How do I convert a large binary String to byte array java?

Question

I have a large binary string \"101101110...\", and I am trying to store it into a byte array. what is the best way of doing it?

Lets say I have largeString = \"01001

Answer 1

Assuming that your binary string module 8 equals 0 binString.lenght()%8==0

 /**
 * Get an byte array by binary string
 * @param binaryString the string representing a byte
 * @return an byte array
 */
public static byte[] getByteByString(String binaryString) {
    int splitSize = 8;

    if(binaryString.length() % splitSize == 0){
        int index = 0;
        int position = 0;

        byte[] resultByteArray = new byte[binaryString.length()/splitSize];
        StringBuilder text = new StringBuilder(binaryString);

        while (index < text.length()) {
            String binaryStringChunk = text.substring(index, Math.min(index + splitSize, text.length()));
            Integer byteAsInt = Integer.parseInt(binaryStringChunk, 2);
            resultByteArray[position] = byteAsInt.byteValue();
            index += splitSize;
            position ++;
        }
        return resultByteArray;
    }
    else{
        System.out.println("Cannot convert binary string to byte[], because of the input length. '" +binaryString+"' % 8 != 0");
        return null;
    }
}

Answer 2

Do it in a loop. Split the string at 8-character chunks and convert them separately. In "pseudocode" it's something like:

byte[] result = new byte[subs.size()];

int i = 0;
int j = 0;
while(i+8 <= s.length){
    result[j] = new Byte.valueOf(largeString.substring(i, i+8), 2);
    i+=8;
    j++;
}

result[j] = new Byte.valueOf(largeString.substring(i, largeString.length));

Answer 3

You can easily build an ArrayList on which you can call toArray if you want an actual array;

List<Integer> list = new ArrayList<>();

for(String str : largeString.split("(?<=\\G.{8})"))
    list.add(Integer.parseInt(str, 2));

System.out.println(list);  // Outputs [78, 187, 96, 17, 21]