Store data in Byte array in java

Question

I am trying to convert a string like "password" to hex values, then have it inside a long array, the loop working fine till reaching the value "6F" (hex value for o char) then I have an exception java.lang.NumberFormatException

String password = "password";
char array[] = password.toCharArray();
int index = 0;
for (char c : array) {
    String hex = (Integer.toHexString((int) c));
    data[index] = Long.parseLong(hex);
    index++;
}

how can I store the 6F values inside Byte array, as the 6F is greater than 1 byte?. Please help me on this

Answer 1

Long.parseLong parses decimal numbers. It turns the string "10" into the number 10. If the input is hex, that is incorrect - the string "10" is supposed to be turned into the number 16. The fix is to use the Long.parseLong(String input, int radix) method. the radix you want is 16, though writing that as 0x10 may be more readable - it's the same thing to the compiler, purely a personal style choice. Thus, Long.parseLong(hex, 0x10) is what you want.

Note that in practice char has numbers that go from 0 to 65535, which doesn't fit in bytes. In effect, you must put a marker down that passwords must not contain any characters that aren't ASCII characters (so no umlauts, snowmen, emoji, funny quotes, etc).

If you fail to check this, Integer.toHexString((int) c) will turn into something like 16F or worse (3 to 4 characters), and it may also turn into a single character.

More generally, converting from char c to a hex string, and then parse the hex string into a number, is completely pointless. It's turning 15 into "F" and then turning "F" into 15. If you just want to shove a char into a byte: data[index++] = (byte) c; is all you need - that is the only line you need in your for loop.

But, heed this:

This really isn't how you're supposed to do that!

What you're doing is converting character data to a byte array. This is not actually simple - there are only 256 possible bytes, and there are way more characters that folks have invented. Literally hundreds of thousands of them.

Thus, to convert characters to bytes or vice versa, you must apply an encoding . Encodings have wildly varying properties. The most commonly used encoding, however, is 'UTF-8'. It represent every unicode symbol, and has the interesting property that basic ASCII characters look the exact same. However, it has the downside that any given character is smeared out into 1, 2, 3, or even 4 bytes, depending on what character it is. Fortunately, java has plenty of tools for this, thus, you don't need to care. What you really want, is this:

byte[] data = password.getBytes(StandardCharsets.UTF8);

That's asking the string to turn itself into a byte array, using UTF8 encoding. That means "password" turns into the sequence '112 97 115 115 119 111 114 100' which is no doubt what you want, but you can also have as password, say, außgescheignet ☃ , and that works too - it's turned into bytes, and you can get back to your snowman enabled password:

String in = "außgescheignet ☃";
byte[] data = in.getBytes(StandardCharsets.UTF8);
String andBackAgain = new String(data, StandardCharsets.UTF8);
assert in.equals(andBackAgain); // true

if you stick this in a source file, make sure you save it in whatever text editor you use to do this as UTF8, and that javac compiles it that way too (javac has an -encoding parameter to enforce this).

If you think this is going to cause issues on whatever you send this to, and you want to restrict it to what someone with a rather USA-centric view would call 'normal' characters, then you want the exact same code as showcased here, but use StandardCharsets.ASCII instead. Then, that line ( password.getBytes(StandardCharsets.ASCII) ) will flat out error if it includes non-ASCII characters. That's a good thing: Your infrastructure would not deal with it correctly, we just posited that in this hypothetical exercise. Throwing an exception early in the process on a relevant line is exactly what you want.