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I have sort of a funky question (that I hope hasn't been asked and answered yet). To start, I'll tell you the order of what I'm trying to do and how I'm doing it and then tell you where I'm having a problem:

  1. Convert a string of characters into ASCII numbers
  2. Convert those ASCII numbers into binary and store them in a string
  3. Convert those binary numbers back into ASCII numbers
  4. Convert the ASCII numbers back into normal characters

Here are the methods I've written so far:

public static String strToBinary(String inputString){

    int[] ASCIIHolder = new int[inputString.length()];

    //Storing ASCII representation of characters in array of ints
    for(int index = 0; index < inputString.length(); index++){
        ASCIIHolder[index] = (int)inputString.charAt(index);
    }


    StringBuffer binaryStringBuffer = new StringBuffer();

    /* Now appending values of ASCIIHolder to binaryStringBuffer using
     * Integer.toBinaryString in a for loop. Should not get an out of bounds
     * exception because more than 1 element will be added to StringBuffer
     * each iteration.
     */
    for(int index =0;index <inputString.length();index ++){

        binaryStringBuffer.append(Integer.toBinaryString
                (ASCIIHolder[index]));
    }


    String binaryToBeReturned = binaryStringBuffer.toString();

    binaryToBeReturned.replace(" ", "");

    return binaryToBeReturned;
}

public static String binaryToString(String binaryString){

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

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

    return returnString;
}

I'm getting a NumberFormatException when I run the code and I think it's because the program is trying to convert the binary digits as one entire binary number rather than as separate letters. Based on what you see here, is there a better way to do this overall and/or how can I tell the computer to recognize the ASCII characters when it's iterating through the binary code? Hope that's clear and if not I'll be checking for comments.

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  • By "ASCII numbers" you mean "UTF-16 code unit" right? Commented Aug 4, 2012 at 21:33
  • You need to make sure each "ASCII character"* in the binary buffer uses 7-bits; that is, use padding. Then you should need to split it along these boundaries and decode each segment into a character individually. * You do realize Java char is a UTF-16 code-point... 7-bit ASCII character encoding is a subset. Make sure you fully understand what you're doing. Commented Aug 4, 2012 at 21:34
  • I'm not sure to be honest -- I only started Java a few months ago and am still learning the particulars. By 'ASCII numbers' I meant whatever value is given from ASCIIHolder[index] = (int)inputString.charAt(index); Commented Aug 4, 2012 at 21:36
  • @veer, I wasn't aware of that. I'll read up on them both and then try rewriting the code. Commented Aug 4, 2012 at 21:38
  • @veer I meant for the text to be UTF - 8. Thanks for the heads up about that, I'm going to go back through my code and try to work it out now that I know that. Commented Aug 4, 2012 at 22:04

2 Answers 2

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So I used OP's code with some modifications and it works really well for me. I'll post it here for future people. I don't think OP needs it anymore because he probably figured it out in the past 2 years.

public class Convert
    {
    public String strToBinary(String inputString){

    int[] ASCIIHolder = new int[inputString.length()];

    //Storing ASCII representation of characters in array of ints
    for(int index = 0; index < inputString.length(); index++){
        ASCIIHolder[index] = (int)inputString.charAt(index);
    }

    StringBuffer binaryStringBuffer = new StringBuffer();

    /* Now appending values of ASCIIHolder to binaryStringBuffer using
     * Integer.toBinaryString in a for loop. Should not get an out of bounds
     * exception because more than 1 element will be added to StringBuffer
     * each iteration.
     */
    for(int index =0;index <inputString.length();index ++){

        binaryStringBuffer.append(Integer.toBinaryString
                (ASCIIHolder[index]));
    }

    String binaryToBeReturned = binaryStringBuffer.toString();

    binaryToBeReturned.replace(" ", "");

    return binaryToBeReturned;
}

public String binaryToString(String binaryString){
    String returnString = "";
    int charCode;
    for(int i = 0; i < binaryString.length(); i+=7)
    {
    charCode = Integer.parseInt(binaryString.substring(i, i+7), 2);
    String returnChar = new Character((char)charCode).toString();
    returnString += returnChar;
    }
    return returnString;
}
}

I'd like to thank OP for writing most of it out for me. Fixing errors is much easier than writing new code.

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Comments

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You've got at least two problems here:

  • You're just concatenating the binary strings, with no separators. So if you had "1100" and then "0011" you'd get "11000011" which is the same result as if you had "1" followed by "1000011".
  • You're calling String.replace and ignoring the return result. This sort of doesn't matter as you're replacing spaces, and there won't be any spaces anyway... but there should be!

Of course you don't have to use separators - but if you don't, you need to make sure that you include all 16 bits of each UTF-16 code point. (Or validate that your string only uses a limited range of characters and go down to an appropriate number of bits, e.g. 8 bits for ISO-8859-1 or 7 bits for ASCII.)

(I have to wonder what the point of all of this is. Homework? I can't see this being useful in real life.)

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9 Comments

Do you think it would be a better idea then to have an array of Strings (or list or something) where each element was its own binary value? I'm writing a DES encryption program in which I convert the input characters (and key) to binary and then run the encryption algorithm.
@Decave: Why do you want it as a string of 1s and 0s? Usually when people talk about converting a string to binary for encryption, it's a matter of calling String.getBytes(encoding). That's not the same thing at all. The input to a raw encryption algorithm should be a byte[], not a String.
Doesn't that only convert it into bytes? I need the actual bits of the text. Again, I'm sorry if this is really nooby but I haven't been programming for that long -- I'm doing this as practice to get better.
@Decave: But why would you need those bits as text themselves? That's a really odd requirement. Unless you know that you're deliberately trying to write an odd approach to encryption, you're missing how it works...
@Decave: I don't actually live on Stack Overflow. While sometimes I'll respond quickly, sometimes I won't. But you really shouldn't be handling binary data as text. You should just use an array of bytes, without using Byte.parseByte - just convert the input text straight to an array of bytes. You should never need to go through a string of the form "100011101" etc.
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