Byte to Integer and then to String conversion in Java

Question

I got a code for MD5 hash generation in Java. It generates the hash in byte array "bytes" and then converts to integer and then to string as follows:

byte[] bytes=md.digest(textToHash.getBytes());

StringBuilder sb=new StringBuilder();
for(int i=0;i<bytes.length;i++)
sb.append(Integer.toString((bytes[i] & 0xff) + 0x100, 16).substring(1));

I understood that bytes[i] & 0xff converts byte to integer of 32 bit length copying the byte to the least significant byte of the integer:

What does value & 0xff do in Java?

However I couldn't understand what + 0x100, 16 does in the parentheses at line 4 of the above code. Your help is appreciated.

Bohemian · Accepted Answer · 2016-04-08 04:53:44Z

3

Breaking down Integer.toString((bytes[i] & 0xff) + 0x100, 16).substring(1):

Adding 0x100 (which is 256 decimal) sets the 9th bit to 1, which guarantees the binary number representation of the result has exactly 9-bits. You could equivalently do & 0x100.
After setting bit 8, the result from the toString() will be 9 chars long (of zeroes and ones).
substring(1) effectively ignores bit 8 and outputs the lower 8 bits

So what?

This code puts leading zeroes on values, so all values are exactly 8 binary characters. There's no way to make Integer.toString() alone do this.

edited Apr 8, 2016 at 4:53

answered Apr 8, 2016 at 4:47

Bohemian♦

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5 Comments

narin.sri Over a year ago

Thanks but whats's that ,16?

Bohemian Over a year ago

16 is the radix - ie base 16, which is hexadecimal. The code basically outputs 1 as "01" instead of "1"

narin.sri Over a year ago

thanks. I also found a simpler way on the forums: sb.append(String.format("%02X", bytes[i]));

Bohemian Over a year ago

@nann yes, using format is the preferred way, but you didn't ask about a better way; you asked "what does this code mean".

narin.sri Over a year ago

you're right. I just wanted to mention this for those who may read this thread later. Thanks again.

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