How to convert binary string to Java String encoded using UFT-8

Question

In order to send a chunk of bits from a 4 words String, I'm doing getting the byte array from the String and calculating the bit string.

StringBuilder binaryStr = new StringBuilder();

byte[] bytesFromStr = str.getBytes("UTF-8");
for (int i = 0, l = bytesFromStr.length; i < l; i++) {
    binaryStr.append(Integer.toBinaryString(bytesFromStr[i]));
}

String result = binaryStr.toString();

The problem appears when I want to do the reverse operation: converting a bit string to a Java String encoded using UTF-8.

Please, Is there someone that can explain me the best way to do that?

Thanks in advance!

I think this is a duplicate of: stackoverflow.com/questions/5499924/…, at the very least I think it will help. — Gavin
– Gavin, Commented Jul 9, 2016 at 17:23
It's impossible to reverse that operation. You can't possibly know if 100011010100110101100100 is the representation of 3 bytes, or 4, or 5, or... What are you trying to achieve? Why are you doing that? — JB Nizet
– JB Nizet, Commented Jul 9, 2016 at 17:27
If you have string "1a" then it is build from characters 1 and a which are placed in Unicode Table at positions 49, 97. In binary form they should be represented as 0110001 1100001. But result of Integer.toBinaryString(49) is 110001 not 0110001 (leading 0 is ignored). So as JB Nizet pointed out, it is impossible to detect if 111 represents 1 1 1 or 11 1 or 1 11 or 111. Anyway what you are doing here looks like XY problem — Pshemo
– Pshemo, Commented Jul 9, 2016 at 17:33
If I have 4 words encoded with UFT-8 means that I have 4 bytes, if I'm not wrong. In that case I think I can reverse the operation. That is for a PoC about steganography and data exfiltration. — osanchezmon
– osanchezmon, Commented Jul 9, 2016 at 17:36
"If I have 4 words encoded with UFT-8 means that I have 4 bytes" what makes you think so? Can you point us to some resource which gave you that idea? What you are saying can be interpreted as "utf-8 writes one word on one byte" but try to think about how many words are out there, and how many numbers byte can hold. — Pshemo
– Pshemo, Commented Jul 9, 2016 at 17:37

Andreas · Accepted Answer · 2016-07-09 17:40:21Z

TL;DR Don't use toBinaryString(). See solution at the end.

Your problem is that Integer.toBinaryString() doesn't return leading zeroes, e.g.

System.out.println(Integer.toBinaryString(1));   // prints: 1
System.out.println(Integer.toBinaryString(10));  // prints: 1010
System.out.println(Integer.toBinaryString(100)); // prints: 1100100

For your purpose, you want to always get 8 bits for each byte.

You also need to prevent negative values from causing errors, e.g.

System.out.println(Integer.toBinaryString((byte)129)); // prints: 11111111111111111111111110000001

Easiest way to accomplish that is like this:

Integer.toBinaryString((b & 0xFF) | 0x100).substring(1)

First, it coerces the byte b to int, then retains only lower 8 bits, and finally sets the 9th bit, e.g. 129 (decimal) becomes 1 1000 0001 (binary, spaces added for clarity). It then excludes that 9th bit, in effect ensuring that leading zeroes are in place.

It's better to have that as a helper method:

private static String toBinary(byte b) {
    return Integer.toBinaryString((b & 0xFF) | 0x100).substring(1);
}

In which case your code becomes:

StringBuilder binaryStr = new StringBuilder();
for (byte b : str.getBytes("UTF-8"))
    binaryStr.append(toBinary(b));
String result = binaryStr.toString();

E.g. if str = "Hello World", you get:

0100100001100101011011000110110001101111001000000101011101101111011100100110110001100100

You could of course just do it yourself, without resorting to toBinaryString():

StringBuilder binaryStr = new StringBuilder();
for (byte b : str.getBytes("UTF-8"))
    for (int i = 7; i >= 0; i--)
        binaryStr.append((b >> i) & 1);
String result = binaryStr.toString();

That will probably run faster too.

Thank's @Andreas. I will do some test with your implementation avoiding 'toBinaryString()' and trying to recover the information.

osanchezmon · Accepted Answer · 2016-07-09 18:42:54Z

0

Thanks @Andreas for your code. I test using your function and "decoding" again to UTF-8 using this:

StringBuilder revealStr = new StringBuilder();
for (int i = 0; i < result.length(); i += 8) {
    revealStr.append((char) Integer.parseUnsignedInt(result.substring(i, i + 8), 2));
}

Thanks for all folks to help me.

answered Jul 9, 2016 at 18:42

osanchezmon

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