Binary to text in Java

Question

I have a String with binary data in it (1110100) I want to get the text out so I can print it (1110100 would print "t"). I tried this, it is similar to what I used to transform my text to binary but it's not working at all:

    public static String toText(String info)throws UnsupportedEncodingException{
        byte[] encoded = info.getBytes();
        String text = new String(encoded, "UTF-8");
        System.out.println("print: "+text);
        return text;
    }

Any corrections or suggestions would be much appreciated.

Thanks!

Answer 1

You can use Integer.parseInt with a radix of 2 (binary) to convert the binary string to an integer:

int charCode = Integer.parseInt(info, 2);

Then if you want the corresponding character as a string:

String str = new Character((char)charCode).toString();

Answer 2

This is my one (Working fine on Java 8):

String input = "01110100"; // Binary input as String
StringBuilder sb = new StringBuilder(); // Some place to store the chars

Arrays.stream( // Create a Stream
    input.split("(?<=\\G.{8})") // Splits the input string into 8-char-sections (Since a char has 8 bits = 1 byte)
).forEach(s -> // Go through each 8-char-section...
    sb.append((char) Integer.parseInt(s, 2)) // ...and turn it into an int and then to a char
);

String output = sb.toString(); // Output text (t)

and the compressed method printing to console:

Arrays.stream(input.split("(?<=\\G.{8})")).forEach(s -> System.out.print((char) Integer.parseInt(s, 2))); 
System.out.print('\n');

I am sure there are "better" ways to do this but this is the smallest one you can probably get.

Answer 3

I know the OP stated that their binary was in a String format but for the sake of completeness I thought I would add a solution to convert directly from a byte[] to an alphabetic String representation.

As casablanca stated you basically need to obtain the numerical representation of the alphabetic character. If you are trying to convert anything longer than a single character it will probably come as a byte[] and instead of converting that to a string and then using a for loop to append the characters of each byte you can use ByteBuffer and CharBuffer to do the lifting for you:

public static String bytesToAlphabeticString(byte[] bytes) {
    CharBuffer cb = ByteBuffer.wrap(bytes).asCharBuffer();
    return cb.toString();
}

N.B. Uses UTF char set

Alternatively using the String constructor:

String text = new String(bytes, 0, bytes.length, "ASCII");

Answer 4

public static String binaryToText(String binary) {
    return Arrays.stream(binary.split("(?<=\\G.{8})"))/* regex to split the bits array by 8*/
                 .parallel()
                 .map(eightBits -> (char)Integer.parseInt(eightBits, 2))
                 .collect(
                                 StringBuilder::new,
                                 StringBuilder::append,
                                 StringBuilder::append
                 ).toString();
}

Answer 5

Here is the answer.

private String[] splitByNumber(String s, int size) {
    return s.split("(?<=\\G.{"+size+"})");
}

Answer 6

The other way around (Where "info" is the input text and "s" the binary version of it)

byte[] bytes = info.getBytes();
BigInteger bi = new BigInteger(bytes);
String s = bi.toString(2); 

Answer 7

Look at the parseInt function. You may also need a cast and the Character.toString function.

Answer 8

Also you can use alternative solution without streams and regular expressions (based on casablanca's answer):

public static String binaryToText(String binaryString) {
    StringBuilder stringBuilder = new StringBuilder();
    int charCode;
    for (int i = 0; i < binaryString.length(); i += 8) {
        charCode = Integer.parseInt(binaryString.substring(i, i + 8), 2);
        String returnChar = Character.toString((char) charCode);
        stringBuilder.append(returnChar);
    }
    return stringBuilder.toString();
}

you just need to append the specified character as a string to character sequence.