String to binary output in Java

java string encoding string-conversion

60,573

Solution 1

Only Integer has a method to convert to binary string representation check this out:

import java.io.UnsupportedEncodingException;

public class TestBin {
    public static void main(String[] args) throws UnsupportedEncodingException {
        byte[] infoBin = null;
        infoBin = "this is plain text".getBytes("UTF-8");
        for (byte b : infoBin) {
            System.out.println("c:" + (char) b + "-> "
                    + Integer.toBinaryString(b));
        }
    }
}

would print:

c:t-> 1110100
c:h-> 1101000
c:i-> 1101001
c:s-> 1110011
c: -> 100000
c:i-> 1101001
c:s-> 1110011
c: -> 100000
c:p-> 1110000
c:l-> 1101100
c:a-> 1100001
c:i-> 1101001
c:n-> 1101110
c: -> 100000
c:t-> 1110100
c:e-> 1100101
c:x-> 1111000
c:t-> 1110100

Padding:

String bin = Integer.toBinaryString(b); 
if ( bin.length() < 8 )
  bin = "0" + bin;

Solution 2

Arrays do not have a sensible toString override, so they use the default object notation.

Change your last line to

return Arrays.toString(infoBin);

and you'll get the expected output.

60,573

Author by

Nick

Updated on August 27, 2020

Comments

Nick almost 4 years
I want to get binary (011001..) from a String but instead i get [B@addbf1 , there must be an easy transformation to do this but I don't see it.
```
public static String toBin(String info){
  byte[] infoBin = null;
  try {
   infoBin = info.getBytes( "UTF-8" );
   System.out.println("infoBin: "+infoBin);
  }
  catch (Exception e){
   System.out.println(e.toString());
  }
  return infoBin.toString();
}
```
Here i get infoBin: [B@addbf1
and I would like infoBin: 01001...

Any help would be appreciated, thanks!
Nick over 13 years

Hi Andrzej, your answer makes a lot of sense but I get [84, 104, 105, 115, 32, 105, 115, 32, 97, 32,..] instead of (01101010...) any suggestions? thank you
stacker over 13 years

maybe you need some padding for your bytes
Nick over 13 years

Thank you very much for your help!
Nick over 13 years

@stacker why is the space endoded on only 6 digits and not 7? (ex: c: -> 100000)
Nick over 13 years

should I manualy add a 0 at the begining? (ex: (ex: c: -> 0100000) )
stacker over 13 years

@Nick that's what I meant whith my first comment, leading zeros are omitted so you need to
stacker over 13 years

@Nick that's what I meant whith my first comment, leading zeros are omitted so you need to add them as you already noticed. Or you write a method which adapt the behaviour Integer.toBinaryString(). If it is not critical I would suggest to pad the strings with leading '0's.
Nick over 13 years

@stacker how can I make it work for a byte that starts with "1" ? If I do it on a byte that starts with "0", I get 8 chars, as expected. If I do it on a byte that starts with "1", I get 32 chars instead.
stacker over 13 years

@Nick check String.length() and use substring(0,8) (not tested)
Pippi over 8 years

Sorry!! If i would like get binary of "%" (it's 00100101) in this way i will receive "0" + "100101" and not "00100101".