简体   繁体   中英

String to binary output in Java

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!

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;

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.

When you try to use + with an object in a string context the java compiler silently inserts a call to the toString() method.

In other words your statements look like

System.out.println("infobin: " + infoBin.toString())

which in this case is the one inherited from Object.

You will need to use a for-loop to pick out each byte from the byte array.

The technical post webpages of this site follow the CC BY-SA 4.0 protocol. If you need to reprint, please indicate the site URL or the original address.Any question please contact:yoyou2525@163.com.

 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM