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.