Binary String to Byte array in Java

Question

I want to do a method to convert binary String to byte array and I wanted to use Java 8 Stream API but I cannot use map to get a byte array. So I have done like this. Do you know any better way to do it?

private static byte[] getBytes(String bitString)
{
    int [] range = IntStream.range(0,  bitString.length() / 8).toArray();
    byte[] byteArray = new byte[range.length];
    for (int i = 0; i < range.length; i++) {
        int pos = range[i];
        byteArray[i] = Byte.parseByte(StringUtils.netSubstring(bitString, pos * 8, 8), 2);
    }
    return  byteArray;
}

Sorry for the confussion. It is a method to do the substring as .Net C# does because I'm translating a lot fo code written in .Net — JuanDYB
– JuanDYB, Commented Sep 2, 2020 at 19:14

Holger · Accepted Answer · 2020-09-01 16:38:28Z

First, it’s important to know that Byte.parseByte(…, 2) expects an optional + or - sign, followed by a positive magnitude. So, if the string has no sign, eight digits, and the leftmost is a one, it will throw an exception, because that positive number wouldn’t fit into the signed byte type. You have to use Integer.parseInt(…,2) instead and cast the result to a byte.

Further, it’s a bit strange to create that int[] range array containing strictly ascending numbers. Within the subsequent loop, pos and i will always have the same value. So the array is entirely obsolete.

Considering these two points, you get the simplified

private static byte[] getBytes(String bitString) {
    byte[] byteArray = new byte[bitString.length()/8];
    for(int i = 0; i < byteArray.length; i++) {
        byteArray[i] = (byte)Integer.parseInt(bitString.substring(i * 8, i * 8 + 8), 2);
    }
    return  byteArray;
}

There is no byte variant of Stream, so the only way to express this via the Stream API would be

private static byte[] getBytes2(String bitString) {
    byte[] byteArray = new byte[bitString.length()/8];
    IntStream.range(0, byteArray.length)
        .forEach(i -> byteArray[i] = (byte)Integer.parseInt(
                                         bitString.substring(i * 8, i * 8 + 8), 2));
    return byteArray;
}

which has no real benefit.

But the whole operation can be seen as an academic exercise anyway, as we can simply use

byte[] ba = new BigInteger(bitString, 2).toByteArray();

This creates an additional first zero byte, if the value would otherwise be negative. If that’s a problem, you can remove it afterwards like

if(ba[0] == 0 && bitString.charAt(0) == '1') ba = Arrays.copyOfRange(ba, 1, ba.length);

and it’s still simpler than doing it manually.

Collectives™ on Stack Overflow

Binary String to Byte array in Java

1 Answer 1

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

Comments

Your Answer

Sign up or log in

Post as a guest

Related