how to find byte index in string in java

Question

I have string and there is 0x80 in it. string presentation is : serialno� and hex presentation is 73 65 72 69 61 6C 6E 6F 80. I want to remove 0x80 from string without convert string to hex string. is it possible in java ? I tried lastIndexOf(0x80). but it returns -1.

my code is (also you can find on https://ideone.com/3p8wKT) :

public static void main(String[] args) {

    String hexValue = "73657269616C6E6F80";
    String binValue = hexStringToBin(hexValue);
    System.out.println("binValue : " + binValue);

    int index = binValue.lastIndexOf(0x80);

    System.out.println("index : " + index);
}

public static String hexStringToBin(String s) {
    int len = s.length();
    byte[] data = new byte[len / 2];
    for (int i = 0; i < len; i += 2) {
        data[i / 2] = (byte) ((Character.digit(s.charAt(i), 16) << 4) + Character.digit(s.charAt(i + 1), 16));
    }
    return new String(data);
}

"and hex presentation is" - how have you determined that? Please provide a minimal reproducible example. I suspect your string actually contains U+FFFD, the Unicode replacement character, but it's not certain at the moment. Additionally, it may well be better to clean the data earlier, before converting it to a string. — Jon Skeet
– Jon Skeet, Commented Mar 22, 2018 at 9:56
You should probably us a charset when you return new String(data); You have some bytes and they'll get translated into characters which could have a different value than the raw byte value. — matt
– matt, Commented Mar 22, 2018 at 10:35
General comment: don't try to create a string like this to represent arbitrary binary data. That's not what it's there for. I'd expect a hexStringToBin method to return a byte[], not a String. — Jon Skeet
– Jon Skeet, Commented Mar 22, 2018 at 11:19

matt · Accepted Answer · 2018-03-22 11:04:35Z

2

Change your hex string method to map directly to characters.

char[] data = new char[len / 2];
for (int i = 0; i < len; i += 2) {
    data[i / 2] = (char) ((Character.digit(s.charAt(i), 16) << 4) + Character.digit(s.charAt(i + 1), 16));
}

return new String(data);

The exchange between a String and byte[] requires an encoding.

Your hex string seems to be a string representation of bytes/characters. It would appear you had an original String -> converted it to your hex string, but we don't know the encoding.

If you want to say that each pair of characters maps to the corresponding character, eg "80" -> char c = 0x80; Then you can achieve that by using a char[], which doesn't get encoded/decoded when creating a string.

If you use a byte[] (as you have done in your example), then it will get decoded and invalid characters get mapped to 0xFFFD, which is unicode replacement character.

edited Mar 22, 2018 at 11:04

answered Mar 22, 2018 at 10:46

matt

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Comments

Bohdan Petrenko · Accepted Answer · 2018-03-22 11:15:23Z

1

It's because you converted � symbol to hex incorrectly (0x80). 1 symbol in UTF-8 can take 1 byte or more. In your case � symbol takes 2 bytes and have the following representation 65533 or 0xFFFD. So, if you replace your code with

int index = variable.lastIndexOf(0xFFFD);
//index will be 8

all will work fine.

Code snippet to proof my words:

String variable = "serialno�";
for (char c : variable.toCharArray())
    System.out.print(((int)c)+ " ");
// 115 101 114 105 97 108 110 111 65533

UPDATE

You've made a mistake in hexStringToBin function. Replace it with

public static String hexStringToBin(String s) {
    int len = s.length();
    char[] data = new char[len / 2];
    for (int i = 0; i < len; i += 2) {
        data[i / 2] = (char) ((Character.digit(s.charAt(i), 16) << 4) + Character.digit(s.charAt(i + 1), 16));
    }

    return new String(data);
}

and all will work fine.

edited Mar 22, 2018 at 11:15

answered Mar 22, 2018 at 10:18

Bohdan Petrenko

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2 Comments

utarid Over a year ago

I tried it for other two bytes variables like 0x81,0x82... all is represent as 65533. but i am looking for only 0x80

matt Over a year ago

@user4757345 it is because the charset you are using to create the string maps all of those values to an invalid character.

Korteby Farouk · Accepted Answer · 2018-03-22 09:55:08Z

0

This is working for me :

int hex = 0x6F;
System.out.println("serialno€".lastIndexOf((char)hex));

Output :

answered Mar 22, 2018 at 9:55

Korteby Farouk

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1 Comment

Korteby Farouk Over a year ago

in char 0x80 is unknown in ASCII table could you tell me which encoding it is ?

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