Tutorial :Convert A String (like testing123) To Binary In Java



Question:

I would like to be able to convert a String (with words/letters) to other forms, like binary. How would I go about doing this. I am coding in BLUEJ (Java). Thanks


Solution:1

The usual way is to use String#getBytes() to get the underlying bytes and then present those bytes in some other form (hex, binary whatever).

Note that getBytes() uses the default charset, so if you want the string converted to some specific character encoding, you should use getBytes(String encoding) instead, but many times (esp when dealing with ASCII) getBytes() is enough (and has the advantage of not throwing a checked exception).

For specific conversion to binary, here is an example:

  String s = "foo";    byte[] bytes = s.getBytes();    StringBuilder binary = new StringBuilder();    for (byte b : bytes)    {       int val = b;       for (int i = 0; i < 8; i++)       {          binary.append((val & 128) == 0 ? 0 : 1);          val <<= 1;       }       binary.append(' ');    }    System.out.println("'" + s + "' to binary: " + binary);  

Running this example will yield:

'foo' to binary: 01100110 01101111 01101111   


Solution:2

A shorter example

private static final Charset UTF_8 = Charset.forName("UTF-8");    String text = "Hello World!";  byte[] bytes = text.getBytes(UTF_8);  System.out.println("bytes= "+Arrays.toString(bytes));  System.out.println("text again= "+new String(bytes, UTF_8));  

prints

bytes= [72, 101, 108, 108, 111, 32, 87, 111, 114, 108, 100, 33]  text again= Hello World!  


Solution:3

A String in Java can be converted to "binary" with its getBytes(Charset) method.

byte[] encoded = "ã"ã‚"にちは、世界!".getBytes(StandardCharsets.UTF_8);  

The argument to this method is a "character-encoding"; this is a standardized mapping between a character and a sequence of bytes. Often, each character is encoded to a single byte, but there aren't enough unique byte values to represent every character in every language. Other encodings use multiple bytes, so they can handle a wider range of characters.

Usually, the encoding to use will be specified by some standard or protocol that you are implementing. If you are creating your own interface, and have the freedom to choose, "UTF-8" is an easy, safe, and widely supported encoding.

  • It's easy, because rather than including some way to note the encoding of each message, you can default to UTF-8.
  • It's safe, because UTF-8 can encode any character that can be used in a Java character string.
  • It's widely supported, because it is one of a small handful of character encodings that is required to be present in any Java implementation, all the way down to J2ME. Most other platforms support it too, and it's used as a default in standards like XML.


Solution:4

Here are my solutions. Their advantages are : easy-understanding code, works for all characters. Enjoy.

Solution 1 :

public static void main(String[] args) {        String str = "CC%";      String result = "";      char[] messChar = str.toCharArray();        for (int i = 0; i < messChar.length; i++) {          result += Integer.toBinaryString(messChar[i]) + " ";      }        System.out.println(result);  }  

prints :

1000011 1000011 100101  

Solution 2 :

Possibility to choose the number of displayed bits per char.

public static String toBinary(String str, int bits) {      String result = "";      String tmpStr;      int tmpInt;      char[] messChar = str.toCharArray();        for (int i = 0; i < messChar.length; i++) {          tmpStr = Integer.toBinaryString(messChar[i]);          tmpInt = tmpStr.length();          if(tmpInt != bits) {              tmpInt = bits - tmpInt;              if (tmpInt == bits) {                  result += tmpStr;              } else if (tmpInt > 0) {                  for (int j = 0; j < tmpInt; j++) {                      result += "0";                  }                  result += tmpStr;              } else {                  System.err.println("argument 'bits' is too small");              }          } else {              result += tmpStr;          }          result += " "; // separator      }        return result;  }  

public static void main(String args[]) {      System.out.println(toBinary("CC%", 8));  }  

prints :

01000011 01000011 00100101  


Solution:5

This is my implementation.

public class Test {      public String toBinary(String text) {          StringBuilder sb = new StringBuilder();            for (char character : text.toCharArray()) {              sb.append(Integer.toBinaryString(character) + "\n");          }            return sb.toString();        }  }  


Solution:6

import java.lang.*;  import java.io.*;  class d2b  {    public static void main(String args[]) throws IOException{    BufferedReader b = new BufferedReader(new InputStreamReader(System.in));    System.out.println("Enter the decimal value:");    String h = b.readLine();    int k = Integer.parseInt(h);      String out = Integer.toBinaryString(k);    System.out.println("Binary: " + out);    }  }     


Solution:7

While playing around with the answers I found here to become familiar with it I twisted Nuoji's solution a bit so that I could understand it faster when looking at it in the future.

public static String stringToBinary(String str, boolean pad ) {      byte[] bytes = str.getBytes();      StringBuilder binary = new StringBuilder();      for (byte b : bytes)      {         binary.append(Integer.toBinaryString((int) b));         if(pad) { binary.append(' '); }      }      return binary.toString();          }  


Solution:8

>       int no=44;  >             String bNo=Integer.toString(no,2);//binary output 101100  >             String oNo=Integer.toString(no,8);//Oct output 54  >             String hNo=Integer.toString(no,16);//Hex output 2C  >       >             String sBNo="101100";  >             no=Integer.parseInt(sBNo,2);//binary to int output 44  >             String sONo="54";  >             no=Integer.parseInt(sONo,8);//oct to int  output 44  >             String sHNo="2C";  >             no=Integer.parseInt(sHNo,16);//hex to int output 44  


Solution:9

public class HexadecimalToBinaryAndLong{    public static void main(String[] args) throws IOException{      BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));      System.out.println("Enter the hexa value!");      String hex = bf.readLine();      int i = Integer.parseInt(hex);               //hex to decimal      String by = Integer.toBinaryString(i);       //decimal to binary      System.out.println("This is Binary: " + by);      }  }  


Solution:10

You can also do this with the ol' good method :

String inputLine = "test123";  String translatedString = null;  char[] stringArray = inputLine.toCharArray();  for(int i=0;i<stringArray.length;i++){        translatedString += Integer.toBinaryString((int) stringArray[i]);  }  

Note:If u also have question or solution just comment us below or mail us on toontricks1994@gmail.com
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