("x"); i got the output which i mentioned.

Peter Lawrey Over a year ago

You can write the code in Java too. Doesn't change the fact that the format is broken.

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