Why does this code have an array of 3 bytes to build a "whole byte"?

Question

I found the following code on the internet, it converts NSString representations such as
@"00F04100002712" into an actual array of bytes. The code works and does generate the correct output; I just don't understand why there is char byte_chars[3] instead of char byte_chars[2] since only the first two positions are used in the code.

void hexStringToBytes(NSString *s, NSMutableData *data)
{
    unsigned char whole_byte;
    char byte_chars[3] = {'\0','\0','\0'};
    int commandLength = (int)[s length];

    // convert hex values to bytes
    for (int i=0; i < commandLength/2; i++)
    {
        byte_chars[0] = [s characterAtIndex:i*2];
        byte_chars[1] = [s characterAtIndex:i*2+1];
        whole_byte = strtol(byte_chars, NULL, 16);
        [data appendBytes:&whole_byte length:1];
    }
}

I think it has something to do with the strtol function call but I am not sure what.

Can someone explain how and why this works?

Answer 1

C style strings have a terminating zero (aka null) character. An ASCII representation of an 8 bit byte in hexadecimal will be two characters plus that terminator.