Java Byte.parseByte() error

Question

I'm having a small error in my code that I can not for the life of me figure out.

I have an array of strings that are representations of binary data (after converting them from hex) for example: one index is 1011 and another is 11100. I go through the array and pad each index with 0's so that each index is eight bytes. When I try to convert these representations into actual bytes I get an error when I try to parse '11111111' The error I get is:

java.lang.NumberFormatException: Value out of range. Value:"11111111" Radix:2

Here is a snippet:

String source = a.get("image block");
int val;
byte imageData[] = new byte[source.length()/2];

try {
    f.createNewFile();
    FileOutputStream output = new FileOutputStream(f);
    for (int i=0; i<source.length(); i+=2) {
        val = Integer.parseInt(source.substring(i, i+2), 16);
        String temp = Integer.toBinaryString(val);
        while (temp.length() != 8) {
            temp = "0" + temp;
        }
    imageData[i/2] = Byte.parseByte(temp, 2);
}

Answer 1

Isn't the problem here that byte is a signed type, therefore its valid values are -128...127? If you parse it as an int (Using Integer.parseInt()), it should work.

By the way, you don't have to pad the number with zeroes either.

Once you parsed your binary string into an int, you can cast it to a byte, but the value will still be treated as signed, so binary 11111111 will become int 255 first, then byte -1 after the cast.