converting hexadecimal string to byte array

Question

I wrote a program that converts a hexadecimal string into a byte array, eg "2e65" produces [2,14,6,5].

Is there a way to do it with less lines of code or without using the ASCII table values.

public static byte[] stringToData( String dataString) {
    byte [] hexStringToByteArr = new byte[dataString.length()];
    for ( int i = 0; i < dataString.length(); i++) {
        char c = dataString.charAt(i);
        if ( c == 48 ) { hexStringToByteArr[i] = 0;}
        if ( c == 49 ) { hexStringToByteArr[i] = 1;}
        if ( c == 50 ) { hexStringToByteArr[i] = 2;}
        if ( c == 51 ) { hexStringToByteArr[i] = 3;}
        if ( c == 52 ) { hexStringToByteArr[i] = 4;}
        if ( c == 53 ) { hexStringToByteArr[i] = 5;}
        if ( c == 54 ) { hexStringToByteArr[i] = 6;}
        if ( c == 55 ) { hexStringToByteArr[i] = 7;}
        if ( c == 56 ) { hexStringToByteArr[i] = 8;}
        if ( c == 57 ) { hexStringToByteArr[i] = 9;}
        if ( c == 97 ) { hexStringToByteArr[i] = 10;}
        if ( c == 98 ) { hexStringToByteArr[i] = 11;}
        if ( c == 99 ) { hexStringToByteArr[i] = 12;}
        if ( c == 100 ) {hexStringToByteArr[i] = 13;}
        if ( c == 101 ) {hexStringToByteArr[i] = 14;}
        if ( c == 102 ) {hexStringToByteArr[i] = 15;}
    }
    return hexStringToByteArr;
}

public static void main(String [] args) {
    String pracString = "2e65";    
    System.out.println(Arrays.toString(stringToData(pracString)));
}

You can use hexStringToByteArr[i]=Character.digit(c,16); instead of all the if statements. — Eran
– Eran, Commented Mar 8, 2020 at 18:32
@ElliottFrisch could you tell me why I need two instead of four? i did what said with the dividing by 2 but it causes my program to have an error. Also I do not understand logically, if my string has 4 values, would I not need the same amount of indexes in my byte array to fit them in? — tomatofighter
– tomatofighter, Commented Mar 8, 2020 at 18:42
One byte is 2 hex digits, if it were one digit it would be constrained to sixteen values (hence hex). — Elliott Frisch
– Elliott Frisch, Commented Mar 8, 2020 at 18:44
Incase I didnt write my question correctly. My code is working as intended with the output. I need every single hexadecimal character to be displayed in its decimal form in a byte array — tomatofighter
– tomatofighter, Commented Mar 8, 2020 at 18:45

Jacob G. · Accepted Answer · 2020-03-08 18:47:18Z

2

"2e65" in hexadecimal represents two bytes (2e, which corresponds to 46 in decimal, and 65, which corresponds to 101 in decimal). To get a byte[] containing two bytes, you can utilize BigInteger:

String hex = "2e65";
byte[] b = new BigInteger(hex, 16).toByteArray();
System.out.println(Arrays.toString(b));

The output of the above snippet is:

[46, 101]

If, instead, you want to convert each of the four hexadecimal digits to a byte and store them in a byte[], then you can use Character#digit while iterating over each char in the String:

String hex = "2e65";
byte[] b = new byte[hex.length()];

for (int i = 0; i < b.length; i++) {
    b[i] = (byte) Character.digit(hex.charAt(i), 16);
}

System.out.println(Arrays.toString(b));

The output of the above snippet is:

[2, 14, 6, 5]

edited Mar 8, 2020 at 18:47

answered Mar 8, 2020 at 18:37

Jacob G.

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1 Comment

tomatofighter Over a year ago

The second output is what I was asking for. Thanks for providing both.

Max Voisard · Accepted Answer · 2020-03-08 18:32:10Z

0

To convert hex string to byte array, you need to first get the length of the given string and include it while creating a new byte array.

string str = "2e65";
byte[] val = new byte[str.length() / 2];

Now, take a for loop until the length of the byte array.

for (int i = 0; i < val.length; i++) {
   int index = i * 2;
   int j = Integer.parseInt(str.substring(index, index + 2), 16);
   val[i] = (byte) j;
}

So, since byte arrays must be half the length, change this line

byte[] hexStringToByteArr = new byte[dataString.length()];

to this:

byte[] hexStringToByteArr = new byte[dataString.length() / 2];

answered Mar 8, 2020 at 18:32

Max Voisard

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2 Comments

tomatofighter Over a year ago

why must byte arrays be half the length? wouldnt I need a byte array of 4 indexes since i have 4 values? If i include byte[] hexStringToByteArr = new byte[dataString.length() / 2]; i get an out of bounds of array error

Max Voisard Over a year ago

@wetmoney You get the ArrayIndexOutOfBounds error because your code tries to access indexes of the byte array that are beyond half the string length. Use my code instead (so the for loop included) and it will work.

Bohemian · Accepted Answer · 2020-03-08 18:43:14Z

0

Your big block o' ifs can be replaced with:

hexStringToByteArr[i] = c < 58 ? c - 48 : c - 87;

answered Mar 8, 2020 at 18:43

Bohemian♦

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